MULTIVARIABLE CALCULUS DIFFERENTIAL EQUATIONS …senthilraani.files.wordpress.com/2019/10/mth201prefinaldraft.pdfv(cx+y) = cT!cv (x)+T!v (y)thanks animesh and Rithik!)- one-one: T!v

MULTIVARIABLE CALCULUS & DIFFERENTIAL EQUATIONSMTH 201

Senthil Raani KS, Department of Mathematical Sciences, IISER [email protected] Office hours: Saturdays (appointments by e-mail)

Contents

1. Vectors in R3 21.1. Geometrical interpretation: dot and cross product of vectors, length of a

vector, orthogonality of vectors; Lines, planes, and quadric surfaces 22. Continuity and differentiability of vector-valued functions 72.1. Functions of two or more variables, Limits and continuity: 72.2. Tangent vectors, Partial derivatives, Gradient and directional derivatives 122.3. Maxima, Minima and saddle points - Lagrange multipliers 202.4. Gist 233. Double and triple integrals 273.1. Change of coordinates 273.2. Vector fields - Line integrals 283.3. Vector fields - Surface integrals 303.4. Green’s, Divergence and Stokes’ theorem 324. First order Ordinary Differential Equations 344.1. Variables separable method; Homogeneous, Linear and Exact equations 345. Exercises and Solutions 356. Appendix: One variable Calculus 466.1. Recall: Limit Concept and Continuity 466.2. Differentiation: 486.3. Integration 52References 54

Rules and Guidelines:

• Academic office regulations: 75% of attendance is compulsory.Course policy: Students are not allowed to give their attendance after 5 min-utes from the scheduled start of the class hours.Schedule: 1500-1600Hrs on Tue, Wed, Thu, FriPlease e-mail me, if you would like to discuss/clarify any doubts personally.• The pass mark is 40. Grading policy: relative

Quizes and/or Group assignments: 20 (best 2 out of 3)Mid Semester Exam: 30End Semester Exam: 50Maximum marks: 100 (for relative grading)

• The draft is prepared with reference to [3, 2].

1

Senthil Raani K S MTH 201

1. Vectors in R3

Formal Definition of Vector in MTH 102: A vector space is a set V that isclosed under vector addition and scalar multiplication, that is, for any u, v, w ∈ Vand for any scalar λ, µ ∈ F (F denotes a field), V is said to be a vector space over Fif it satisfies the following axioms:

• Commutativity: u+ v = v + u• Associativity of vector addition: (u+ v) + w = u+ (v + w)• additive identity: 0 + v = v + 0 = v• Existence of additive inverse: For all v ∈ V , there exists −v ∈ V such thatv + (−v) = 0 = (−v) + v• Associativity of scalar multiplication: λ(µv) = λµv• Distributivity of scalar sums: (λ+ µ)v = λv + µv• Distributivity of vector sums: λ(u+ v) = λu+ λv• Scalar multiplication identity: 1v = v

Then the elements v in the vector space V are called the vectors (also denoted as−→v ). Throughout this course, we are going to restrict ourselves to the following vectorspace.

V = Rn, where F = R, for all n = 1, 2, · · ·(we will mostly discuss R2 and R3,

while we recall R, that was done in MTH 101 now and then)

1.1. Geometrical interpretation: dot and cross product of vectors, lengthof a vector, orthogonality of vectors; Lines, planes, and quadric surfaces.In this subsection we will recall the Geometrical Perceptions of the terms that arefamiliar in MTH 102:

A vector may be depicted as the translation map from Rn to Rn:

y

x

12x+ y

12x

(0, 0)

−→v = ( 12 ,12 )

T−→vT−→v (y)

T−→v (x)

T−→v (12x+ y)

12T−→v (x)

(0, 0)

2


−→v

T−→v (y)

y

−→v

T−→v (x)

x −→v

T−→v (12x+ y)

12x+ y

−→v

Let us write these mathematically. Suppose−→v = (v1, · · · , vn) denote a vector in Rn. Thenconsider the map T−→v : Rn → Rn defined asT−→v (x) = (x1 + v1, · · · , xn + vn) when x =(x1, · · ·xn). This map is, in fact, one-one, ontomap (recall! It is not a Linear transformation:Tv(cx+y) = cT−→c v(x)+T−→v (y) thanks animesh andRithik!)- one-one: T−→v (x) = T−→v (y) =⇒ x = y,onto: for every y ∈ Rn, there exists x ∈ Rn suchthat T−→v (x) = y. Observe (0, 0) is mapped to −→v ;T−→v (x) − x = −→v ∀x ∈ Rn. That is, −→v can be represented in uncountable ways as−→PQ where P represents a co-ordinate point in Rn and Q represents the point mappedunder T−→v .

Hence a vector at any position may be viewed as a vector from the origin. In otherwords, one may move the vector to the space without disturbing the direction andthe magnitude. So a vector ’is’ a co-ordinate a = (a1, · · · an) in the sense that thedirection is viewed from the origin to a and its magnitude is the length between aand the origin.

γ

−→v

−→u

−→v −−→u

−→v +−→u

The sum and difference of two vectors in R2 aregiven by the diagonals of the parallelogram formedby the vectors.

Fun exercises: Geometrically interpret the fol-lowing:

• Commutativity and Associativity of vectoraddition• additive identity and existence of additive

inverse• Associativity of scalar multiplication• Distributivity of scalar and vector sums• Scalar multiplication identity

Length of a vector: What is the magnitude (or) length of a vector−→PQ? - distance

between the points P and Q. Hence if for a vector −→v = (v1, · · · , vn), the length isthe distance between the origin and the point (v1, · · · , vn):

|v| = |−→v | =√v21 + · · ·+ v2n

L a,v

a −→v

−→v

Lines: may be of finite length or infinitelength. A line la,b in R2: is a set of thepoints {(x1, x2); x2 = ax1 + b} for fixedpoints a, b ∈ R. A line is associated witha point a ∈ R2 and a direction of a vector−→v , that is, the set of points {a + t−→v : t ∈R}.

3


Hence the analogue in Rn is: For a point a ∈ Rn

and a vector −→v ∈ Rn, La,−→v = {a + t−→v : t ∈R}.

γ

P

b

R

a

Q

c

Direction of a vector: The direction isdetermined by the angle made by the vec-tor with the line L1 (line that is parallel tothe x − axis and is passing through the ori-gin of the vector; the x − axis is presumedto be the initial observation (or) basic refer-ence, in the language of Physics and one ofthe basis element in the language of Mathemat-ics.)

Remember that the polar co-ordinates in R2 isgiven by (r cos θ, r sin θ) and in R3 is given by (r cos θ cosφ, r cos θ sinφ, r sin θ). Howdo we measure an angle between two lines? Recall the cosine formula: given a trianglewith sides a, b, c the angle γ between the sides a and b is given by

cos γ =a2 + b2 − c2

2ab.

In particular, angle γ between two vectors v = (v1, · · · , vn) and u = (u1, · · · , un) isgiven by

cos γ =

∑ni=1 uivi√∑n

i=1 v2i

√∑ni=1 u

2i

Note that the angles γi formed by the vector −→v with the ith-co-ordinate axis satisfy

n∑i=1

cos2 γi = 1, where cos γi =vi√∑ni=1 v

2i

.

cos γi’s are called the directional cosines of the vector −→v . (Picturize the above inR2 and R3. Relate the directional cosines of two vectors and the angle between them.)

One of the very useful operation that branches out from the study of angles be-tween two vectors is dot product. Mathematically, dot product is a function fromthe Cartesian product of Vector space with itself to the scalar field ((·) : Rn×Rn → R)defined by x · y =

∑ni=1 xiyi where x = (x1, · · · , xn) and y = (y1, · · · , yn). From the

above discussion we have (u · v) = |u||v| cos γ where γ denote the angle between thetwo vectors u and v.

If La,−→u denotes a line for a fixed point a ∈ Rn and for a fixed vector −→u of unitmagnitude, then note that vector joining any two points a + t1

−→u and a + t2−→u is

(t2− t1)−→u which is parallel to −→u , that is (t2− t1)u ·u = t2− t1. The vector associatedto a line lies parallel to it and represents the direction of the line. {x : (x− a).v = 0}- represents the line passing through the point a in the direction perpendicular to v.

4


Orthogonality of vectors: Since u·v = |u||v| cos γ, two vectors are perpendicular(orthogonal) if and only if u · v = 0. The vectors

ei = (0, · · · , 0, 1ith−co−ordinate

, 0, · · · , 0), i = 1, · · · , n

are othogonal to each other. (interested may recall inner product, learnt during MTH102, to understand how this basic view is helpful in a wide area of study in Math)

Planes: Note that the equation of hyperplane was briefly discussed. Let us recallequation of planes. For fixed scalars a, b, c ∈ R the plane is the set {(x1, x2, x3) : x3 =ax1 + bx2 + c}.

Note that the lines are one-dimensional in R2, planes are two-dimensional in R3.The planes that are one less dimensional in the given Euclidean space Rn are calledhyperplanes. Naturally, the hyperplane in Rn is given by {(x1, · · · , xn) : xn =a1x1 + · · · + an−1xn−1 + an} for fixed scalars a1, a2, · · · , an ∈ R - not all are zero.In other words, {(x1, · · · , xn) : a1x1 + · · · + an−1xn−1 + anxn = c} for fixed scalarsa1, a2, · · · , an, c ∈ R - not all are zero. (interested may recall Null spaces as a part(or) an advancement in the course MTH102) How do you represent it in terms ofvectors? For a fixed vector v = (v1, · · · , vn) and a constant c the plane is the set

Pc,−→v := {x = (x1, · · · , xn) : x · v = c}.

90o

−→u

x y

If Pc,u denotes a plane for a fixed point c ∈ R anda fixed vector u of unit magnitude, then note thatvector joining any two points x and y in the planesatisfy x ·u = c and y ·u = c which indicates that(x − y).u = 0, that is, the vector joining x andy is perpendicular to u. The vector associated toa plane is normal to it and gives the direction ofthe plane. One may deduce that the plane passingthrough the point a = (a1, · · · , an) and normal tothe vector v is

P va := {x = (x1, · · · , xn) : v · (x− a) = 0}

Cross product of vectors: Recall that the cross product of two vectors v =(v1, v2) and u = (u1, u2) is given by

u× v = u1v2 − u2v1.

Geometrically, the cross product of two vectors in R2 gives twice the area of the tri-angle formed by all the points of the two vectors, in other words, it gives the area ofthe parallelogram spanned by the two vectors. Also note that the formula is just thedeterminant of 2× 2 matrix formed by two vectors.

Like the determinant of the 2 × 2 matrix, do we have an interpretation of thedeterminant of 3 × 3 matrix formed by three vectors in R3? Yes. Recall that, thevector product of two vectors in R3 is a vector again, that is, for u = (u1, u@, u3) and

5


v = (v1, v2, v3),

u× v = (u2v3 − u3v2, u3v1 − u1v3, u1v2 − u2v1).(each co-ordinate is just the cross product of two 2D vectors)

Observe that these are the minors of the determinant of matrix formed by threevectors u, v, w and hence det(u, v, w) = (u× v) ·w, where w = (w1, w2, w3). This (de-terminant) magnitude of three vectors in the three dimensional space is the volumeof the parallelopiped spanned by those three vectors.

A simple computation shows that |u× v|2 = |u|2|v|2 − (u · v)2 in R3 (do it yourselffor R2 as well), which implies that |u× v| = |u||v| sin γ where γ is the angle betweenthe two vectors u and v.

Let us have a closer look at what does the cross product of two vectors in three(and higher) dimensional space represent? The vector u × v is orthogonal to boththe vectors u = (u1, u2, u3) and v = (v1, v2, v3). (Justify yourself via dot (scalar)product.) This observation leads to many interesting solutions for various problems.(For example, to find a vector parallel to the lines of intersection of two planes, it isenough to find the cross product of the normal vectors of those two planes.)

u× v = |u||v| sin γ n,

where γ is the angle between the vectors and n represents the unit normal vector ofthe plane spanned by the vectors.

In 4D: ∣∣∣∣∣∣∣∣t1 t2 t3 t4u1 u2 u3 u4v1 v2 v3 v4e1 e2 e3 e4

∣∣∣∣∣∣∣∣ .You then see (e1, e2, e3, e4) as the canonical basis of R4. Then the previous determi-nant is (α1, α2, α3, α4) with

α1 = t4u3v2 − t3u4v2 − t4u2v3 + t2u4v3 + t3u2v4 − t2u3v4α2 = −t4u3v1 + t3u4v1 + t4u1v3 − t1u4v3 − t3u1v4 + t1u3v4

α3 = t4u2v1 − t2u4v1 − t4u1v2 + t1u4v2 + t2u1v4 − t1u2v4α4 = −t3u2v1 + t2u3v1 + t3u1v2 − t1u3v2 − t2u1v3 + t1u2v3

It’s a vector orthogonal to the other three.Parallelopipeds in higher (> 3D) dimensions: Seminar by interested students

to the interested audience in extra time.

Vector Projections: The vector projection projuv or puv or Puv of a vector

v = (v1, v · · · , vn) =−→PQ onto a non-zero vector u = (u1, · · · , un) =

−→PR is the

perpendicular line from Q to the line segment PR and is given by(v·uu·u

)u. (Picturize)

|v| cos γ = v · u|u| is called the scalar component of v in the direction of u. (interested

may read on how physical terms are highly depending on the terminologies, like howwork is defined)

6


Quadric surfaces: Solution of second degree polynomial in multi-variable, likesphere, ellipsoid, paraboloid, curved surfaces, · · · . How do we find the angle betweentwo (nice) curves? - what do I mean by nice here! The angles are calculated as theangle between two tangents of the respective curves at the point of intersection of thecurves. Now, I said nice, because, if we cannot find the tangent at the intersectionpoints then we are doomed. So nice - mean differentiable curve. We will see in detailas required in the upcoming chapters.

2. Continuity and differentiability of vector-valued functions

2.1. Functions of two or more variables, Limits and continuity: Functionsof two or more variables: Let f : Rn → Rm be a function. (Note that exampleof such functions have been dealt in MTH 102, where we have learnt linear transfor-mation from one vector space to vector space. In this course we will stick to basicfunctions - which need not satisfy linearity. Also, if n = 1 = m, then we have dis-cussed the continuity, differentiability of such functions in MTH 101, which will alsobe recalled then and there in the upcoming chapters. So we will focus on n > 1 andm ≥ 1.) When n > 1,m = 1 we call f a real valued function of a vector variable ora scalar field.

