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7/24/2019 Chapter 2 Geometrical Applications of Calculus
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Anti-differentiation: The process of nding a primitive(original) function from the derivative. It is the inverseoperation to differentiation
Concavity: The shape of a curve as it bends around (it canbe concave up or concave down)
Differentiation: The process of nding the gradient of atangent to a curve or the derivative
Gradient of a secant: The gradient (slope) of a linebetween two points that lie close together on a curve
Gradient of a tangent: The gradient (slope) of a line thatis a tangent to a curve at a point on a function. It is thederivative of the function
Horizontal point of inexion: A stationary point (wherethe rst derivative is zero) where the concavity of thecurve changes
Instantaneous rate of change: The derivative of a function
Maximum turning point: A local stationary point (where
the rst derivative is zero) and where the curve isconcave down. The gradient of the tangent is zero
Minimum turning point: A local stationary point (wherethe rst derivative is zero) and where the curve isconcave up. The gradient of the tangent is zero
Monotonic increasing or decreasing function: A function isalways increasing or decreasing
Point of inexion: A point at which the curve is neitherconcave upwards nor downwards, but where theconcavity changes
Primitive function: The original function found byworking backwards from the derivative. Found by anti-differentiation
Rate of change: The rate at which the dependent variablechanges as the independent variable changes
Stationary (turning) point: A local point at which thegradient of the tangent is zero and the tangent ishorizontal. The rst derivative is zero
TERMINOLOGY
GeometricalApplications
of Calculus
2
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51Chapter 2 Geometrical Applications of Calculus
DID YOU KNOW?
Although Newtonand Leibnizare said to have discovered calculus, elements of calculus were
around before then. It was Newton and Leibniz who perfectedthe processes and developed the
notation associated with calculus.
Pierre de Fermat(160165) used coordinate geometry to nd maximum and minimum
values of functions. His method is very close to calculus. He also worked out a way of nding the
tangent to a curve.
The 17th-century mathematicians who developed calculus knew that it worked, but it was
not fully understood. Limits were not introduced into calculus until the nineteenth century.
INTRODUCTION
YOU LEARNED ABOUTdifferentiation in the Preliminary Course. This is the
process of finding the gradient of a tangent to a curve. This chapter looks at
how the gradient of a tangent can be used to describe the shape of a curve.Knowing this will enable us to sketch various curves and find their maximum
and minimum values. The theory also allows us to solve various problems
involving maximum and minimum values.
Gradient of a Curve
To learn about the shape of a curve, we first need to revise what we know
about the gradient of a tangent. The gradient (slope) of a straight line measures
the rate of change of ywith respect to the change in x.
Since the gradient of a curve varies, we find the gradient of the tangent at
each point along the curve.
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EXAMPLES
1.
2.
3.
4.
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53Chapter 2 Geometrical Applications of Calculus
In the examples on the previous page, where the gradient is positive, the
curve is going up, or increasing (reading from left to right).
Where the gradient is negative, the curve is going downwards, or
decreasing.
The gradient is zeroat particular points of the curves. At these points
the curve isnt increasing or decreasing. We say the curve is stationaryat thesepoints.
If 0,f x >l] g the curve is increasingIf 0,f x l
] g for all x(monotonic increasing)or f x 0f x
l
] g for increasing curve. xx
x
2 4 0
2 4
2
i.e >
>
>
So the curve is increasing for x 2> .
This function is a
parabola.
CONTINUED
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2. Find the stationary point on the parabolay x x6 32= + .
Solution
dx
dy
x2 6=
For stationary points,dx
dy
x
x
x
0
2 6 0
2 6
3
i.e.
=
=
=
=
When ,x y3 3 6 3 3
6
2= = +
=
] g
So the stationary point is ,3 6^ h.
3. Find any stationary points on the curvey x x48 73= .
Solution
y x3 482= l
`
For stationary points,
y
x
x
x
x
0
3 48 0
3 48
16
4
i.e. 2
2
2
=
=
=
=
=
l
When , ( )x y4 4 48 4 7
135
3= =
=
When , ( )x y4 4 48 4 7
121
3= =
=
] g
So the stationary points are ,4 135^ hand ,4 121^ h.You will use stationary points
to sketch curves later in this
chapter.
