Assignment 2, SolutionsProblem 10/ p. 684. Sketch the region in the plane consisting of all points whose polarcoordinates (r, θ) satisfy the given conditions:

2 < r ≤ 5, 3π/4 < θ < 5π/4.

Solution. The sector is bounded by the black curve that is a part of the circle of radius 5centered at the origin, by the two blue line-segments, and by the blue curve that is a partof the circle of radius 2 centered at the origin; not including the blue boundary because ofthe strict inequalities there. See the plot:

Problem 26/page 684. Find a polar equation for the curve represented by the Cartesianequation xy = 4.Solution. x = r(θ) cos(θ), y = r(θ) sin(θ) and

r2(θ) cos(θ) sin(θ) = 4.

From here, taking into account that x and y must have the same sign (both positive orboth negative) we conclude that the polar angle θ must be restricted to0 ≤ θ ≤ π/2 andπ ≤ θ ≤ 3π/2. Assuming that r ≥ 0 we have

r(θ) =2√

cos(θ) sin(θ)=

2√

2√sin(2θ)

.

In order to verify the result first by using CAS Maple parametric plot

plot([[cos(t) ∗ 2 ∗ 20.5/(sin(2 ∗ t)0.5), sin(t) ∗ 2 ∗ 20.5/(sin(2 ∗ t)0.5), t = 0..P i/2],

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[cos(t)∗2∗20.5/(sin(2∗t)0.5), sin(t)∗2∗20.5/(sin(2∗t)0.5), t = Pi..3∗Pi/2]], color = [black]);

we draw the graph of the curve with polar coordinates

r(θ) =2√

2√sin(2θ)

, x = r(θ) cos(θ), y = r(θ) sin(θ), 0 ≤ θ ≤ π/2, π ≤ θ ≤ 3π/2.

and next, by using CAS Maple implicit plot we draw the graph of the same curve but withthe Cartesian equation xy = 4:

implicitplot(x ∗ y = 4, x = −30..30, y = −30..30);

We see that both graphs are the same so, this verifies the correctness of our answer.Problem 30/ p.684. Sketch the curve with the given polar equation

r2 − 3r + 2 = 0.

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Solution. We haver2 − 3r + 2 = (r − 1)(r − 2) = 0

and from here r = 1 or r = 2. Hence, the curve consists of two circle centered at the originof radius 1 and 2, correspondingly. Here is the graph of the curve by using CAS Maplepolar-plot:

polarplot([1, 2], t = 0..2 ∗ Pi);

Problem 38/ p.684. Sketch the curve with the given polar equation

r = cos(5θ).

Solution. We have | cos(5θ)| ≤ 1 and let us call the points from the curve with θ such thatcos(5θ) = ±1 the pick-points of the curve.(a) For k = 0, 2, 4, 6, 8 at 5θk = kπ ⇒ θk = kπ

5 we have cos(5θk) = 1. Hence, the pointswith polar coordinates (1, 0), (1, 2π/5), (1, 4π/5), (1, 6π/5), (1, 8π/5) are pick-points of thecurve.(b) For k = 1, 3, 5, 7, 9 at 5θk = kπ ⇒ θk = kπ

5 we have cos(5θk) = −1. Hence, thepoints with polar coordinates (−1, π/5) = (1, 6π/5), (−1, 3π/5) = (1, 8π/5), (−1, 5π/5) =(1, 0), (−1, 7π/5) = (1, 2π/5), (−1, 9π/5) = (1, 4π/5) are from the curve but these are thesame points described in (a).From here the curve will contain only 5 leaves with pick-points given in (a).(c) In order to see better the shape of the leaves we are to determine the polar coordinatesof the points with r = 0, i.e., r = cos(5θs) = 0 ⇒ θs = (2s+1)π

10 , s = 0, 1, 2, 3, 4, 5, 6, 7, 8, 9.From here we obtain 10 points with polar coordinates:

(0, π/10), (0, 3π/10), (0, 5π/10), (0, 7π/10), (0, 9π/10),

(0, 11π/10), (0, 13π/10), (0, 15π/10), (0, 17π/10), (0, 19π/10)

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all of them coinciding with the pole (the origin) of the polar coordinate system. May beyou remember that we agree the pole to be considered as a set of all pointswith polar coordinates (0, θ) with r = 0 and any angle θ.Description of the sketch of the curve:(1) Between θ = 0 and θ = π/10 we sketch a half-leaf with a pick point (1, 0).(2) Between θ = π/10 and θ = 3π/10 we sketch one leaf with a pick point (−1, 2π/10) =(1, 6π/5).(3) Between θ = 3π/10 and θ = 5π/10 we sketch one leaf with a pick point (1, 2π/5).(4) Between θ = 5π/10 and θ = 7π/10 we sketch one leaf with a pick point (−1, 3π/5) =(1, 8π/5).(5) Between θ = 7π/10 and θ = 9π/10 we sketch one leaf with a pick point (1, 4π/5).(6) Between θ = 9π/10 and θ = π we sketch the second-half to the half-leave sketched in(1). Note that the curve is entirely traced once for 0 ≤ θ ≤ π. Here is the sketch of thefive leaves curve r = cos(5θ):

Remark. it In order to verify the sketch we can use CAS Maple polar-plot:

polarplot(cos(5 ∗ t), t = 0..P i, color = [blue]);

or CAS Maple parametric plot:

plot([cos(5 ∗ t) ∗ cos(t), cos(5 ∗ t) ∗ sin(t), t = 0..P i], color = [red]);

Problem 56/ p.684. Match the polar equations with the graphs labeled I-VI. Give reasonsof your choice. Do not use a graphing device.Solution. (a) Consider first the polar equations (a) and (b). These two curves are spirals

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and for θ > 1 the spiral (b) will be more stretched than the spiral (a); and for θ < 1 thespiral (b) will shrink more than (a). From here the polar equation (a) corresponds tothe curve V and the polar equation (b) corresponds to the curve II.From the other 4 curves only one, that is I, does not have the origin (0,0) as its point. Fromhere, the polar equation (e) corresponds to the curve I because concerning the polarequation r(θ) = 2 + sin(3θ) in (e) we have

r = 2 + sin(3θ) ≥ 2− 1 = 1 > 0.