Note that a function f : Rn → Rm at every point x = (x1, · · · , xn) ∈ Rn has avector value in Rm, that is, f(x) = (f1(x), · · · , fm(x)), where we call f1, · · · , fm thecomponents of f and each are scalar valued functions on Rn.

Generally, when m > 1 we call f a vector valued function. When n = 1,m > 1we call f a vector valued function of a real variable. Observe that f : R → Rm

can be written as f(t) = (f1(t), f2(t), · · · fm(t)). (Also referred as parametrization ofcurves/surfaces on Rm).

————Weekend————

7


For a fixed point a ∈ Rn and a positive number r, the set Br(x) or B(a; r) of allpoints x ∈ Rn that satisfies |x − a| < r, where |x| denote the magnitude / norm√∑n

i=1 x2i is called the open n-ball or open ball of radius r and center a.

Let S be a subset of Rn and a ∈ S. a is called an interior point of S if there is anopen n-ball with center at a all of whose points belong to S. b is called an exteriorpoint of S if there is an open n-ball with center at b with no points in S. A point issaid to be a boundary point if it is neither interior nor exterior.

boundary pointinterior point

exterior point

—Open triangle—

A set S ∈ Rn is called Open if all its points areinterior points.

Why do we need these terminology. Recall thatby the definition 6.1 of continuous function, wehave the concept of approaching a point in R.Since in R we have only two side of a point, viz,left and right, the concept of approaching a pointwas both left and right. In fact we had left limitand right limit. However, if we have a point in R2

or R3, the point can be ’approached’ in uncount-able direction. We will address initially what is the analogue of limits initially andthen we will explore if there are any directional limits like left or right limit.

(2.1.1)8


(2.1.2)

9


Limit of a function: Let f : Rn → Rm be a given function. Let a ∈ Rn andb ∈ Rm. We write x→ a if and only if |x−a| → 0. hence lim

x→af(x) = b makes sense as

lim|x−a|→0

|f(x)− b| = 0. Can you say it is equivalent to limx→a|f(x)| = |b|?-easy to see-do

it yourself- that the limit definition implies this. How about the converse? - we willshortly see

Remark 2.1. For proofs WITHOUT ε − δ, PLEASE HAVE A LOOK AT[3]. Some results analogous to what we have seen in MTH 101:

If a ∈ Rn, b ∈ Rm, λ ∈ R, limx→a

f(x) = b and limx→a

g(x) = c, then

• limx→a

(f(x) + g(x)) = b+ c

Proof. Since limx→a

f(x) = b and limx→a

g(x) = c, we have the following:

Let ε > 0 be given. Then there exists δ1, δ2 such that |x−a| < δ1 =⇒ |f(x)−b| < ε/2and |x − a| < δ2 =⇒ |g(x) − c| < ε/2. Hence for δ = min{δ1, δ2}, we have|f(x)− b|+ |g(x)− c| < ε. Thus

by Cauchy’s inequality, |(f(x) + g(x))− (b+ c)| ≤ |f(x)− b|+ |g(x)− c|< ε, ∀ x : |x− a| < δ.

Hence the proof. �

• limx→a

λf(x) = λb (Proof similar to the above one)

• limx→a

f(x) · g(x) = b · c

Proof. Let 0 < ε. Choose 0 < ε1 such that ε21 + (|b| + |c|)ε1 < ε. Since limx→a

f(x) = b

and limx→a

g(x) = c, we have the following: there exists δ1, δ2 such that

|x− a| < δ1 =⇒ |f(x)− b| < ε1 and |x− a| < δ2 =⇒ |g(x)− c| < ε1.

Hence for δ = min{δ1, δ2}, we have,

|f(x) · g(x)− b · c| = |(f(x)− b) · (g(x)− c) + b · (g(x)− c)− (f(x)− b) · c|(by Cauchy’s inequality) ≤ |f(x)− b||g(x)− c|+ |b||g(x)− c)|+ |f(x)− b||c|

< ε21 + (|b|+ |c|)ε1 < ε

�

Similar to the definition 6.3 of continuous function on R, we define continuity formultivariable functions as well. A function f : Rn → Rm is said to be continuous ata point a ∈ Rn if lim

x→af(x) = f(a). As we have seen in Remark 2.1, we can similarly

prove that

• the sum and dot product of two continuous functions is continuous;• scalar multiple of continuous function is again continuous.

Hence we have10


• Polynomial functions in n variables, that is, functions of the form

f(x) =n∑i=1

li∑ki=0

ck1···knxk11 · · · xknn

are continuous at every point in Rn.• Rational functions of two scalar valued continuous functions is again a scalar

field and a continuous function.

Theorem 2.2. f : Rn → Rm is continuous at a point if and only if each componentfk is continuous at that point.

Proof. Let ei = (0, · · · , 1ith−position

, · · · , 0) denote the unit vector along the ith axis.

Suppose f : Rn → Rn is continuous at a point a = (a1, · · · , an). Then for a givenε > 0, there exists δ > 0 such that |x − a| < δ =⇒ |f(x) − f(a)| < ε. Note thatfi(x) = f(x) · ei. Hence for every x with |x− a| < δ we have

|fi(x)− fi(a)| = |f(x) · ei − f(a) · ei| = |(f(x)− f(a)) · ei|= |f(x)− f(a)||ei| = |f(x)− f(a)| < ε

Hence fi’s are continuous at a.

Conversely, suppose fi is continuous for all i at a. Then for a given ε > 0, thereexists δi > 0 such that |x − a| < δi =⇒ |fi(x) − fi(a)| < ε/n. Choose δ =min{δ1, · · · , δn. Since f(x) =

∑ni=1 fi(x)ei we have

|f(x)− f(a)| ≤n∑i=1

|(fi(x)− fi(a))ei| (by Cauchy’s inequality)

≤n∑i=1

ε/n = ε (since |ei| = 1 ∀i).

Hence the proof. �

Remark 2.3. Note that, in the view of the above Theorem 2.2, it is enough to un-derstand the continuity properties for function f : Rn → R.

Example 2.4. • Identity function is continuous at all points.• Linear transformation is a continuous function

(Proof is to be done by yourself)

Example 2.5. Composition of two continuous functions is continuous.

Proof. Let g : Rm → Rn be a continuous function at a ∈ Rm and f : Rn → Rl becontinuous function at g(a) ∈ Rn. Then to prove f ◦ g : Rm → Rl is continuous at a.

From the hypothesis, for ε > 0 there exists δ1 > 0 such that |y − g(a)| < δ1 =⇒|f(y)− f(g(a))| < ε and there exists δ such that |x− a| < δ =⇒ |g(x)− g(a)| < δ1.Thus

|x− a| < δ2 =⇒ |g(x)− g(a)| < δ1

=⇒ |f(g(x))− f(g(a))| < ε

=⇒ |f ◦ g(x)− f ◦ g(a)| < ε.11


�

A scalar valued multi variable function may be continuous at all single variablebut not continuous as a multivariable function. To be precise, if there different limitswhen you approach the point a in different direction, we say that the function isdiscontinuous at a.

Example 2.6. The function f : R2 → R defined as

f(x1, x2) =x1x2x21 + x22

, ∀(x1, x2) 6= (0, 0) and f(0, 0) = 0

is not continuous at (0, 0)

Proof. (Those who are interested: Click here to check out the picturization for different values of x,y)

Note that when x2 = 0, |x−0| = |(x1, 0)−(0, 0)| < δ =⇒ |f(x1, 0)−f(0, 0)| = 0 < ε.However along x2 = x1 we have f(x1, x2) = 1

2for all x1 6= 0 which implies not

continuous. �

2.2. Tangent vectors, Partial derivatives, Gradient and directional deriva-tives. Recall the definition 6.6 of derivative in one variable. How can we interpret in3D? For a given curve, we know what does a tangent mean. ’Tangent vector’ denotesthe tangent line along with its direction on a given curve.

In the given picture at the given point, are there only onetangent? In fact we can rotate the direction of this tangentvector along a plane.

Hence we can speak of tangent plane on a givenpoint in the surface in R3. So how about R4 andother higher dimensional spaces? We have higher di-mensional tangential body on a given point in the givensurface. We will see how to interpret these math-ematically, shortly. One may recall that the tan-gent line was given by another one variable func-tion.

At first we see how we extend the notion of tangent vectorto curves in a general dimensional space, that is, image offunctions/parametrizations of the form f : R → Rn - thisidealogy is equivalent to what we have seen in MTH 101.Secondly, we will see how to picturize the derivatives of thefunctions of the form f : Rn → R - this is termed as de-rivative of scalar field-, thereby we will see how and whatare the different ways (directional, partial, total) to picturizethe tangent planes on Rn for n ≥ 3. From the Example 2.6,the question that arises is: will we be able to speak deriva-tives of surfaces when we approach in one direction and noderivatives in certain other directions?

12

https://academo.org/demos/3d-surface-plotter/


Case 1: f : R→ Rn

Recall the picturization of continuity of parametrized curves in (2.1.2). The image ofthe function will be always a curve. Hence we have only tangent vectors.

f : R→ Rn is said to be differentiable at a ∈ R if and only if

f ′(a) = limh→0

f(a+ h)− f(a)

h

exists. Note that f(x) is a vector and h is a scalar. Hence limh→0

f(a+h)−f(a)h

makes sense

as a scalar (1/h) multiple of a vector. f ′(a) is as well a vector. Also we may speakleft derivative and right derivative as in the previous course.

Note that the limit exists if and only if |f(a+h)−f(a)−hf ′(a)h

| → 0 whenever h → 0.However we have seen that when f : R→ Rn, f(x) = (f1(x), · · · fn(x)), where fi’s arecomponents. Analogue to Theorem 2.2, one may check that, for functions f : R→ Rm

to be differentiable, it is sufficient and necessary that each component is differentiable.( - Proof Submission is encouraged from interesting students) Since every componentthen is a single variable function, this case is easily dealt.

Case 2: f : Rn → Rm

(Note that similar to Theorem 2.2, if we know how to deal f : Rn → R, it is enough.)

Recall the picturization of continuity of scalar fields in (2.1.1). The image of thefunction will not be curves (always). First we speak of a derivative along a givenvector.

Definition 2.7. Derivative of a scalar field with respect to a vector.Given a scalar field f : S → R where S ⊂ Rn. Let a be an interior point of S and lety be an arbitrary point in Rn. The derivative of f at a with respect to y is denotedby the symbol f ′(a; y) and is defined as

f ′(a; y) = limh→0

f(a+ hy)− f(a)

h,

whenever the limit exists.13


(Note that here as well, we can speak of left and limit.)

(If the vector v1 is parallel to say x-axis, then the curve obtained from intersectingthe surface with a plane parallel to yz plane gives a tangent vector which we callthe partial derivative along x-axis.) When the vector y is one of the unit vectors ei,then we call them partial derivatives. When a = (a1, · · · , an) denote the partialderivatives as

f ′(a; ej) = Djf(a) =∂f

∂xj(a1, · · · , an) and f ′xj(a1, · · · , an).

Some easy computations done in the class· · ·

(*) what is the derivative along zero vector?(*) what is the derivative of linear transformation?(*) If g(t) = f(a + ty), then g is differentiable if and only iff ′(a+ ty; y) exists. Also g′(t) = f ′(a+ ty; y).(*) what is gradient? (you might have seen in Physics:) Let fbe a given scalar field on Rn. For a = (a1, · · · , an), ∇f(a) =(∂x1f(a), · · · , ∂xnf(a)).(Note that collecting all the tangent vectors, -whenever exists- give us a tangent plane that sits on the point on the surface.How do we speak of the existence of tangent plane in general.Implicitly it is clear that the derivative should exists alongall the direction. But is that enough?)

Consider the example f(x, y) defined as xy2

x2+y4whenever x 6= 0

and 0 whenever x = 0. As we did in the previous week exercise, it is clear thatthis is not continuous at (0, 0) along x = y2. However, check that the directionalderivative along any vector at origin exists for this function. (done in the class) (Ifyou understood well at the origin, the curve is behaving uncontrollable. If we haveto draw a tangent, then around the origin, there should be nice surface to say thatthe tangent exists. So then exactly when do we say the total derivative exists to afunction? Theoretically,)

14


Definition 2.8. Ta : Rn → R, a linear transformation is called the total derivativeof f : Rn → R at a ∈ Rn if the map Ta and a scalar function E(a, v) exist such that

f(a+ v) = f(a) + Ta(v) + |v|E(a, v),

for all v in a neighborhood of a and E(a, v) → 0 as v → 0. (It turns out to be thenthat the map Ta is a linear transformation)

(Where did we get such an equation? How did we arrive at this definition? Howdoes this give the total derivative or why does this make sense as the derivative?)Recall the Taylor’s theorem 6.8. The equation in the above definition makes moresense now, seeing as an analogue of the equation in the Taylor’s theorem. (However,still it is not clear to us, how are we getting the derivative. Before we look into this indetail, let us recall Mean value theorem:) As an application of Mean value theorem6.7, we have

Theorem 2.9. Mean value theorem for derivatives of scalar fields: Assumethe derivative f ′(a+ ty; y) exists for each t in the interval 0 ≤ t ≤ 1. Then for someteal t0 ∈ (0, 1) we have f(a+ y)− f(a) = f ′(z; y) where z = a+ t0y.

Back to understanding the definition of Total derivative: Let us restrict ourselvesto R2. Consider a function f : R2 → R whose partial derivatives exists. To find thederivative we usually would like to see the rate of change of the function:

f(x0, y0) + (h, k))− f(x0, y0)

(difference),

however the difference here is between the origin and a vector which would implythe distance. Instead of the rate, let us try to see how the difference of changeof the function, that is L = f(x0, y0) + (h, k)) − f(x0, y0) behave through everyvariable one by one: Let g1(x) = f(x, y0 + k) and g2(y) = f(x0, y). Then we haveg′1(x) = ∂xf(x, y0 + k) and g′2(y) = ∂yf(x0, y)

L = f((x0 + h, y0 + k))− f((x0, y0 + k)) + f(x0, y0 + k)− f(x0, y0)

= g1(x0 + h)− g1(x0) + g2(y0 + k)− g2(y0)= hg′1(x) + kg′2(y), for some x ∈ (x0, x0 + h) and y ∈ (y0, y0 + k)

(by the Mean value theorems 6.7 and 2.9)

= h∂xf(x, y0 + k) + k∂yf(x0, y)

= h∂xf(x0, y0) + k∂yf(x0, y0)

+h(∂xf(x, y0 + k)− ∂xf(x0, y0)) + k(∂yf(x0, y)− ∂yf(x0, y0))

= h∂xf(x0, y0) + k∂yf(x0, y0) + |(h, k)|E(x0, y0, h, k)

= (h, k) · ∇f(x0, y0) + |(h, k)|E(x0, y0, h, k)

whereE(x0, y0, h, k) = h|(h,k)|(∂xf(x, y0+k)−∂xf(x0, y0))+

k(h,k)

(∂yf(x0, y)−∂yf(x0, y0)).