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55Chapter 2 Geometrical Applications of Calculus
PROBLEM
What is wrong with this working out?
Find the stationary point on the curvey x x2 12= + .
Solution
y x4 1= +l
.
y
x
x
x
0
4 1 0
4 1
0 25
For stationary points,
i.e.
=
+ =
=
=
l
When . , ( . )x y0 25 4 0 25 1
1 1
0
= = +
= +
=
So the stationary point is . ,0 25 0
^ h.
Can you nd the
correct answer?
1. Find the parts of each curve
where the gradient of the tangent
is positive, negative or zero. Label
each curve with +, or 0.
(a)
(b)
(c)
(d)
2. Find all values of xfor which the
curvey x x2 2= is decreasing.
3. Find the domain over which
the function f x x4 2= ] g isincreasing.
4. Find values of xfor which the
curvey x x3 42= is
decreasing(a)increasing(b)
stationary.(c)
5. Show that the function
f x x2 7= ] g is always(monotonic) decreasing.
2.1 Exercises
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6. Prove thaty x3= is monotonic
increasing for all 0x .
7. Find the stationary point on the
curve ( ) xf x 3= .
8. Find all x values for which the
curvey x x x2 3 36 93 2= + + is
stationary.
9. Find all stationary points on the
curve
(a) y x x2 32=
(b) f x x9 2= ] g (c) y x x x2 9 12 43 2= +
(d) .y x x2 14 2= +
10. Find any stationary points on thecurvey x 2 4= ] g .
11. Find all values of xfor which
the curve ( ) x xf x 3 43= + is
decreasing.
12. Find the domain over which the
curvey x x x12 45 303 2= + + is
increasing.
13. Find any values of xfor which the
curvey x x x2 21 60 3
3 2= +
isstationary(a)
decreasing(b)
increasing.(c)
14.The function f x x px2 72= + +] g has a stationary point at x 3= .
Evaluatep.
15. Evaluate aand bif
3y x ax bx3 2= + has stationary
points at x 1= and x 2= .
16. (a) Find the derivative of
.y x x x3 27 33 2= +
(b) Show that the curve is
monotonic increasing for all
values of x.
17. Sketch a function with 0f x >l] g for x< 2, 0f 2 =l] g and 0f x .
18. Draw a sketch showing a curve
withdxdy 0< for x 4< ,
dxdy 0= when
x 4= anddx
dy0> for x 4> .
19. Sketch a curve withdx
dy0> for all
1x anddx
dy0= when x 1= .
20. Draw a sketch of a function
that has 0f x >l] g for x 2< ,x 5
>, 0f x
=l
] g for ,x 2 5=
and0
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57Chapter 2 Geometrical Applications of Calculus
Types of Stationary Points
There are three types of stationary points.
Local minimum point
The curve is decreasing on the left and increasing on the right of the
minimum turning point.
x LHS Minimum RHSf xl] g < 0 0 > 0
Local maximum point
The curve is increasing on the left and decreasing on the right of the
maximum turning point.
x LHS Maximum RHS
f xl] g > 0 0 < 0
Local maximum and minimum points are also called turning points,
as the curve turns around at these points. They can also be called relative
maxima or minima.
Point of horizontal inflexion
The curve is either increasing on both sides of the inflexion or it is decreasing
on both sides. It is not called a turning point as the curve does not turn
around at this point.
The derivative has the
same sign on both sides
of the inexion.
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x LHS Inflexion RHS
f xl] g > 0 0 > 0or < 0 0 < 0
The stationary points are important to the shape of a curve. A reasonablyaccurate sketch of the curve can be made by finding these points, together
with the intercepts on the axes if possible.
EXAMPLES
1. Find the stationary point on the curvey x3= and determine which
type it is.
Solution
dx
dyx3 2=
dx
dy
x
x
0
3 0
0
For stationary points,
i.e. 2
=
=
=
0, 0x y
0
When 3= =
=
So the stationary point is ,0 0^ h.To determine its type, check the curve on the LHS and the RHS.
x 1 0 1
dx
dy3 0 3
Since the curve is increasing on both sides, (0, 0) is a point of inflexion.