Observe that as h and k tend to 0, E(x0, y0, h, k)→ 0 because

h→ 0 and k → 0 =⇒ x→ x0, y0 + k → y0 and y → y0

andh

|(h, k)| ,k

(h, k)are bounded.

Thus f(x0, y0) + (h, k))− f(x0, y0) = (h, k) · ∇f(x0, y0) + |(h, k)|E(x0, y0, h, k).15


Substituting a = (x0, y0), v = (h, k) we have

(2.2.3) f(a+ v)− f(a) = ∇f(a) · v + |v|E(a, v),

which is precisely as in the definition. The left out fact that to be addressed in thedefinition is v in a neighborhood of a. Note that we have arrived at this only withthe existence of partial derivatives; and we have predominantly used only h→ 0 andk → 0 rather than |v| = |(h, k)| → 0. This is possible only when we consider all thedirectional derivatives and do the above machinery again, - in that case, then whatdo we mean by ∇ if we are not taking partial derivatives. In simple words, if we havethe existence of the equation in the definition true for all v in the neighborhood of awith a map Ta instead of ∇, that will take care of this as well.

Important: We have seen an example whose directional derivative exists in all di-rection but not continuous. That serves as the example of a function whose partialderivatives exists but the total derivative dont exist.

f : Rn → R f : R→ RAt a ∈ Rn, f ′(a) = Ta : Rn → R is a function. Ta existsimplies all partial derivatives exist and Ta(v) = ∇f(a)·v

At a ∈ R, f ′(a) is ascalar and f ′ : R→ R

• Ta exists implies the function f is continuous at a• All partial derivatives exist does not imply Ta exist.

But if f is continuous at a and has all partial derivatives at all points aroundsome neighborhood of a, then Ta exists and Ta(v) = ∇f(a) · vIn fact Ta(v) = f ′(a; v)

Hence the differentiability of a scalar field may be checked by initially checking thecontinuity and the existence of the partial derivatives.

• Picturize that the gradient is the normal vector to the surface

Some important applications: (self reading - for interested)

• For f : Rn → R, the equation∑n

i=1∂2f∂x2i

= 0 is called the Laplace equations.

(Steady-state temperature distributions)

• The equation ∂2w∂t2

= c2 ∂2w∂x2

where w represents the wave height, x representsthe distance variable, t represents the time variable and c denotes the velocitywith which the waves are propagated is called Wave equation.

Theorem 2.10. Chain rule: If f : S → R is totally differentiable at γ(t0), whereS is an open suset of Rn and γ : R → S is a differentiable curve, that is, γ is adifferentiable at all the points. Then g := f ◦ γ : R → R is also differentiable at t0and g′(t0) = ∇f(γ(t0)) · γ′(t).

For example suppose a particle moves along a curve in the given time. Mathematically,as t denotes the time, γ(t) denotes the position of the particle in R3. Now say,we would like to check out the temperature of the particle, mathematically, f(γ(t))denotes the temperature of the particle along the curve path that it travels. Howdo we calculate the rate of change of the temparature? While f : Rn → R denotesthe temperature of particle at any position, the rate of change that we would like toobserve is along the change that the particle made, which is along the curve path.Hence the terminology differentiation along the curve.

16


Proof. Let a = γ(t0) ∈ S. Since S is an open set, there exists ε > 0 such thatBε(a) ⊂ S. Now γ is a differentiable function and hence for the ε > 0 above, thereexists a δ > 0 such that

(2.2.4) |t− t0| < δ1 =⇒ |γ(t)− γ(t0)

t− t0− γ′(t0)| < ε.

Since f is differentiable at a, there exists r > 0 such that we have

(2.2.5) f(a+ v)− f(a) = ∇f(a) · v + |v|E(a, v), ∀ |v| < r

and E(a, v)→ 0 as v → 0, that is,

(2.2.6) ∃δ1 > 0 : |E(a, v)| < ε.

For δ0 = min{r, δ, δ1}, substituting a = γ(t0) in the above equation along with v =γ(t)− γ(t0) we have

g(t)− g(t0)

t− t0=

1

t− t0(f(γ(t))− f(γ(t0)))

=1

t− t0(∇f(γ(t0)) · (γ(t)− γ(t0)) + |γ(t)− γ(t0)|E(γ(t0), γ(t)− γ(t0)))

(by 2.2.5)

= ∇f(γ(t0)) ·(γ(t)− γ(t0))

t− t0+|γ(t)− γ(t0)|

t− t0E(γ(t0), γ(t)− γ(t0))

= ∇f(γ(t0)) · γ′(t0) +∇f(γ(t0)) ·(

(γ(t)− γ(t0))

t− t0− γ′(t0)

)+|γ(t)− γ(t0)|

t− t0E(γ(t0), γ(t)− γ(t0))

17


This implies∣∣∣∣g(t)− g(t0)

t− t0−∇f(γ(t0)) · γ′(t0)

∣∣∣∣=

∣∣∣∣∇f(γ(t0)) ·(

(γ(t)− γ(t0))

t− t0− γ′(t0)

)+|γ(t)− γ(t0)|

t− t0E(γ(t0), γ(t)− γ(t0))

∣∣∣∣≤∣∣∣∣∇f(γ(t0)) ·

((γ(t)− γ(t0))

t− t0− γ′(t0)

)∣∣∣∣+ |γ′(t0)||E(γ(t0), γ(t)− γ(t0))|

+

∣∣∣∣ |γ(t)− γ(t0)− (t− t0)γ′(t0)|t− t0

E(γ(t0), γ(t)− γ(t0))

∣∣∣∣< ε by (2.2.6)

As t→ t0, |γ(t)− γ(t0)| → 0 which implies E(γ(t0), γ(t)− γ(t0))→ 0. Hence lettingt→ t0 in the above equation, we have the proof. �

• Note that suppose we have differentiable functions f : S → R and γ : R→ S.Consider the surface P = {x ∈ S : f(x) = 0}. What is the derivative alongthis P?

f(x) = 0 =⇒ f ◦ γ(s) = 0 ∀s=⇒ (f ◦ γ)′(s) = 0 ∀s=⇒ ∇f(γ(s)) · γ′(s) = 0 ∀s

This implies that for all s, γ′(s) and ∇f(γ(s)) are perpendicular. This saysthat the vector ∇f(γ(s)) acts as the normal to the curve γ(s), since γ′(s)denotes the tangent to the curve γ(s).

From the above note can prove the following:

Example 2.11. Let f be a non-constant scalar field, differentiable everywherein the plane and let c be a constant. Suppose f(x1, x2) = c describes a curveC having a tangent at each of its points. Prove that

– the gradient vector ∇f is normal to C.– the directional derivative of f is zero along C.– the directional derivative of f has its largest value in a direction normal

to C - do it yourself -

If f denotes the temperature of the particle on a plane, then the levelsurfaces(or)curvesare called isothermals and the perpendiculars denote theheat flow (lines of flow). Hence the tangent plane to the level surface P at ais given by the equation ∇f(a) · (x− a) = 0.• Recall that a differentiable function f : R2 → R can be written as f(x, y) =f(x0, y0) + ∇f(x0, y0) · (x − x0, y − y0) + |(x − x0, y − y0)|E(x, y, x0, y0) forsome (x, y). We call L(x, y) = f(x0, y0) +∇f(x0, y0) · (x − x0, y − y0) as thelinearization of the function f and also called standard linear approximationof f at a point (x0, y0);E is called the error term. How do we find the error term? If not the exacterror term, can we find how fast it goes to zero?One way to see is to find the second order partial derivatives and find the

18


maximum value M obtained and using the inequality |E(x, y)| ≤ 12M(|x −

x0| + |y − y0|)2. (The exact method to obtain the least upper bound of theerror term will be discussed a little later.)

Differentiation of Vector fields: Let f : S ⊂open

Rn → Rm. Analogous to the

results we have seen above, we can conclude:

• Suppose f(x) = (f1(x), · · · , fm(x)). For a an interior point of S and v anyvector in Rn, we have the directional derivative if the following limit exists

f ′(a; v) = limh→0

f(a+ hy)− f(a)

h

Note that f ′(a; v) is a vector in Rm and f ′(a, v) = (f ′1(a; v), · · · , f ′m(a; v)) =∑mk=1 f

′k(a; v)ek.

• f is differentiable at a with total derivative Ta : Rn → Rm if there exists r > 0and a function E(a, v) : Rn × Rn → Rm such that

f(a+ v)− f(a) = Ta(v) + |v|E(a, v)

for all |v| < r and E(a, v)→ 0 as v → 0.• Total differentiable implies continuity and existence of directional deriva-

tive with Ta(v) = f ′(a, v). If f = (f1, · · · fm) and v = (v1, · · · vn) we haveDf(a)(v) = Ta(v) =

∑mk=1 (∇fk(a) · y) ek = (∇f1(a) · y, · · · ∇fm(a) · y). Note

that Df(a) denotes an m×n Jacobian matrix with coefficients Dkfi(b) = ∂fi(b)∂xk

where i = 1, · · ·m and k = 1, · · ·n:

Ta = Df(a) =

D1f1(a) D2f1(a) · · · Dnf1(a)D1f2(a) D2f2(a) · · · Dnf2(a)· ·· ·· ·

D1fm(a) D2fm(a) · · · Dnfm(a)

.• Chain Rule: Let f and g be two vector fields such that the compositionh = f ◦ g is defined in a neighborhood of a point a. Assume that g is differ-entiable at a with total derivative Dg(a) = Ta. Let b = g(a) and assume thatf is differentiable at b with total derivative Df(b) = T ′b. Then h is differen-tiable at a and the total derivative Dh(a) is given by Dh(a) = T ′b ◦ Ta, thecomposition of transformations.Since Transformations are given by matrices, this boils down to matrix op-erations. Composition of transformations are simply the respective matrixmultiplications.

Example 2.12. Represent the partial derivatives of φ(r, θ) = f(r cos θ, r sin θ)in terms of the partial derivatives of f .

(might have done a different example in the class) Let h(r, θ) = (h1(r, θ), h2(r, θ))be a vector field from R2 to R2, where h1(r, θ) = r cos θ and h2(r, θ) = r sin θare components scalar fields of h. Then for f : R2 → R, we have φ = f ◦ h :R2 → R. By Chain rule we have

∂φ

∂r=∂f

∂xcos θ +

∂f

∂ysin θ,

∂φ

∂θ= −r∂f

∂xsin θ + r

∂f

∂ycos θ

19


2.3. Maxima, Minima and saddle points - Lagrange multipliers.

Recall that when do you say that a function f : R→ R attains maximum or minimum:

(the tangent is parallel to the x-axis, in other words that angle of the tangent withthe x-axis is zero whenver the function attains maxima or minima, locally or globally.We will see the analogue statements in the multi-variable set-up. The concept is sameas single variable. To speak about tangents, in general derivative tests we need tohave differentiable functions, but just to speak about maxima or minima, we dontneed the differentiability nature)A scalar field f : S ⊂ Rn → R is said to have an absolute maxima at a point aof a set S if f(x) ≤ f(a) (for minima: f(x) ≥ f(a)). f is said to have a relativemaxima at a point a if f(x) ≤ f(a) (for relative minima: f(x) ≥ f(a)) for all x insome Br(a). (Note that relative maxima or minima points contains global maxima orminima points, but not the otherwise) A number which is either a relative maximaor a relative minima of f is called an extremum value of f . Let f be a differentiablescalar field. A point a is called a stationary point if ∇f(a) = 0 and a saddle pointif for every r > 0, Br(a) contains points x such that f(x) < f(a) and also other pointssuch that f(x) > f(a).

Theorem 2.13. Second order Taylor formula for scalar fields: Let f be ascalar field with continuous second order partial derivatives Dijf in a Br(a). Thenfor all y in Rn such that a+ v ∈ Br(a) (or |v| ≤ r) we have

f(a+ v)− f(a) = ∇f(a) · y +1

2!vH(a)vt + |v|2E2(a, v),

20


where E2(a, v) → 0 as v → 0 and H(a) denotes the Hessian matrix with coefficientsDijf(a) for all i, j = 1, · · · , n:

H(a) =

D11f(a) D12f(a) · · · D1nf(a)D21f(a) D22f(a) · · · D2nf(a)· ·· ·· ·

Dn1f(a) Dn2f(a) · · · Dnnf(a)

.

Theorem 2.14. Derivative tests for extrema of functions of two variables:Let ∆(a) denote the determinant of the Hessian matrix H(a).

• If ∆(a) < 0, f has saddle point at a.• If ∆(a) > 0 and fxx(a) > 0, then f has a relative minimum at a.• If ∆(a) > 0 and fxx(a) < 0, then f has a relative maximum at a.• If ∆(a) = 0, then the test is inconclusive.

Let us see in detail, why the above theorem, with R2 in mind: Recall (2.2.3).Suppose F : [0, 1] → R is defined as F (t) = f(a + th, b + tk), where f : R2 → R isthrice differentiable functions. Then by Chain rule (the computation was done duringthe class) we have

F ′(t) = ∇f · (h, k) = hfx + kfy

and since fx and fy are differentiable again, we have

F ′′(t) = h2fxx + 2hkfxy + k2fyy.

From the continuity of the partial differentials, we have the continuity of F on [0, 1]and also differentiability on (0, 1). By using Mean value Theorem 6.7 for single

variable functions we have F (1) − F (0) = F ′(0)(1 − 0) + F ′′(c) (1−0)2

2. By replacing

the definition of F (t) = f(a+ th, b+ tk), we have

(2.3.7) f(a+h, b+k)−f(a, b) = ∇f |(a,b) ·(h.k)+1

2(h2fxx+2hkfxy+k2fyy)|(a+ch,b+ck).