Substitute x =1 intodx
dy.
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59Chapter 2 Geometrical Applications of Calculus
2. Find any stationary points on the curve f x x x x2 15 24 73 2= + ] g anddistinguish between them.
Solution
6 30 24f x x x2= +l] g ( )
( )( )
f x
x x
x x
x x
x
0
6 30 24 0
5 4 0
1 4 0
1 4
For stationary points,
i.e.
or
2
2
`
=
+ =
+ =
=
=
l
( )f 1 2 1 15 1 24 1 7
4
= +
=
3 2] ] ]g g g
So ,1 4^ his a stationary point.
x 0 1 2f xl] g 24 0 12
,1 4` ^ his a maximum stationary point( ) ( )f 4 2 4 15 4 24 4 7
23
= +
=
3 2] ]g g
So ,4 23^ his a stationary point.
x 2 4 5
f xl] g 12 0 24
,4 23` ^ his a minimum stationary point.
Substitutex 0= andx 2=
into (x)f . Take care that
there is no other stationary
point between the x values
you choose.
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1. Find the stationary point on the
curvey x 12= and show that it
is a minimum point by checkingthe derivative on both sides of it.
2. Find the stationary point on the
curvey x4= and determine its
type.
3. Find the stationary point on the
curvey x 23= + and determine its
nature.
4. The function f x 2x x7 4= ] g has one stationary point. Find itscoordinates and show that it is a
maximum turning point.
5. Find the turning point on
the curvey x x3 6 12= + + and
determine its nature.
6. For the curvey x4 2= ] g find theturning point and determine its
nature.
7. The curvey x x6 53 2
= + has 2turning points. Find them and
use the derivative to determine
their nature.
8. Show that the curve f x x 15= +] g has a point of inflexion at ,0 1^ h.
9. Find the turning points on
the curvey x x3 53 2= + and
determine their nature.
10. Find any stationary points on thecurve f x 4 2 .x x2 3= ] g Whattype of stationary points are they?
11. The curvey x x3 23= + has
2 stationary points. Find their
coordinates and determine
their type.
12. The curvey x mx x7 55 3= + +
has a stationary point at x 2= .
Find the value of m.
13. For a certain function,
.f x x3= +l] g For what valueof xdoes the function have a
stationary point? What type of
stationary point is it?
14. A curve has ( 1).f x x x= +l] g Forwhat xvalues does the curve have
stationary points? What type are
they?
15. For a certain curve,
( ) ( ) .dx
dyx x1 22= Find the x
values of its stationary points and
determine their nature.
16. (a) DifferentiateP x x250
= + with
respect to x.
Find any stationary points(b)
on the curve and determine their
nature.
17. For the function ,h h
A82 52 +
=
find any stationary points and
determine their nature.
18. Find any stationary points for
the function 40 V r r3= and
determine their nature (correct to
2 decimal places).
19. Find any stationary points on the
curve rS r120
2= + (correct to 2
decimal places) and determine
their nature.
20. (a) Differentiate .A x x3600 2=
Find any stationary points for(b)
A x x3600 2= (to 1 decimal
place) and determine their nature.
2.2 Exercises
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61Chapter 2 Geometrical Applications of Calculus
Higher Derivatives
A function can be differentiated several times:
differentiating f x] ggives f xl] gdifferentiating f xl] ggives f xm] gdifferentiating f xm] ggives ,f xn] g and so onthe other notation is ,
dx
d y
dx
d y
2
2
3
3
and so on
EXAMPLES
1. Find the first 4 derivatives of ( )f x x x x4 3 23 2= + .
Solution
( )x x x3 8 3= +
( )
( )
x x
x
6 8
0
=
=
( )x 6=
f
f
f
2
f
l
m
n
mm
2. Find the second derivative ofy x2 5 7= +] g .Solution
dx
dyx
x
dx
d yx
x
7 2 5 2
14 2 5
14 6 2 5 2
168 2 5
6
6
2
25
5
$
$ $
= +
= +
= +
= +
] ]]
]
g gg
g
3. Find f 1m] gif f x x 14= ] g Solution
( )
( )( )
f x x
f x xf
4
121 12 1
12
3
2
2`
=
=
=
=
l
m
m ] g
Only the rst 2 derivatives
are used in sketching
graphs of curves.
Differentiating further will
only give 0.
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2.3 Exercises
1. Find the first 4 derivatives of
x x x x2 37 5 4 + .
2. If f x x 59 =] g , find .f xm] g 3. Find f xl] gand f xm] gif
f x x x2 15 3 +=] g .4. Find f 1l] gand f 2m] g, given
f t t t t 3 2 5 44 3= + ] g .5. Find the first 3 derivatives of
x x x2 4 77 6 4 + .
6. Differentiatey x x2 3 32= +
twice.
7. If f x x x x x2 5 14 3 2= + ] g , findf 1l] gand f 2m] g.
8. Differentiate x 4 twice.
9. If ,g x x=] g find .g 4m] g 10. Given h t t t 5 2 53 2= + + , find
dt
d h2
2
when t 1= .
11. Find the value of xfor which
dxd y 3
2
2
= giveny x x x3 2 53 2= + .
12. Find all values of xfor
which f x 0>m] g given thatf x 3 2x x x 9= + +
] g.
13. Differentiate x4 3 5] g twice.14. Find f xl] gand f xm] gif
.f x x2= ] g 15. Find the first and second
derivatives of .f xxx3 1
5=
+] g
16. Finddt
d v2
2
if ( ) .v t t3 2 1 2= + ] g17. Find the value of bin
y bx x x2 5 43 2= + + ifdx
d y2
2
2
=
when .x21
=
18. Find the exact value of f 2m] gif( )f x x x3 4= .
19. Find f 1m] gif ( )f t t t 2 1 7= ] g .20. Find the value of bif
f x bx x5 42 3= ] g and .f 1 3 = m] g
Sign of the Second Derivative
The second derivative gives extra information about a curve that helps us to
find its shape.
Since f xm] gis the derivative of f xl] g, then f xm] gand f xl] ghave the samerelationship as f xl] gand f x] g.
( ) ( )
( ) ( )
( ) ( )
f x f x
f x f x
f x f x
0
0
0
That is if then is increasing
if then is decreasing
if then is stationary
>
m] g then f xl] gis increasing. This means that the gradient of thetangentis increasing, that is, the curve is becoming steeper.
Notice the upward shape of these curves. The curve lies above the
tangents. We say that the curve is concave upwards.
If f x 0
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Class Investigation
How would you check that concavity changes?
The sign must change for
concavity to change.
What type of point is it?
If ,f x 0>m] g the curve is concave upwards.If ,f x 0
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65Chapter 2 Geometrical Applications of Calculus
2. Find all values of xfor which the curve xf x x x2 7 5 43 2= +] g isconcave downwards.
Solution
( )
( )
f x x x
f x x
6 14 5
12 14
2=
=
l
m
For concave downwards, ( )f x
x
x
0
12 14 0
12 14
2
2
As the curve is decreasing, the number of unemployed people is(b)
decreasing.
Since the curve is concave upwards, the gradient is increasing. This(c)
means that the unemployment rate is increasing.
Remember that the gradient,
or rst derivative, measures
the rate of change.
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1. For each curve, describe the sign
ofdx
dyand
dx
d y
2
2
.
(a)
(b)
(c)
(d)
(e)
2. The curve below shows the
population of a colony of seals.
Describe the sign of the first(a)
and second derivatives.
How is the population rate(b)
changing?
3. Inflation is increasing, but the
rate of increase is slowing. Draw a
graph to show this trend.
4. Draw a sketch to show the shape
of each curve:
(a) andf x f x0 0< l m] ]g g (d) andf x f x0 0> >l m
] ]g g
5. The size of classes at a local TAFE
college is decreasing and the rate
at which this is happening is
decreasing. Draw a graph to show
this.
6. As an iceblock melts, the rate at
which it melts increases. Draw a
graph to show this information.
7. The graph shows the decay of a
radioactive substance.
Describe the sign ofdt
dMand .
dt
d M2
2
2.5 Exercises
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8. The populationPof fish in a
certain lake was studied over
time, and at the start the number
of fish was 2500.
During the study,(a) .dt
dP0<
What does this say about the
number of fish during the study?