From this we can conclude the Hessian matrix as mentioned earlier

H(a, b) =

[fxx(a, b) fxy(a, b)fxy(a, b) fyy(a, b)

]and also conclude the proof of the Theorem 2.14. Also note that the last term in theequation (2.3.7) can be written as

1

2(h, k)H(a, b)(h, k)t =

1

2

[h k

] [fxx(a, b) fxy(a, b)fxy(a, b) fyy(a, b)

] [hk

]at the point (a + ch, b + ck) and hence we may also observe the second term in theconclusion of the Theorem 2.13.Self-read:—the relation between the above test and eigen values of the Hessian matrix—

Note that the Hessian matrix is a real symmetric matrix. Hence there exists an or-thogonal matrix P such that H(a) = CACt where A is a diagonal matrix with eigenvalues λi (i = 1, 2, · · · , n) of H(a) as the elements in the diagonal of the matrix H(a).

21


(- recall from MTH 102 course). Hence vH(a)vt = vCACtvt and denoting y = vC,we have vH(a)vt = yAyt =

∑ni=1 λiy

2i . Thus we have for all y 6= 0

• vH(a)vt > 0 if and only if all eigenvalues λi’s are positive• vH(a)vt < 0 if and only if all eigenvalues λi’s are negative

Exercise: Figure out the connection between ∆(a) and vH(a)vt.Recall the Mid sem question: If exists, find the maxima and minima of f(x, y, z) = z2,subject to the constraint, x2 + y2− z = 0 with justifications. The function f by itselfis not what we are focussing on, to find the maximum and minimum, it is subject tothe given constraint.

Theorem 2.15. Lagrange’s Multipliers: If a scalar field f(x1, · · · , xn) has a rela-tive extremum subject to m constraints, say g1(x1, · · · , xn) = 0, · · · , gm(x1, · · · , xn) =0, where m < n, then there exists m scalars λ1, · · · , λm such that ∇f = λ1∇g1 + · · ·+λm∇gm at each extremum point.

The examples done in the class are from the reference books.

• Finding a point closest to the origin on a plane P : ax + by + cz = d isequivalent finding the minimum value of the function

√x2 + y2 + z2 (how?-as

we saw in the class that it denotes the distance between the origin and anypoint) subject to the constraint ax + by + cz = d. This kind of problems areeasy to observe and do computations: We consider f(x, y, z) = x2 + y2 + z2,substitute the value of z from ax + by + cz = d, then use derivative tests tofind the minimum of the function f .• However, we saw that, it is difficult when we had tried the same method to

find a point closest to the origin on a cylinder.• We observed that the function f(x, y, z) = x2 + y2 + z2 represents sphere and

the point that we have to find is the intersection of the blowing sphere andthe cylinder (given by the function g) where the normals of the sphere andthe cylinder are (scalar multiples) lying in the same line: ∇f = nablag• For more details on two given constraints, look for the problems in the classes

(17th to 19th September). Notes will be distributed then after.

Extreme value theorem: Refer Theorem 9.9 in page 321. We saw the explana-tion of this statement in the class. (will update in detail, if time permits before 20th)What I expect from you is, you should be able to solve problems using the statementof this theorem.

22


2.4. Gist.

1. Vector v = (v1, · · · , vn); Magnitude |v| =√∑n

i=1 v2i ;

Direction - given by the angle γ between the vectorand the x-axis

Scalar/Dot Product -inner product

cos γ = v·e1|v| v · u =

∑ni=1 viui

2. Sum and difference of two vectors: γ

−→v

−→u

−→v −−→u

−→v +−→u

3. Lines in R2: {x : x = a + tv, t ∈ R} - this representsthe line passing through the point a in the direction v.However {x : (x − a).v = 0} - this represents the linepassing through the point a in the direction perpendic-ular to v.

((a + t1v) − (a +t2v)).v =scalar, thatis, a line segment withtwo ends in the line isparallel to v

Since we are in R2, there are only two directions. Soknowing anyone of the two makes sense

4. Planes: {x ∈ R3 : (x − a).v = 0} - this represents theplane passing through a with v as a normal

5. Equation of hyperplane: {x ∈ R3 : (x−a).v = 0} - thisrepresents the hyperspace passing through a with v asa normal

6. Polar Co-ordinates in R2: (r cos θ, r sin θ) (or)(r, θ)

Cylindrical co-ordinates in R3: (r cos θ, r sin θ, x3) or(r, θ, x3)

Spherical (analogous to polar) co-ordinates in R3: (r cos θ cosφ, r cos θ sinφ, r sin θ)

(or) (r, θ, φ)Spherical Co-ordinates in Rn: (r, η1, · · · , ηn−1)The change of co-ordinates are more helpful in integra-tion.

7. Cross product u × v - normal vector to the plane ob-tained by spanning the vectors u and v, (area of theprallelogram spanned by u and v in R2)

8. Functions of several variablesCase (i): f : Rn → R:-f(x1, · · · , xn) ∈ R scalar fieldCase (ii): f : R→ Rm:-f(t) = (f1(t), · · · , fm(t)) parametrization.Case (iii): f : Rn → Rm:-∀x = (x1, · · · , xn) ∈Rn, f(x) = (f1(x), · · · fm(x))

Vector fields

9. Limits and continuity of multivariable functions & Ex-amples

Component wise con-tinuity

- A vector x approach a if and only if |x− a| → 0- We retain then the properties of the continuity ana-logue to one variable

23


10. discontinuity at points and sets- Caution: Curves/sets in higher dimension might lookcontinuous only in certain ways

Example: x1x2x21+x

22

11. Derivative ofparametriccurve

Directional derivative Total derivative

f : R → Rn

is differentiableif and only if ev-ery component isdifferentiable

f : Rn → R has a directionalderivative at a along a vector

v if f ′(a; v) = limh→0

f(a+hv)−f(a)h

exists

f : Rn → R is totally dif-ferentiable a if Df(a) :=Ta : Rn → R exists suchthat f(a + v) − f(a) =Ta(v) + |v|E(a, v) for all|v| < r for some r andE(a, v)→ 0 as |v| → 0

For v = ej, the unit vectoralong the co-ordinate axes, we

call f ′(a; v) = ∂if(a) = ∂if(a)∂xi

the partial derivative for all i =1, · · · , ngradient is given by ∇f(x) =(∂1f(x), · · · , ∂nf(x))

Ta(v) = ∇f(a) · v

Higher order partial deriva-tives

∂xy = ∂yx

Existence of all directionalderivatives does not imply ex-istence of total derivative

Existence of total deriva-tive at a implies existenceof all directional deriva-tives at a

Existence of all directionalderivatives at all points arounda point a and continuity at aimplies existence of total deriv-ative

Existence of total deriva-tive at a implies continuityat a

Exercise problems for the con-verse statements

12. Mean value the-orem for deriva-tives of scalarfields

Assume the derivative f ′(a + ty; y) exists for each t in theinterval 0 ≤ t ≤ 1. Then for some teal t0 ∈ (0, 1) we havef(a+ y)− f(a) = f ′(z; y) where z = a+ t0y.

13. Linearization L(x, y) = f(x0, y0) +∇f(x0, y0) · (x− x0, y − y0)with Error E |E(x, y)| ≤ 1

2M(|x− x0|+ |y − y0|)2

M is the largest value of |∂xxf |, |∂xyf |, |∂yyf |14. Taylor’s theorem Suppose f and its n+1 - order partial derivatives are contin-

uous throughout an open neighborhood S of a point (x0, y0).Then for every (x, y) ∈ S there exists (x1, y1) ∈ S such thatf(x, y) = f(x0, y0) + (h∂x + k∂y)f(x0, y0) + 1

2(h2∂xx +

2hk∂xy + k2∂yy)f(x0, y0) + · · ·+ 1n!

(h∂x + k∂y)nf(x1, y1)

24


15. Chain Rule If f : S → R is totally differentiable at γ(t0), where Sis an open suset of Rn and γ : R → S is a differentiablecurve, that is, γ is a differentiable at all the points. Theng := f ◦ γ : R→ R is also differentiable at t0 and g′(t0) =∇f(γ(t0)) · γ′(t).(Observe the use of chain rule in changing the co-ordinatesystem)

16. Differentiationalong a curve(similar to alongdirection)

When the function γ describes the curve C, the derivativeγ′ is the tangent to the curve. We apply chain rule then toconclude.

17. Equation of tan-gent plane

at the point a = (a1, · · · an), ∇f(a) · (x− a) = 0

Vector Fields:18. f : Rn → Rm the directional derivative if the following limit exists

f(x) is written f ′(a; v) = limh→0

f(a+hy)−f(a)h

(f1(x), · · · , fm(x)) f ′(a; v) is a vector in Rm

f ′(a, v) = (f ′1(a; v), · · · , f ′m(a; v)) =∑m

k=1 f′k(a; v)ek

19. Total derivative Ta(v) = f ′(a, v) =∑m

k=1 (∇fk(a) · y) ek= (∇f1(a) · y, · · · ∇fm(a) · y).Jacobian Matrix:

Ta = Df(a) =

D1f1(a) D2f1(a) · · · Dnf1(a)D1f2(a) D2f2(a) · · · Dnf2(a)· ·· ·· ·

D1fm(a) D2fm(a) · · · Dnfm(a)

.Examples: derivatives of linear transformationHigher order partial derivatives of vector fieldsSimilar to 12.

20. Chain Rule Let f and g be two vector fields such that the composi-tion h = f ◦ g is defined in a neighborhood of a point a.Assume that g is differentiable at a with total derivativeDg(a) = Ta. Let b = g(a) and assume that f is differen-tiable at b with total derivative Df(b) = T ′b. Then h isdifferentiable at a and the total derivative Dh(a) is givenby Dh(a) = T ′b ◦ Ta, the composition of transformations(matrix multiplication).

21. Critical Points The points a for which Taf = 0Note that when f is a scalar field Taf = 0 means that∇f(x) = 0, that is a vector is zero vector at the pointx = a. Solve for x to obtain the critical points.

22. Extrema valuesand saddle points

Step 1: To find critical points

Step 2: To find Hessian matrix and analyse

25


H(a) =

D11f(a) D12f(a) · · · D1nf(a)D21f(a) D22f(a) · · · D2nf(a)· ·· ·· ·

Dn1f(a) Dn2f(a) · · · Dnnf(a)

.∆(a) =the determinant of the Hessian matrix H(a)

Derivative tests In two variables:

• If ∆(a) < 0, f has saddle point at a.• If ∆(a) > 0 and fxx(a) > 0, then f has a relative

minimum at a.• If ∆(a) > 0 and fxx(a) < 0, then f has a relative

maximum at a.• If ∆(a) = 0, then the test is inconclusive.

23. Lagrange’s Mul-tipliers:

If a scalar field f(x1, · · · , xn) has a relative ex-tremum subject to m constraints, say g1(x1, · · · , xn) =0, · · · , gm(x1, · · · , xn) = 0, where m < n, then there existsm scalars λ1, · · · , λm such that ∇f = λ1∇g1+ · · ·+λm∇gmat each extremum point.Solve for x1, · · · , xn, λ1, · · · , λm

24. Extreme valuetheorem:

If f is continuous on a closed rectangle (or cuboid)S inR2 (in R3), then there exists pints c and d in S such thatf(c) = supx∈S f(x) and f(d) = infx∈S f(x).

25. Tangent Linesand Planes:

a) For a single variable function f : I ⊂ R → R, f ′(t)denotes the tangent at the point tb) For a vector valued parametric equation f : I ⊂ R →Rn, f ′(t) = (f ′1(t), · · · , f ′n(t)) denotes the tangent vector.c) For a scalar valued function, f : Rn → R, ∇f |x =(∂1f |x, · · · , ∂nf |x) is normal vector (or line) at the pointx = (x1, · · · , xn) and note that we have tangent plane orline or surfaces depending on the value of n.

— When n = 2, we have tangent line at the point z =(t1, t2)given by the equation

∂f∂x|z(x− t1) + ∂f

∂y|z(y − t2) = 0

— When n = 3, we have tangent plane at the pointz = (t1, t2, t3)given by the equation

∂f∂x1|z(x1 − t1) + ∂f

∂x2|z(x2 − t2) + ∂f

∂x3|z(x3 − t3) = 0

d) Suppose you have a function f : R2 → R3 (recall thequestion in the mid-sem: Let f : [0, 2π]× [0, 2π] defined byf(s, t) = ((2 + cos s) cos t, (2 + cos s) sin t, sin s). Computethe normal: Partial derivatives will give you directionaltangents and hence the normal is given by the determinantof the jacobian, that is, by the 7) in this gist, we have thecross product of the partial derivatives gives the normal.

26


3. Double and triple integrals

3.1. Change of coordinates.

• One may read Apostol to see the details on upper integral and lower integralas mentioned in the class, and further properties of the integrals - the detailswill be observed during professional courses

26. Single integral To find area under the curve f : [a, b]→ R in R2:

Note that the interval I = [a, b] has length∫ badx

Area =∫ baf(x)dx

27. Double Integrals Area of the region R is given by∫ ∫

Rdxdy

To find the volume of the vertical rectangular portion thatis beneath the surface (f : S → R2) above the region Sis given by V olume =

∫ ∫Sf(x, y)dA, where A denote the

area of the region R

• If S = [a, b] × [c, d] is a rectangle, then∫ ∫Sf(x, y)dA =

∫ ba

∫ dcf(x, y)dydx

• If S = {(x, y) : a ≤ x ≤ b; g1(x) ≤ y ≤ g2(y)}, then∫ ∫Sf(x, y)dA =

∫ ba

∫ g2(x)g1(x)

f(x, y)dydx

28. Triple Integrals Volume of the solid S is given by∫ ∫ ∫

Sdxdydz

To find the four dimensional volume we use∫ ∫ ∫Sf(x, y)dV where V denote the volume of the

solid S.29. Change the order

of integration*Sketch the region of integration and figure out the order:Line parallel to the x-axis gives you the starting and endingcurve of x-variable and similarly line parallel to the y-axisgives you the starting and ending curve of y-variable.*

30. Change the co-ordinate intodifferent co-ordinate system

R denotes the region on x− y-plane. Let x = X(u, v) andy = Y (u, v) and T denote the transformed region. Then:∫ ∫

Rf(x, y)dxdu =

∫ ∫Tf(X(u, v), Y (u, v))|J(u, v)|dudv,

where |J(u, v)| denotes the modulus value of the determi-nant of the Jacobian matrix [

∂X∂u

∂Y∂u

∂X∂v

∂Y∂v

](and similarly for the triple integral with 3× 3 Jacobian)In particular, dxdy = rdrdθ for polar co-ordinates.