If at the same time,(b) ,d
d P0>
2
2
what can you say about the
population rate?
Sketch the graph of the(c)
populationPagainst t.
9. The graph shows the level of
education of youths in a certain
rural area over the past 100 years.
Describe how the level of
education has changed over this
period of time. Include mention
of the rate of change.
10. The graph shows the number ofstudents in a high school over
several years.
Describe how the schoolpopulation is changing over time,
including the rate of change.
dx
dy0>
dx
dy0
2
2
dx
d y0m] g there is a minimumturning point.
If f x 0=l] g and ,f x 0=l m] ]g g Maximumturning point: , .f x f x0 0
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71Chapter 2 Geometrical Applications of Calculus
EXAMPLES
1. Find the stationary points on the curve f x x x x2 3 12 73 2= +] g anddistinguish between them.
Solution
f x x x6 6 122= l] g For stationary points, f x 0=l] g
( ) ( )
x x
x x
x x
6 6 12 0
2 0
2 1 0
i.e. 2
2
=
=
+ =
( ) ( )
x
f
2 1
2 2 2 3 2 12 2 7
13
or2
` =
= +
=
3] ]g g
So ( , )2 13 is a stationary point.
( ) ( )f 1 2 1 3 1 12 1 7
14
3 2 = +
=
] ]g g
So ,1 14^ his a stationary point.Now ( )
( ) ( )
(concave upwards)
f x x
f
12 6
2 12 2 6
18
0>
=
=
=
m
m
So ,2 13^ his a minimum turning point.( ) ( )
(concave downwards)
f 1 12 1 618
0
=
=
m
,3 26` ^ his a minimum turning point( ) ( )
0 (concave wards)
1 6 1 6
12
down
2= =
=
m ] g
So (1, 0) is a minimum turning point.
When ,
(concave upwards)
x y1 12 1 48
0>
= =
=
2m
] g
So (1, 0) is a minimum turning point.
When ,
(concave downwards)
x y0 12 0 4
4
02
2
=
=
=
=
=
x 3` = gives a minimum value
So 300 items manufactured each year will give the minimum expenses.
(c) When ,x E3 3 6 3 12
3
2= = +
=
] g
So the minimum expense per year is $30 000.
2. The formula for the volume of a pond is given by
,V h h h2 12 18 503 2= + + where his the depth of the pond in metres.
Find the maximum volume of the pond.
Solution
V h h6 24 182= +l
For stationary points, 0
i.e. 6 24 18 0
( )( )
V
h h
h h
h h
4 3 0
1 3 0
2
2
=
+ =
+ =
=
l
E is the expense in units of
ten thousand dollars.
This will give any maximum
or minimum values of E.
x is the number of items inhundreds.
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So h = 1 or 3
V h12 24= m
When , ( )h V1 12 1 24= = m
(concave downwards)12
0
=
So h= 3 gives a minimum V.
When ,h V1 2 1 12 1 18 1 50
85
= = + +
=
3 2] ] ]g g g
So the maximum volume is 58 m3
.
In the above examples, the equation is given as part of the question.
However, we often need to work out an equation before we can start
answering the question. Working out the equation is often the hardest part!
EXAMPLES
1. A rectangular prism has a base with length twice its breadth. The volume
is to be 300 cm3. Show that the surface area is given by .S x x49002= +
Solution
Volume:
2
( )
V lbh
x x h
x h
x h
300 2
2
3001
2
2`
=
=
=
=
Surface area:
( )
( )
( )
S bh lh
x xh xh
x xh
x xh
lb2
2 2 2
2 2 3
4 6
2
2
2
= + +
= + +
= +
= +
CONTINUED
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Now we substitute (1) into this equation.
S x xx
x
x
4 62
300
4900
22
2
$= +
= +
2. ABCDis a rectangle withAB= 10 cm andBC= 8 cm. LengthAE= xcm
and CF=ycm.
Show that triangles(a) AEBand CBFare similar.
Show that(b) xy= 80.
Show that triangle(c) EDFhas area given by .A x x80 5320
= + +
Solution
(a) (given)
(corresponding , )(similarly, )
EAB BCF
BFC EBABEA FBC
AB DCAD BC
90
s
+ +
+ +
+ +
+