31. Pappus’ theorem Let S be the solid of revolution generated by rotating theregion R about x-axis, where R is the region obtained be-tween two curves f(x) and g(x) on [a, b]. If a(R) denotesthe area of the region and y = 1

a(R)

∫ ∫Rydxdy denotes

the y-co-ordinate of the centroid of the region R, then

V olume(S) = 2πya(R)[=∫ ba

12(f 2(x)− g2(x))dx].

————Mid Sem————

27


32. Discontinuousfunctions andintegration

Suppose f is a bounded function on the rectangle R ⊂ R2.If f is discontinuous on the subset A = {x ∈ R} andcontinuous on R\A, then

∫ ∫Rfdxdy exists.

33. Double integralsand density

To find the mass distributed over a region (if it is overdiscrete points, we use summation), double integrals canbe usedA thin plate of having a region R with matter distributedover it with density function f on R has mass m(R) =∫ ∫

Rf(x, y)dxdy and hence the average density of the plate

is given by m(R)a(R)

, where a(R) denotes the area of the region

R.Center of the mass: (x, y), with xm(S) =∫ ∫

Rxf(x, y)dxdy, ym(S) =

∫ ∫Ryf(x, y)dxdy

— When the density is constant, we get the above as thecentroid of the region as done in the assignments 6.— SIMILAR CONCEPTS FOR TRIPLE INEGRAL

3.2. Vector fields - Line integrals. (As stated above, the mass M of a region withdistributed matter of density f and area A is given by

∫ ∫f dA)

• How do you find the mass M of a wire (in a plate or in a space) with lengthL in which the matter is distributed with density f? Naturally it seems to be∫f dL. We will make sense of this dL as ds in the following.

• Similarly, suppose if we have a force field f (note that this is a vector fieldand not scalar field) acts on a particle of mass M to move along a curve Cgiven by the parametrization γ. How do you measure the work done by theforce field? We make sense of this as

∫Cf · dγ in the following.

34. Arc length: Let I denote a closed interval. A continuous function γ :I → Rn is called a continuous path in n-space. It is saidto be smooth if the derivative γ′ exists and is continuousin the open interval of I. It is said to be piecewise smoothif the interval I can be partitioned into a finite number ofsubintervals where the path is smooth.

• length of an arc in plane: Partition the arc intosegments Pi = [xi, xi+1]’s, i = 1, 2 · · ·m; the sumof the lengths of these segments, that is, the sumof√

∆x2i + ∆y2i will be close to the length of thearc as the partition is finer and finer. Since bymean value theorem, ∆yi = f ′(x∗i )∆xi and thus

limm→∞∑m

i=1

√∆x2i + (γ′(xi))2∆x2i

• in general,∫ ba||γ′(t)||dt

The arc length function s(u) =∫ ua||γ′(t)||dt and its deriv-

ative s′(t) = ‖γ′(t)‖(Read Apostol for further integral properties)

28


A set S is said to be connected if every pair of points in S can be joined by apiecewise smooth path whose graph lies in S. The line integral of f is independentof the path in S if it is independent of the path from a to b for every pair of pointsa and b in S. A field F on R is said to be conservative if

∫F.dγ is path independent

on R35. Path and line in-

tegralsLine integral of a vector field f defined and bounded on

the image of γ:∫f · dγ =

∫ baf(γ(t)) · γ′(t)dt

—If a particle moves along a curve γ under the action ofthe force field f , the work done by the force f is given bythe above integral. Also the above integral represents thecirculation/flow of the vector field along the curve. So onemay call the above integrals asa flow integral.—A wire C (parametrized as γ on an interval I with lengthfunction s) made of a material with density f has the total

mass M =∫cf(x, y, z)ds =

∫ baf(γ(t))‖γ′(t)‖dt

—Center of mass: (x, y, z) given by xM =∫Cxf(x, y, z)ds,

yM =∫Cyf(x, y, z)ds and zM =

∫Czf(x, y, z)ds

—Flux across a plane curve: -to find the rate at which afluid is entering or leaving a region enclosed by a smoothcurve C: (note that here we deal with closed curve) IfF (x, y) = (F1(x, y), F2(x, y)) denotes the vector field alongthe curve C, with n denoting as the outward-pointing unitnormal vector on C, the flux is given by

∫CF · nds.

Suppose the curve C is parametrized as γ(t) = (γ1(t), γ2(t)) for a ≤ t ≤ b. Theoutward-pointing normal vector is given by either (−γ′2(t), γ′1(t)) or (γ′2(t),−γ′1(t))depending on the direction of the curve. (recall that the unit normal vector nis the vector that satisfies n · γ′(t) = 0, since γ′(t) denotes the tangent vector -so direction of n is also given by the above said vectors). If the curve is clock-wise, then ‖γ′(t)‖n = (−γ′2(t), γ′1(t)) and and if the curve is anti-clockwise, then‖γ′(t)‖n = (γ′2(t),−γ′1(t)). Suppose F (x, y) = (F1(x, y), F2(x, y)). Then the flux isgiven by

∫C

[F1γ′2(t)− F2γ

′1(t)]dt or

∫C

[−F1γ′2(t) + F2γ

′1(t)]dt depending on the circu-

lation of the curve.[Note that ± γ”(t)

‖γ”(t)‖ is also a unit normal vector - where the sign depends

on the direction of the curve. One may use this also to compute the flux.]36. Potential func-

tionIf a vector field h is a gradient of a scalar field f then fis called a potential function for h. The level sets of f arecalled equipotential lines/surfaces/· · ·Let F be a continuous differential vector field. F is a gradi-ent on S if and only if partial derivatives Difj(x) = Djfi(x)for all i, j and x ∈ S.A continuous vector field on an open connected set S is agradient of some potential if and only if f is conservativeon S, that is

∫CF.dγ = 0 for every closed loop C.

—HOW TO CONSTRUCT POTENTIAL FUNCTIONFROM THE GIVEN VECTOR FIELD?——–explained via an example!

29


Example 3.1. Find the potential of f(x, y, z) = (3y4z2, 4x3z2,−3x2y2)

Solution:f is not a gradient Since, note that

∂f

∂x= (0, 12x2z2,−6xy2)

∂f

∂y= (12y3z2, 0,−6x2y)

=⇒ ∂f2∂x6= ∂f1

∂y

Example 3.2. Find the potential: f(x, y, z) = (x+ z,−y − z, x− y)

Solution: (One may quickly check that ∂f1∂y

= ∂f2∂x

= 0, ∂f1∂z

= ∂f3∂x

= 1 and∂f2∂z

= ∂f3∂y

= −1). However, be cautious that the above facts says that a continu-

ous differentiable field f is a gradient if and only if the partial derivatives coincides.Hence there might be a vector field which is a gradient but not differentiable.)Let φ(x, y, z) be the potential of f . Then

∂φ

∂x= x+ z

∂φ

∂y= −y − z

∂φ

∂z= x− y

=⇒ φ(x, y, z) =

∫(x+ z)dx =

x2

2+ xz + φ1(y, z)(3.2.1)

=⇒ ∂φ

∂y=∂φ1(y, z)

∂yand

∂φ

∂z= x+

∂φ1(y, z)

∂z

=⇒ −y − z =∂φ1(y, z)

∂yand − y =

∂φ1(y, z)

∂z

=⇒ φ1(y, z) = −y2

2− yz + φ1(z) and φ1(y, z) = −yz + φ3(y)

=⇒ φ2(z) = c1 and φ3(y) = −y2

2+ c2

Hence (3.2.1) becomes φ(x, y, z) =x2

2− y2

2+ xz − yz + c

3.3. Vector fields - Surface integrals. Note that the following are all three differ-ent ways of expressing the equation of the unit sphere {(x, y, z) : x2 + y2 + z2 = 1}:

• Set of all (x, y, x) that satisfies F (x, y, z) = 0 where F (x, y, z) = x2+y2+z2−1.

• The graph of z = −√

1− x2 − y2.• Spherical co-ordinates or transforming into different co-ordinate systems. For

example, the vector equation of the sphere is given byr(u, v) = (cosu cos v, sinu cos v, sin v) for 0 ≤ u ≤ 2π and 0 ≤ v ≤ 2π.

30


37. — Sphere: r(u, v) = (a cosu cos v, a sinu cos v, a sin v)– Parametricrepresentationand vectorequation

— Cone: r(u, v) = (v sin θ cosu, v sin θ sinu, v cos θ), θ isfixed angle that represents half the vertex angle of the cone,0 ≤ u ≤ 2π and 0 ≤ v ≤ h where h denotes the height ofthe cone.— Recall Torus: (Mid-sem question)Note that we have only two variables here, since we arespeaking about two-dimensional surfaces in 3D. (for theinitial parametric curves γ(t), there was only one variable,since it represents one dimensional curve in 2D or 3D.)

38. – on cross prod-uct

(According to the question in the mid-sem on Torus, howdid we compute the normal? The partial derivatives gavethe tangential vectors on the torus. The vector normalto these tangent vectors, according to the first week lec-ture is the cross product of the two partial derivatives.)So if a surface is given by the vector equation r(u, v) =(X(u, v), Y (u, v), Z(u, v)), where (u, v) belongs to some re-gion R.— For fixed (u, v), r(u, v) is called a regular point of r,if the partial derivatives ∂r/∂u, ∂r/∂v are continuous andthe fundamental vector product ∂r/∂u×∂r/∂v is non-zero.— A point that is not regular is called singular point.—- Regular and singular points depend on the parametriza-tion chosen— The plane spanned by the vectors ∂r/∂u|(u0,v0) and∂r/∂v|(u0,v0) is the tangent plane at the point (u0, v0)—- ∂r/∂u|(u0,v0)×∂r/∂v|(u0,v0) is the normal to the surfaceat (u0, v0).

39. – area of para-metric surfaces

— Recall that a horizontal line v = c in R is transformed asa u−curve in X−Y plane and ∂r/∂u denotes the velocityvector of this curve. Similarly ∂r/∂v denotes the velocityvector of v−curve transformed from the vertical line u = cin R. Hence the amount of change ∆u in u in R is sameas the ‖∂r/∂u‖∆u approximately.hence the area of the small rectangle in u − v plane ismapped in x − y plane as a curved rectangle with areanearly equal ‖ ∂r

∂u× ∂r

∂v‖∆u∆v.

40. Definition of sur-face integral

—surface integrals are nothing but double integrals on thevector fields as the analogue of the line integrals on thevector fields∫ ∫

r(R)fdR =

∫ ∫Sf(r(u, v))‖ ∂r

∂u× ∂r

∂v‖dudv.

41. Mass of the fluidflow

∫ ∫r(R)

F · ndS =∫ ∫

RF · n

∥∥ ∂r∂u× ∂r

∂v

∥∥ dudvwhere

n =∂r∂u× ∂r

∂v

‖ ∂r∂u× ∂r

∂v‖

31


denotes the unit vector of the surface S = r(R), a parametrized surface r(u, v) =(X(u, v), Y (u, v), Z(u, v)) that denotes a collection of points. (Think of S as a fluidwith collection of particles) Let V (x, y, z) denote the velocity of the vector at (x, y, z)(velocity field of the fluid flow. Let us consider that the velocity depends only on theposition of the particle and not on the time, that is we are considering the steady stateflow). Let ρ(x, y, z) denotes the density of the surface (fluid) at the point (x, y, z)(Density is a scalar field). The flux density F (x, y, z) = ρ(x, y, z)V (x, y, z) is a vectorfield. The mass of the fluid (flux is the mass per unit area per unit volume) is givenby∫ ∫

r(R)

F · ndS =

∫ ∫R

F · n∥∥∥∥∂r∂u × ∂r

∂v

∥∥∥∥ dudv = ±∫ ∫

T

F (r(u, v)) · ∂r∂u× ∂r

∂vdudv

where ± is chosen according to the given orientation of the surface. Note that if we

denote ∂(Y,Z)∂(u,v)

= ∂Y∂u

∂Z∂v− ∂Y

∂v∂Z∂u

and similarly other analogue terms, we have

∂r

∂u× ∂r

∂v=

(∂(Y, Z)

∂(u, v),∂(Z,X)

∂(u, v),∂(X, Y )

∂(u, v)

).

-this is called Fundamental vector product.

42. change of co-ordinate systems

Suppose r maps a region A in the uv-plane onto a para-metric surface r(A). Suppose that A is the image of aregion B in the st-plane under a one − to − one continu-ously differentiable mapping G(s, t) = (U(s, t), V (s, t)), if(s, t) ∈ B. Consider the function R defined on B by theequation R(s, t) = r(G(s, t)). Then if the surface integralsexists, then they are equal, that is∫ ∫

r(A)fdS =

∫ ∫r(B)

fdS

43. Notations If F (x, y, z) = (P (x, y, z), Q(x, y, z), R(x, y, z)) andr(u, v) = (X(u, v), Y (u, v), Z(u, v)), then we have∫ ∫

SF · n dS =

∫ ∫TP (r(u, v))∂(Y,Z)

∂(u,v)dudv

+∫ ∫

TQ(r(u, v))∂(Z,X)

∂(u,v)dudv +

∫ ∫TR(r(u, v))∂(X,Y )

∂(u,v)dudv

=∫ ∫

SP dy Λ dz +

∫ ∫SQ dz Λ dx+

∫ ∫SR dx Λ dy

3.4. Green’s, Divergence and Stokes’ theorem. - Curl represents the infinites-imal rotation of the vector field. Suppose F represents the velocity field of a fluidflow, the curl of the field represents the strength and the direction of the rotation orswirling at each point in the space. While the divergence captures the expansion orthe contraction of the fluid flow.(36. can be restated as:) A continuously differentiable vector field is a gradient on agiven open convex set S if and only if we have curl of it as a zero vector on S. Workout the details/proof as an exercise

32


Let F (x, y, z) = (P (x, y, z), Q(x, y, z), R(x, y, z)) be a differentiable vector fieldin the following.

44. Curl of a vectorfield

curl F = ∇× F =(∂R∂y− ∂Q

∂z, ∂P∂z− ∂R

∂x, ∂Q∂x− ∂P

∂y

)45. Divergence of a

vector fielddiv F = ∇ · F = ∂P

∂x+ ∂Q

∂y+ ∂R

∂z.

46. Properties of curland divergence

- trace of the Jacobian matrix of the vector fieldF (x, y, z) = (P (x, y, z), Q(x, y, z), R(x, y, z)) is its diver-gence- Jacobian matrix J can be written as the sum of a sym-metric matrix A = 1

2(J+J t) and a skew-symmetric matrix

B = 12(J − J t). The non-zero elements of B is the compo-

nents of the curl vector.- Divergent of the gradient of a potential function is alsocalled the Laplacian of φ, ∇2φ =div(∇φ). φ is called har-monic if its Laplacian is zero.- A field with curl zero is called irrotational.

Recall the first fundamental theorem 6.9 of Calculus. Analogue to that we can alsostate the second fundamental theorem of calculus for line integrals.47. Second funda-

mental theoremof Calculus

Let φ be a differentiable scalar field with a continuous gra-dient ∇φ on an open connected set S in Rn. Then for anytwo points a and b joined by a piecewise smooth path γ in

S we have∫ ba∇φ · dγ = φ(b)− φ(a).

Green’s theorem is the two-dimensional analogue of the second fundamental theo-rem of calculus - Expressing a double integral over a planar region as a line integralalong the boundary of the region. (In the class, we will try to see why this theoremholds in a special case of R = {(x, y) : a ≤ x ≤ b and f(x) ≤ y ≤ g(x)}.)48. Green’s theorem Let P and Q be scalar fields that are continuously differ-

entiable on an open set S in the xy-plane. Let C be apiecewise smooth Jordan curve and R denote the union ofC and its interior. Assume R is a subset of S. Then wehave∫ ∫

R

(∂Q∂x− ∂P

∂y

)dxdy =

∫CPdx+Qdy,

where the direction of C is in the counter-clockwise direc-tion. (Jordan curve - simply closed curves are called Jordancurves)

Note that... If F (x, y) = (P (x, y), Q(x, y)),(∂Q∂x− ∂P

∂y

)represents the curl of F and(

∂P∂x

+ ∂Q∂y

)represents the divergent of F .

Circulation:∫CF.dγ =

∫ ∫curl Fdxdy

Outward Flux:∫CF.nds =

∫ ∫R

div FdxdyStokes’ theorem is the extension of Green’s theorem. In particular, Stokes’ theorem

and divergence theorem are the generalization of the second fundamental theorem of

33


calculus involving surface inetegrals.49. Stokes’ theorem Assume that S is a smooth simple parametric surface say

S = r(T ) where T is a region in the uv-plane bounded bya piecewise smooth Jordan curve Γ. Assume also that r isa one-to-one continuous differentiable map on some openset containing T ∪ Γ. Let C denote the image of Γ underr and let P,Q and R be continuously differentiable scalarfields on S. Then we have∫ ∫

S

(∂R∂y− ∂Q

∂z

)dy Λ dz +

(∂P∂z− ∂R

∂x

)dz Λ dx

+(∂Q∂x− ∂P

∂y

)dx Λ dy =

∫CPdx+Qdy +Rdz.

can be restatedas:

If F = (P,Q,R) and γ denotes the parametrization of C,then

∫ ∫S(curl F ) · ndS =

∫CF · dγ.

Thus we have seen the relationship between a surface integral on the region andthe line integral on the boundaries of the region in Stokes’ theorem. To relate thetriple integral and the surface integral, we use the divergence of the vector field. Inparticular, the outward flux of a vector field is the same as the integration of theamount of the compression or expansion. In other words, the density within a regionis affected only be changed by its flow through or away from the region from itsboundary.50. Divergence theo-

rem - Gauss’ the-orem

Let V be a solid in 3D by an orientable closed surfaceS and n be the unit outer normal to S. If F is a con-tinuously differentiable vector field defined on V we have∫ ∫ ∫

F(divF )dxdydz =

∫ ∫SF · n dS.

4. First order Ordinary Differential Equations

4.1. Variables separable method; Homogeneous, Linear and Exact equa-tions.

—

34


5. Exercises and Solutions

Week 1

Group number and Exercises: (As long as you understand, you may ignore therigour involved in the solutions. Interested may submit rigorous solutions to me forcorrections.)

(1) Let a,b ∈ Rn, u be a unit vector in Rn and c ∈ R. Find the angle between theplane P := {x ∈ Rn : a·x = c} and the line L := {x ∈ Rn : x = b+tu, t ∈ R}.Hint: The angle between the plane and the line is given as 90o− θ, where θ isthe angle between the normal vector to the plane and the line L. Note thatwe had discussed in the class why a is a normal vector to the plane P and thedirection of the line L is given by the vector u. Hence θ is given by cos θ = a.u

|a| .

(2) Express v = (v1, v2, v3) as the sum of a vector parallel to u = (u1, u2, u3) anda vector orthogonal to u.Hint: Note that sum of two vectors w, z is given as one of the diagonals of theplane spanned by the vectors w, z.Solution: Let w be a vector orthogonal to u(6= 0), that is, w · u = 0 such thatw = v − tu for some t ∈ R. Then

w · u = 0 =⇒ (v − tu) · u = 0

=⇒ v · u = t|u|2

=⇒ w = v − v · u|u|2 u.

Thus v =(v − v·u

|u|2u)

+ v·u|u|2u where w = v − v·u

|u|2u is othogonal to u and v·u|u|2u

is a scalar multiple of u, that is arallel to u.(Other form of solutions are also welcomed)

(3) Justify that the distance from a point A = (a1, a2, a3) to a line through

B = (b1, b2, b3) parallel to a vector (v1, v2, v3) is |−→AB×v||v| .

Solution: A line through B = (b1, b2, b3) parallel to v = (v1, v2, v3) is givenby L := {x : x = B + tv}. The distance between a point A and a line L isgiven by the distance d of the normal vector (perpendicular distance) from Ato the line L. The vector joining two points x = B + v,B ∈ L is given by v.

Now the area of the parallelogram spanned by−→AB and v is given by |−→AB×v|;

also given by the formula base× height - the base is |v| and the height is the

required perpendicular distance d. Hence |−→AB × v| = |v|d. Hence the proof.

(4) Find a vector parallel to the line of intersection of two given planes.Hint: The line L of intersection lies in both the planes. Hence both the normalvectors to the planes are perpendicular to the line L. The cross product of thenormal vectors will be perpendicular to both the normal and hence parallelto the line L.

35


(5) Let PQ denote a diameter of a unit circle. If R denotes any other point in thecircle, show that PR and QR are orthogonal. What can you say about suchscenarios in unit sphere?Hint: For a circle, the proof is done in your mid-schools. In unit sphere, adiameter is the line joining two points A,B in the circle whose distance istwice the radius. If we have another point C in the sphere, there is a circleformed by all these points. - How do you write this rigorously?

For a given scalar field f , that is, f : Rn → R the set of all points x ∈ Rn

such that f(x) = c is called a level set of f . Make a sketch to describe thelevel sets corresponding to the given values of c:

(6) f(x, y) = x2 − y2, c = −2,−1, 0, 1, 2Hint: Except for c = 0, the rest forms hyperbolas. Hence you will be drawingvarious hyperbolas centered at origin in one graph.

(7) f(x, y) = exy, c = e−2, e−1, 1, e, e2 and draw {(x, y, f(x, y)) : x, y ∈ R}Hint: exy = e−2 if and only if xy = −2 which is an equation of hyperbola.So these level sets are again parabolas in one graph (2D graph). However the{(x, y, f(x, y)) : x, y ∈ R} is a challenging.

(8) f(x, y, z) = x2 + y2 + 3z2; c = 0, 6, 12.Hint: Ellipsoid

(9) Draw a curved surface obtained by spanning the lines parallel to the z− axisand passing through the parabola x2 = x21, that is, draw {(t, t2, s) : t, s ∈ R}.

(10) Draw elliptic paraboloid: {(x1, x2, x3) :x214

+x224

= x3}.For all the above five problems, one may check here:https://academo.org/demos/3d-surface-plotter/ to see how the 3D images willlook like.

36

https://academo.org/demos/3d-surface-plotter/


Week 2Solutions will be sent separately

Group number and Exercise problems:

1. Let f : R2 → R.Suppose lim

(x1,x2)→(a1,a2)f(x1, x2) = L, lim

x1→a1f(x1, x2) and lim

x2→a2f(x1, x2) exist.

Then prove that limx1→a1

[limx2→a2

f(x1, x2)

]= lim

x2→a2

[limx1→a1

f(x1, x2)

]= L. - by

Ayesha?

2. State the converse of the above statement. Prove or disprove the same. - byDeepjyothi

Determine the set of points at which the following functions (wherever is well defined)is continuous.

3. f : R2 → R, f(x1, x2) = tanx21x2

- by Chitrak

4. f : R2\{(0, 0)} → R2, f(x1, x2) =

(x1√x21+x

22

, x2√x21+x

22

)- by OmDwivedi

Answer with justifications: What should be the value of f(0, 0) for the follow-ing function (defined everywhere except at the origin) to be continuous at (0, 0)

5. f : R2 → R, f(x1, x2) =sin(x21+x

22)

x21+x22

- By Manish

6. f : R2 → R, f(x1, x2) = x1 sin 1x2

- by Punit

7. f : R2 → R, f(x1, x2) = x1−x2x1+x2

- by Prerna

8. f : R3 → R, f(x1, x2, x3) =x21+x

22−x23

x21+x22+x

23

- by Bhaibhav

9. f : R2 → R, f(x1, x2) =x41x

42

(x21+x42)

3 - By Tanisha

10. Prove that the given curve f : R→ Rn, f(t) = (cos t, sin t, sin 2t) is continuousat t0 = π

2- by Tasneem

37


Week 3problems are from the reference books

Group number and Exercise problems:Solutions shared separately

Find the gradient vector at each point at which it exists for the scalar fieldsdefined by f :

10. f(x, y, z) = log(x2 + 2y2 − 3z2) - by Saurabh9. f(x, y, z) = xy

z- by Swadha

8. Show for f(x, y) = 3√xy that f is continuous and that the partial derivatives

exist at the origin but that the directional derivatives do not exist in all otherdirections - by Hardik

Evaluate the directional derivatives of the following scalar fields for the pointsand directions given:

7. f(x, y, z) = (x/y)z at (1, 1, 1) in the direction of (2, 1,−1) - by Chinmay6. f(x, y, z) = xz2 + y2 + z2 at (1, 0,−1) in the direction of (2, 1, 0) - by Lingaraj

Find all the second order partial derivatives of the following functions:5. f(x, y) = tan−1(y/x) - by Nisha4. f(x, y) = log(x+ y) - by Soumyadeep

Find all the first order partial derivatives of the following functions:3. f(x, y, z) = tanh(x+ y + 3z) - by Bighnesh

2. f(x, y, z) = 1/(√

1 + x+ y2 + z2 - by Banhi Shikha

1. Find ∂2f∂x2

+ ∂2f∂y2

of f(x, y) =√x+

√x2 + y2 - by Amrit

38




Suppose f : R2 → R is differentiable at each point of a ball Br(a) := {x =(x1, x2) : |x− a| < r} of radius r > 0.

2. If ∇f(x) = 0 for every x ∈ Br(a), prove that f is constant on B(a).4. If f(x) ≤ f(a) for all x ∈ Br(a), prove that ∇f(a) = 0.

6. In R3, let f(x, y, z) = |(x, y, z)|. Show that ∇f(x, y, x) is a unit vector in thedirection of (x, y, z).

8. Assume that f ′(x; v) = 0 for every x ∈ Br(a) for fixed r > 0 and for every v.Use the Mean value theorem to prove that f is constant on Br(a).

10. Give an example of a scalar field f such that f ′(x; v) > 0 for a fixed vector vand every vector x. (You may example of f : R3 → Rorf : R2 → R)

1. Let f, g be two scalar field whose partial derivatives exists. Prove that∇(fg)(x) = f(x)∇g(x) + g(x)∇f(x). Also prove that ∇f(x) = 0 for all xif f is a constant function.

We had seen in the class that a differentiable function f : R2 → R can bewritten as f(x, y) = f(x0, y0) + ∇f(x0, y0) · (x − x0, y − y0) + |(x − x0, y −y0)|E(x, y, x0, y0). We call L(x, y) = f(x0, y0)+∇f(x0, y0)·(x−x0, y−y0) as thelinearization of the function f and also called standard linear approximationof f at a point (x0, y0) and E is called the error term. (Refer to [1] page 936and 937 for an example)

3. Find the linearization of f(x, y) = ex cos y at (0, π/2)5. Find an upper bound of the error in the approximation of f(x, y) = x2−3xy+5

at (2, 1) and the rectangle : |x−2| ≤ 0.1 and |y−1| ≤ 0.1 using the inequality|E(x, y)| ≤ 1

2M(|x− x0|+ |y − y0|)2

Find the equation of the tangent plane to the surface defined by f(x, y) at thegiven point:

7. f(x, y) = ax2 + bx2y + cxy2 + dy2, at (1, 1) (a, b, c, d are constants)

9. f(x, y) =∫ x2+y20

e−t2dt at (1, 1).

39




Find the total differentials of the following functions:2. f(x, y) = cos(x+ log y)4. f(x, y) = euv where u(x, y) = cx and v(x, y) = cos y

6. Prove that the functions h(x, y, z) = 1√x2+y2+z2

satisfy the respective Laplace’s

equations: ∂2xxf + ∂2yyf + ∂2zzf = 0.

Let f(x, y, z) = (x+ 2y2 + 3z3, 2y − x2, 0) and g(u, v) = (eu+2v, sin(v + 2u))8. Find the (Jacobian) matrix Df for f and g10. Find the (Jacobian) matrix for h = g ◦ f

1. Find the value of t such that1

2

∂f

∂x(x+

1

2t, y +

1

4t) +

1

4

∂f

∂y(x+

1

2t, y +

1

4t)

= f(x+1

2, y +

1

4)− f(x, y)

where f(x, y) = xy + y2, x = y = 0

Find the critical points of the surfaces and classify them (when does f attainmaxima/minima if existing):

3. f(x, y) = x− 2y + log√x2 + y2 + 3 arctan y

xvwhenever x > 0

5. f(x, y) = (5x+ 7y − 25)e−(x2+xy+y2)

7. Compute the Hessian matrix for f(x, y, z) = x4 + y4 + z4 − 4xyz. What arethe eigen values of the Hessian Matrix. Relate the eigen values and the natureof the critical point, that is, points at which the function attains maxima orminima or saddle points.

9. For what values of c, does the second derivative test guarantee that f(x, y) =x2 + kxy + y2 will have a saddle point, local minimum at (0, 0)?

40


Assignment 6(problems are from the reference books)

Remark: Let A(S) denote the area of the region S ⊂ R2. The centroid (x, y) of theregion S is given by the formula

xA(S) =

∫ ∫S

xdxdy, yA(S) =

∫ ∫S

ydxdy.

We have observed in the class that if S is a region between the graphs of two contin-uous functions f and g over an interval and R is the solid of revolution generated byrotating S about the x− axis we have

V olume(R) = 2πyA(S).

Group number and Exercise problems:6. Make a sketch of the ordinate set (recall that the ordinate set means the graph)

of the following function define on Q = [1, 2]× [1, 4] and compute the volumeof this ordinate set by double integration:f(x, y) = (x+ y)−2 if x ≤ y ≤ 2x and 0 elsewhere

7. Compute the volume of the ordinate set of f(x, y) = y + 2x + 20 over S ={(x, y) : x2 + y2 ≤ 16}.Draw the sketch of the region bounded by the following and find its centroid:

8. y = log x, y = 0, 1 ≤ x ≤ a9. x2/3 + y2/3 = 1, x = 0, y = 0, in first quadrant

(Note that it is not about the x− axis in the following questions.)

10. Calculate the volume of the solid of revolution of the region bounded by y =2√x− 1 and y = x− 1 about the line x = −1.

5. Calculate the volume of the solid of revolution of the region bounded by y = 3√x

and y = x/4 about the y − axis.4. An isosceles triangle T has base 1 and altitude h. The base of T coincides with

one edge of a rectangle R of base 1 and altitude 2. Find the value of h so thatthe centroid of R ∪ T will lie on the edge common to both R and T .

3. Transform the following to iterated integrals in polar co-ordinates:

a)∫ 1

0

∫ √1−x21−x f(x, y)dydx b)

∫ 2

0

∫ 3

xf(√x2 + y2)dydx.

2. Find the linear transformation u = ax+ by, v = cx+ dy that maps the paral-lelogram S in the xy-plane with vertices (0, 0), (2, 10), (3, 17) and (1, 7) ontoa rectangle R in the uv-plane with opposite vertices (0, 0) and (4, 2). Thereby,calculate

∫ ∫Sxydxdy by transforming it into the rectangle as mentioned.

1. Find a suitable change of variables such that∫ ∫

Sf(x+ y)dxdy =

∫ 1

−1 f(u)du,where S = {(x, y) : |x|+ |y| ≤ 1}.

Note: I am not giving assignments on triple integrals. I insist self-assignments onExercises 13.4 in Thomas and Finney’ book: Calculus and analytic geometry.

41




Test if the following are conservative vector fields and find the potential1. F (x, y, z) = (y sin z, x sin z, xy cos z)2. F (x, y, z) = (ey+2z, xey+2z, 2xey+2z)

3. Show that the values of the integral∫ BAz2dx + 2ydy + 2xzdz do not depend

on the path taken from A to B.

4. Integrate f(x, y, z) = (x+ y + z)/(x2 + y2 + z2) over the pathγ(t) = (t, t, t), 0 < a ≤ t ≤ b.

5. Find the flow of F (x, y, z) = (−4xy, 8y, 2) along the curveγ(t) = (t, t2, 1), 0 ≤ t ≤ 2

6. Integrate f(x, y, z) = −√x2 + z2 over the circle

γ(t) = (0, a cos t, a sin t), 0 ≤ t ≤ 2π.

7. Find a field F (x, y) = (F1(x, y), F2(x, y)) with the property that at any point(a, b) 6= 0, F is a unit vector tangent to the circle x2+y2 = a2+b2 and pointingin the clockwise direction.

8. Integrate f(x, y, z) = x+√y− z2 over the path from (0, 0, 0) to (1, 1, 1) given

by the curve following curves:C1 : γ1(t) = (0, 0, t), 0 ≤ t ≤ 1C2 : γ2(t) = (0, t, 1), 0 ≤ t ≤ 1C3 : γ3(t) = (t, 1, 1), 0 ≤ t ≤ 1

Find the flow and the flux of the following field F along the union of the curves:C1 : (a cos t, a sin t) with 0 ≤≤ π and C2 : (t, 0) with −a ≤ t ≤ a

9. F (x, y) = (−y, x)10. F (x, y) = (x2, y2)

42




1. Integrate g(x, y, z) = x√y2 + 4 over the surface cut from the parabolic cylinder

y2 + 4z = 16 by the planes x = 0, x = 1 and z = 0.2. Find the area of the portion of the paraboloid x2 + z2 = 2ay which is cut off

by the plane y = a.3. Compute the area of that portion of the conical surface x2 +y2 = z2 which lies

between the two planes z = 0 and x+ 2z = 3.

4. Show that the parametric surface r(u, v) = (u cos v, u sin v, u2), 0 ≤ u ≤ 4 and0 ≤ v ≤ 2π is a portion of a surface of revolution. Find ∂r/∂u × ∂r/∂v andthe area of S.

5. Find ∂r/∂u × ∂r/∂v and the area of the surface of revolution r(u, v) =(u cos v, u sin v, f(u)). (Write the integration without limits)

6. Find the flux of the field F (x, y, z) = (4x, 4y, 2) outward (away from the z-axis)through the surface cut from the bottom of the paraboloid z = x2 + y2 by theplane z = 1.

7. Let S be the portion of the cylinder y = ex in the first octant that projectsparallel to the x-axis onto the rectangle Ryz : 1 ≤ y ≤ 2, 0 ≤ z ≤ 1 in theyz-plane. Let n be the unit normal vector to S that points away from theyz-plane. Find the flux of the field F (x, y, z) = (−2, 2y, z) across S in thedirection of n.

8. Let S be the level surface of f(x, y, z) = g(x, y) − z whose projection is R.Prove that the area of the surface is

∫ ∫R

√f 2x + f 2

y + 1dxdy.

9. If r = (x, y, z) and A is a constant vector, show that curl(A× r) = 2A and div(‖r‖mr) = 0 for all integers m.

10. Let F be a continuous differentiable vector field. Prove that curl(curlF )=grad(div F )−∇2F .

43




(1) Suppose f is a continuously differentiable function such that it satisfies theLaplace equation fxx + fyy = 0, then

∫fydx − fxdy = 0. What can you say

about the converse? Does∫fydx− fxdy = 0 satisfy the Green theorem?

(2) Verify the Green’s theorem for F (x, y) = (−x2y, xy2), R : x2 + y2 ≤ a2 andC : (a cos t, a sin t), 0 ≤ t ≤ 2π.

(3) Use Green’s theorem to find the outward flux and the counter-clockwise cir-culation flow for the field F (x, y) = (tan−1 y

x, log(x2 + y2)) and C with polar

co-ordinates for 1 ≤ r ≤ 2 and 0 ≤ θ ≤ π.

(4) Show that the value of∫Cxy2dx + (x2y + 2x)dy around any square depends

only on the area of the square and not on its location in the plane.

(5) Use Green’s theorem to find the outward flux and the counter-clockwise circu-lation flow for the field F (x, y) = (− sin y, x cos y) around and over the squarecut from the first quadrant by the lines x = π/2 and y = π/2.

(6) Find the work done by F (x, y) = (4x − 2y, 2x − 4y) in moving a particlecounter-clockwise around C : (x− 2)2 + (y − 2)2 = 4.

(7) Compute∫C−y3dx + x3dy along any circle C. What can you say about any

other closed curve C.

(8) Apply Green’s theorem to find the line integral∫y2dx + x2dy along the tri-

angle bounded by x = 0, x+ y = 1 and y = 0.

(9) Apply Green’s theorem to find the line integral∫

3ydx+2xdy along the bound-ary given by 0 ≤ x ≤ π, 0 ≤ y ≤ sinx.

(10) Find the area of the region enclosed by the curve γ(t) = (t2, t3/3− t), −√

3 ≤t ≤√

3 using Green’s theorem.

44




(1) Let C be any simple closed curve in the plane 2x+2y+z = 2 oriented counter-clockwise. Show that

∫C

2ydx + 3zdy − xdz depends only on the area of theregion enclosed by C and not on the position or shape of C.

(2) Using Stokes’ theorem, calculate the circulation of the field F (x, y, z) = (x2y3, 1, z)along C : the intersection of the cylinder x2 + y2 = 4 and the hemispherex2 + y2 + z2 = 16, z ≥ 0.

(3) Using Stokes’ theorem, calculate the circulation of the field F (x, y, z) = (y2 +z2, x2 + z2, x2 + y2) along the boundary C : of the triangle cut from the planex+ y + z = 1 by the first octant counter-clockwise when viewed from above.

(4) Show that the curl of F (x, y, z) = ( −yx2+y2

, xx2+y2

, z) is zero but∫CF · dγ 6= 0 if

γ denotes the parametrization of the unit circle C. Why Stokes’ theorem isfailing here?

(5) Let F (x, y, z) = (−z+ 12+x

, tan−1 y, x+ 14+z

). Find the value of∫ ∫

S∇×F ·ndA

where n is the outer unit normal (away from the origin) of the parabolic shellS : 4x2 + y + z2 = 4, y ≥ 0

(6) Let S be the cylinder x2+y2 = a2, 0 ≤ z ≤ h together with its top x2+y2 ≤ a2,z = h. Use Stokes’ theorem to calculate the flux of ∇ × F outward throughS, where F (x, y, z) = (−y, x, x2).

(7) Prove that ∇ × ∇φ = 0 for any scalar field φ. Use this identity and Stokes’theorem to prove that the circulation of the field F = ∇ × (x, y, z) aroundany smooth closed oriented surface in space is zero.

(8) Let f(x, y, z) = (x2 + y2 + z2)−1/2. Show that the clockwise circulation of thefield F = ∇f around the circle x2 + y2 = a2 in the xy-plane is zero by findingthe line integral and also by Stokes’theorem.

(9) Find the flux of the curl of the field F (x, y, z) = (3y, 5− 2x, z2− 2) across thesurface S : r(u, v) = (

√3 sinu cos v,

√3 sinu sin v,

√3 cosu), 0 ≤ u ≤ π/2 and

0 ≤ v ≤ 2π in the direction of the outward unit normal n.

(10) Find the flux of the curl of the field F (x, y, z) = (y− z, z−x, x+ z) across thesurface S : r(u, v) = (u cos v, u sin v, 9 − u2) in the direction of the outwardunit normal n.

45


6. Appendix: One variable Calculus

6.1. Recall: Limit Concept and Continuity.Limit concepts in functions:

Definition 6.1. For a function f : A → R a real number is said to be a limit of fat c if given any ε > 0, there eixsts a δ > 0 such that for x ∈ A with 0 < |x − c| <δ =⇒ |f(x)− L| < ε.

• If f : A→ R then f can have only one limit point at c.• Sequential Criterion: Let f : A→ R be a function.

limx→c

f(x) = L

if and only if

for every sequence (xn) in A that converges to c such that xn 6= c for alln ∈ N, the sequence (f(xn)) converges to L.

Definition 6.2. A function f : A→ R is said to be bounded on a neighborhood of c,if there exists a δ-neighborhood Vδ(c) of c and a constant M > 0 such that |f(x)| ≤Mfor all x ∈ A ∩ Vδ(c).

• For A ⊂ R, limx→c

f(x) exists, then f is bounded on some neighborhood of c.

• (algebra of limit of functions)• Squeeze theorem: Let A ⊂ R and f, g, h : A → R. If f(x) ≤ g(x) ≤h(x) ∀x ∈ A, x 6= c and f

x→c(x) = L = h

x→c(x), then g

x→c(x) = L

Definition 6.3. Continuous functions: Let A ⊂ R and f : A → R. At c ∈ A,we say that f is continuous at c, if for given ε > 0, there exists δ > 0 such that|x− c| < δ =⇒ |f(x)− f(c)| < ε(Note that the choice of δ depends on ε and at c)

• Sequential Criterion: Let f : A→ R be a function.

limx→c

f(x) = f(c)

if and only if

for every sequence (xn) in A that converges to c such that xn 6= c for alln ∈ N, the sequence (f(xn)) converges to f(c).

• (algebra of continuous functions)• Boundedness Theorem: Let I := [a, b] be a closed bounded interval and

let f : I → R be continuous on I. Then f is bounded on I.

Definition 6.4. Let A ⊂ R and let f : A → R. We say that f has an absolutemaximum on A if there is a point x0 ∈ A such that f(x0) ≥ f(x) for all x ∈ A.Similarly, we say that f has an absolute minimum on A if there is a point x0 ∈ Asuch that f(x0) ≤ f(x) for all x ∈ A.

• Maximum-minimum theorem: Let I = [a, b] be a closed bounded intervaland let f : I → R be continuous on I. Then f has an absolute maximum andan absolute minimum on I.

46


• Location of Roots: Let I = [a, b] and let f : I → R be continuous on I. Iff(a) < 0 < f(b) or if f(b) < 0 < f(a), then there exists a number c ∈ (a, b)such that f(c) = 0.(The proof of this is called bisection method)• Bolzano’s Intermediate value theorem: Let I be an interval and letf : I → R be continuous on I. If a, b ∈ I with f(a) < d < f(b), then thereexists a point c ∈ (a, b) such that f(c) = d. Figure 1:

x

y

(0, 0) a b

f(a)

f(b)

d = f(c)

c

f(bn)

bn

f(an)

an

• A continuous function on a closed and bounded interval is a closed boundedinterval. In fact, f(I) is an interval

Definition 6.5. A function f : A→ R is said to be uniformly continuous on A,if for any given ε > 0 there exists δ > 0 such that |x − y| < δ =⇒ |f(x) − f(y)| <ε ∀ x, y ∈ A.(note that the δ is choosen only depending on ε and not on x unlike continuitydefinition)

• (Note that it is easy to check uniform continuity on a set implies continuityat point in the set).Uniform Continuity theorem: Let I be a closed bounded interval andf : I → R be continuous on I. Then f is uniformly continuous on I.• Continuous inverse theorem: Let I ⊂ R be an interval and f be a real

valued continuous function defined on I such that it is strictly monotone. Thenthe inverse of f (what is the inverse of f here mean?) is strictly monotone andcontinuous on J = f(I).• Let f : I → R be a monotone function defined on the interval I. The set of

points at which f is discontinuous is a countable set.47


Let f be a given real valued function on set in R.Continuity of f at a point x0 Uniform Continuity of f on a set AFor a given ε > 0 For a given ε > 0there exists δ > 0 such that there exists δ > 0 such that(δ depends on f, ε, x0) (δ depends on f, ε, A)|x− x0| < δ =⇒ |f(x)− f(x0)| < ε. |x− y| < δ =⇒ |f(x)− f(y)| < ε, ∀x, y ∈ A.

6.2. Differentiation: Motivation: The key point to study about the derivative ofa function is since there is a connection between the derivative and the average slope;recall the slope formula - y−y′

x−x′ .

Definition 6.6. A real-valued function f [a, b]→ R is differentiable at x0 ∈ (a, b)if

limh→0

f(x0 + h)− f(x0)

h= lim

x0→x

f(x)− f(x0)

x− x0= f ′(x0) exists.

For a function on closed interval, it is differentiable at the end points if the appro-priate one-sided limit exists. When the function is differentiable at all points of theinterval I we say that is differentiable on I.When f is differentiable at x0 we define the tangent line to f at x0 to be the affinefunction (affine functions f(x) = l(x)+c where l(x) a linear function - Seen in Linearalgebra:) T (x) = f(x0) + f ′(x0)(x − x0). Note that if f is differentiable at a pointthen the tangent line passes through the point (x0, f(x0)) with slope f ′(x0).

slope m

P

Q1

slope m1

Q2

slope m2

Q3

slope m3

Q4

slope m4

Note that ’locally’ theslope of the tangent lineequals the slope of thefunction. We see thatthe function is ’approxi-mated’ by tangent lines:Let f be a function on[a, b] that is differentiableat x0. Let T (x) be thetangent line to f at x0.Then T is the unique affine(line) function with prop-

erty that limx→x0

f(x)−T (x)x−x0 =

0.

Let I ⊂ R be an interval and f be a real valued function defined on I. Let c ∈ I.

We say that f is differentiable at c if limx→c

f(x)−f(c)x−c exists and the existing limit is

called derivative (-denoted as f ′(c)) of f at c.

Trivia: The function f ′ is defined on I if the function f is differentiable at all pointsof I. f ′ is defined on I by finding f ′(x) for all x ∈ I.

48


• (differentiability implies continuity) If f is differentiable at x0 then it is con-tinuous at x0.How about the converse? Not true. Consider (the standard example) f(x) =|x| on any interval that does not contain 0. When we say that a functionis not differentiable, we say that the limit in the definition of differentiablefunctions is not exiting. In the example the right limit in the definition or theright derivative is 1 and the left derivative is -1.Example of a function that is differentiable except at one point butcontinuous everywhere: f(x) = xsin(1/x) for x 6= 0 and 0 for x = 0.(continuous at 0 since |sin(1/x)| ≤ 1 and x→ 0 for x very small)• Combinations of differentiable functions - refer pages 163 and 164.• Caratheodory’s Theorem: Let f be defined on an interval I containing

the point c. f is differentiable at c if and only if there exists a function φ onI that is continuous at c and satisfies f(x)− f(c) = φ(x)(x− c) for x ∈ I.In this case, we have φ(c) = f ′(c).• Chain Rule: Let I, J be intervals in R. Let g : I → R and f : J → R be

functions such that f(J) ⊂ I. Let c ∈ J . If f is differentiable at c and if g isdifferentiable at f(c), then the composite function g ◦ f is differentiable at cand

(g ◦ f)′(c) = g′(f(c))f ′(c).

• Derivative of the inverse of a function: Let I be an interval in R and fbe strictly monotone continuous function on I. Let J = f(I) and g denotethe inverse of the function. (Note that by 31), g is strictly monotone andcontinuous) If f is differentiable at c ∈ I and f ′(c) 6= 0, then g is differentiableat d = f(c) and

g′(d) =1

f ′(c)=

1

f ′(g(d)).

A real valued function f defined on an interval I is said to have a relative (local)maximum at c ∈ I, if there exists a neighborhood Vδ(c) such that f(x) ≤ f(c) forall x ∈ Vδ(c)∩I. Similarly it is said to have a relative (local) minimum at c ∈ I, ifthere exists a neighborhood Vδ(c) such that f(x) ≥ f(c) for all x ∈ Vδ(c)∩ I. f has arelative (local ) extremum at c if it has either local maximum at c or local minimumat c.

• Let c be an interior point of the interval I at which f : I → R has a relative(local) extremum.• Fermat’s theorem: Let f be a continuous function on an interval [a, b] that

takes its maximum or minimum value at a point x0. Then(1) x0 is an endpoint, that is x0 = a or x0 = b (or)(2) f is not differentiable at x0 (or)(3) f is differentiable at x0 and f ′(x0) = 0.

In other words, if real function f defined on (a, b) has a local maximum ata point x0 ∈ (a, b) and if f ′(x0) exists then f ′(x0) = 0• Rolle’s Theorem: Suppose that f is continuous on a closed interval I = [a, b]

and the derivative f ′ exists at every point in the open interval (a, b) withf(a) = f(b) = 0. Then there exists at least one point c ∈ (a, b) such that

49


f ′(c) = 0

x

y

(0, 0) (a, f(a))

(b, f(b))

In fact Rolle’s theorem statesthat there exists a point c between a and b such that the slope at c is same asthe slope of the line joining the points (a, f(a)) and (b, f(b)) of graph of f .Assignment: What happens if f(a) = f(b) = 0 is replaced in the above state-ment with f(a) = f(b)? Will the statement be still true?(Differentiability at every interior point of [a, b] is necessary: Exam-ple: f(x) = 1−|x| on [−1, 1]. It does not have any point at which f ′(x) = 0. fis differentiable at every point except the point x = 0 at which the maximumoccurs.)

Theorem 6.7. Mean Value theorem: Suppose that f is continuous on a closedinterval I = [a, b] and f has a derivative in the open interval (a, b). Then there existsat least one point c ∈ (a, b) such that f(b)− f(a) = f ′(c)(b− a)

• Let f be differentiable function on the interval I = [a, b].a) f is incrasing on I if and only if f ′(x) ≥ 0 for all x ∈ I.b) f is decreasing on I if and only if f ′(x) ≤ 0 for all x ∈ I.

• First Derivative Test for Extremum points: Let f be a continuous func-tion on the interval I = [a, b] and c be an interior point of I. Suppose f isdifferentiable on (a, c) and (c, b). Then

a) If there is a neighborhood (c − δ, c + δ) ⊂ I such that f ′(x) ≥ 0 forc − δ < x < c and f ′(x) ≤ 0 for c < x < c + δ, then f has a relativemaximum at c.

b) If there is a neighborhood (c − δ, c + δ) ⊂ I such that f ′(x) ≤ 0 forc − δ < x < c and f ′(x) ≥ 0 for c < x < c + δ, then f has a relativeminimum at c.

• Darboux’s Theorem: If f is differentiable on I = [a, b] and if k is a numberbetween f ′(a) and f ′(b), then there is at least one point c ∈ (a, b) such thatf ′(c) = k.

If f is differentiable on [a, b] then f ′ cannot have any simple discontinuitieson [a, b].The function f on R that assumes 0 for irrational and 1 for rational is anexample of everywhere discontinuous limit function which is not Riemann-integrable.

50


a) Suppose that limx→a+

f(x) = 0 = limx→a+

g(x). If limx→a+

f ′(x)g′(x)

= L ∈ R ∪{∞,−∞}, then

limx→a+

f(x)

g(x)= L.

b) Suppose that limx→a+

g(x) = ±∞. If limx→a+

f ′(x)g′(x)

= L ∈ R ∪ {∞,−∞}, then

limx→a+

f(x)

g(x)= L.

• L’Hospital’s Rules: Let −∞ ≤ a < b ≤ ∞ and let f, g be differentiable on(a, b) such that g′(x) 6= 0 ∀ x ∈ (a, b).

Theorem 6.8. Taylor’s theorem: Let n ∈ N and let I := [a, b]. Suppose f : I → Rbe such that f and its derivatives f ′, f”, · · · , f (n) are continuous on I and f (n+1) existson (a, b). If x0 ∈ I, then for any x ∈ I, there exists a point c between x and x0 suchthat

f(x) = f(x0) + f ′(x0)(x− x0) +f”(x0)

2!(x− x0)2

+ · · ·+ f (n)(x0)

n!(x− x0)n +

f (n+1)(c)

(n+ 1)!(x− x0)n+1.

The following is a collection of derivative formulas by Animesh Nayak:Let gi : [0,∞)→ R be a differentiable functions for i = 1, 2, 3 with g3(x) 6= 0∀x. Letc ∈ R and a 6= 0 ∈ R.

Let f : [0,∞)→ R Then the derivative f ′

be defined as: wherever it is definedis given by:

f(x) = c f ′(x) = 0 ∀xf(x) = xr f ′(x) = nxr−1 ∀x,∀r ∈ R,

(if r < 0, x 6= 0)f(x) = g1 ◦ g2(x) f ′(x) = g′1(g2(x))g′2(x) ∀x,∀n ∈ Nf(x) = cg1(x) f ′(x) = cg′1(x) ∀x

f(x) = (g1 + g2)(x) f ′(x) = g′1(x) + g′2(x) ∀xf(x) = (g1 − g2)(x) f ′(x) = g′1(x)− g′2(x) ∀xf(x) = g1(x)g2(x) f ′(x) = g1(x)g′2(x) + g′1(x)g2(x) ∀xf(x) = g1(x)

g3(x)f ′(x) =

g3(x)g′1(x)−g1(x)g′3(x)(g3(x))2

∀xf(x) = loga(x) f ′(x) = 1

(log a)x∀x 6= 0

f(x) = ex f ′(x) = ex ∀xf(x) = ax f ′(x) = ax log a ∀x

f(x) = sin(x) f ′(x) = cos(x) ∀xf(x) = cos(x) f ′(x) = − sin(x) ∀xf(x) = tan(x) f ′(x) = sec2(x) ∀xf(x) = cot(x) f ′(x) = −cosec2(x) ∀xf(x) = sec(x) f ′(x) = sec(x)tan(x) ∀xf(x) = cosec(x) f ′(x) = −cosec(x)cot(x) ∀x

51


6.3. Integration. MTH101:Integration, Riemann integral viewed as anarea, partitions, upper and lower integrals, existence of the Riemannintegral, basic properties, fundamental theorem of integral calculus,integration by parts, applications

Riemann Integral

(Throughout, we assume f to be real and bounded)Partition: For a closed bounded interval in R, a partition of I is a finite orderedset P := {x0, x1, · · · , xn} of points where a = x0 < x1, · · · , xn−1, xn = b. The normof P is the number |P| := max{x1 − x0, x2 − x1, · · ·xn − xn−1}.A partition P ∗ is said to be a refinement of the partition P if P ∗ ⊃ P . A partitionP ∗ is said to be common refinement of the given two partitions P1 and P2 ifP ∗ = P1 ∪ P2.

Let P ∗ be a refinement of P . Then L(P, f) ≤ L(P ∗, f) and U(P ∗, f) ≤ U(P, f).

Notation:

• ∆xi = xi − xi−1.• Mi = supx∈[xi−1,xi]

f(x).• mi = infx∈[xi−1,xi] f(x).• Upper sum: U(P, f) =

∑ni=1Mi∆xi.

• Lower sum: L(P, f) =∑n

i=1mi∆xi.

• Upper integral:∫ bafdx = inf U(P, f).

• Lower integral:∫ bafdx = supL(P, f).

We say that a function f is Riemann integrable over [a, b] or belongs to R (class ofRiemann integral function) if the lower and upper integral are equal and we denote

the value by∫ baf(x)dx.

R[a, b] denotes the class of Riemann integrable functions.Step function: I(x) = {0 if x ≤ 0 and 1 if x > 0}.• Suppose f, g ∈ R[a, b]. Then

a) For all k ∈ R, the function kf ∈ R[a, b] and∫ bakf(x)dx = d

∫ baf(x)dx.

b) f + g ∈ R[a, b] and∫ ba(f + g)(x)dx =

∫ baf(x)dx+

∫ bag(x)dx.

c) If f(x) ≤ g(x) ∀ x ∈ [a, b], then∫ baf(x)dx ≤

∫ bag(x)dx.

• If f ∈ R[a, b], then f is bounded on [a, b].• Squeeze Theorem Let f : [a, b]→ R. f ∈ R[a, b] if and only if for every ε > 0

there exists function f1,ε, f2,ε ∈ R[a, b] with f1,ε(x) ≤ f(x) ≤ f2,ε(x) ∀x ∈ [a, b]

and∫ ba(f1,ε(x)− f2,ε(x))dx < ε.

• A continuous function on [a, b] is Riemann integrable on [a, b].So what functions are integrable?Answer: Suppose f is bounded on [a, b]. f ∈ R if and only if f is continuousalmost everywhere on [a, b].• If a function is monotone on [a, b] then f ∈ R[a, b].• Let f : [a, b] → R and let c ∈ (a, b). f ∈ R[a, b] if and only if the restriction

of f to [a, c] and [c, b] are both Riemann integrable. In this case∫ baf(x)dx =∫ c

af(x)dx+

∫ bcf(x)dx.

52


• Why integration and differentiation can be considered as inverse operations?Let f ∈ R on [a, b]. For a ≤ x ≤ b, let F (x) =

∫ xaf(t)dt. Then F is

continuous on [a, b]. Furthermore, if f is continuous at a point x0 of [a, b],then F is differentiable at x0 and F ′(x0) = f(x0).

Theorem 6.9. Suppose there is a finite set E in [a, b] and functions f, F :[a, b]→ R such that

– F is continuous on [a, b]– F ′(x) = f(x) for all x ∈ [a, b]\E– f ∈ R[a, b].

Then∫ baf = F (b)− F (a).

• Integration by parts: Suppose F and G are differentiable functions on [a, b],F ′ = f ∈ R and G′ = g ∈ R. Then∫ b

a

Fg = F (b)G(b)− F (a)G(a)−∫ b

a

fG.

53


References

[1] G. B. Thomas and R. L. Finney, Calculus and Analytic Geometry, 9th edition, Indian studentedition, Addison-Wesley, 1998

[2] R. Courant and F. John, Introduction to Calculus and Analysis, Vol. 2, Classics in Mathematics,Springer, 1989

[3] T. M. Apostol, Calculus, Volumes 1 and 2, 2nd edition, Wiley Eastern, 1980

Department of Mathematical SciencesIISER Berhampur, India.E-mail: [email protected]

54

Documents

MULTIVARIABLE CALCULUS DIFFERENTIAL EQUATIONS …senthilraani.files.wordpress.com/2019/10/mth201prefinaldraft.pdfv(cx+y) = cT!cv (x)+T!v (y)thanks animesh and Rithik!)- one-one: T!v