Now, consider what is left: (c), (d), (f) and III, IV, VI. We observe that between the curvesIII,IV,VI, only the curve III has two points with strictly positive cartesian x-coordinate. Thepoints on the x-axis distinct from (0,0) of one curve has a polar angle θ = kπ, k = 0, 1, 2, . . ..With θ = 0 in (d) we obtain a point with cartesian coordinates (3,0) and with θ = π in(d) we obtain a point with polar coordinates (−1, π) = (1, 2π) = (1, 0) that is a point withcartesian coordinates (1,0). Hence, (d) corresponds to III.Now, consider what is left: (c), (f) and IV, VI. The points with cartesian x-coordinate0, excluding the origin (0, 0), correspond to a polar angle θ = (2k+1)π

2 , k = 0, 1, 2, . . ..Consider the polar equation (f). For θ = π/2 we obtain a point with polar coordinates(−1, π/2) = (1, 3π/2) and for θ = 3π/2 we obtain a point with polar coordinates (3, 3π/2)hence, 2 points with negative cartesian y-coordinates on the y-axis. Obviously, choosingfrom IV and VI, this corresponds to IV. Hence, the polar equation (f) corresponds tothe curve IV. In addition to our observation we can add that, concerning the curve witha polar equation (c) and all its points on the y-axis (possibly excluding the origin (0,0)) wehave

cos(

(2k + 1)π2

13

)= cos

((2k + 1)π

6

)= {−

√3/2, 0,

√3/2}

and from here again we can conclude that the polar equation (c) does not correspond tothe curve IV hence, the polar equation (f) corresponds to the curve IV.Finally, from the previous considerations, the polar equation (c) corresponds to thecurve VI.Remark. By using CAS Maple polar-plot we plot the curves with polar equation(a),(b),(c),(d),(e),(f) in order to verify the solution:

5

(a)

(b)

(c)

(d)

(e)

6

(f)

Problem 12/ page 689. Sketch the curve and find the area A that it encloses:

r(θ) = 2− sin(θ).

Solution.

A =∫ 2π

0

(2− sin(θ))2

2dθ =

9π2.

Problem 32/ page 689. Find the area A of the region that lies inside both curves:

r(θ) = 3 + 2 cos(θ), r(θ) = 3 + 2 sin(θ).

Solution. 3 + 2 cos(θ) = 3 + 2 sin(θ) for θ = π/4, 5π/4. From here considering for conve-nience a negative angle in the first integral we have

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A =∫ π/4

−3π/4

(3 + 2 sin(θ))2

2dθ +

∫ 5π/4

π/4

(3 + 2 cos(θ))2

2dθ.

Substituting v = π + θ in the first integral we obtain∫ π/4

−3π/4

(3 + 2 sin(θ))2

2dθ =

∫ 5π/4

π/4

(3− sin(v))2

2dv

and in view of this

A =∫ 5π/4

π/4

[(3− sin(θ))2

2+

(3 + 2 cos(θ))2

2

]dθ =

∫ 5π/4

π/4

18 + 12 cos(θ)− 12 sin(θ) + 42

dθ

= 11π + 6 (sin(θ) + cos(θ))|5π/4π/4 = 11π − 12√

2.

Remark. 1. Instead of the substitution v = π + θ in the first integral, we can substitutev = π/2− θ, dv = −dθ:∫ π/4

−3π/4

(3 + 2 sin(θ))2

2dθ = −

∫ π/4

5π/4

(3 + cos(v))2

2dv

=∫ 5π/4

π/4

(3 + cos(v))2

2dv

and from here:

A = 2∫ 5π/4

π/4

(3 + 2 cos(θ))2

2dθ = 11π − 12

√2.

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Remark 2. Denote r1(θ) = 3 + 2 cos(θ) and r2(θ) = 3 + 2 sin(θ). Another solution tothe problem can be obtained by observing that r1(π/2 − θ) = r2(θ) which means that thecurves are symmetric with respect to the line determined by the directions θ = π/4, 5π/4.From here:

A = 2∫ 5π/4

π/4

(3 + 2 cos(θ))2

2dθ = 11π − 12

√2.

Problem 46/ page 689. Find the exact length L of the polar curve

r = e2θ, 0 ≤ θ ≤ 2π.

Solution. We have

L =∫ 2π

0

√(e2θ)2 + (2e2θ)2 =

∫ 2π

0

√e4θ + 4e4θdθ

=∫ 2π

0

√5e4θdθ =

√5∫ 2π

0e2θdθ =

√5

(e2θ

2

)∣∣∣∣∣2π

0√5

2

(e4π − 1

).

Problem 6/ page 696. Find the vertex, focus, and the directrix of the parabola x − 1 =(y + 5)2 and sketch its graph.Solution. Consider first x = y2 or equivalently 4px = y2. From here x = −1/4 is thedirectrix of this parabola, (1/4, 0) is its focus, and (0, 0) is its vertex. The graph of thegiven parabola is 1 to the right and 5 below if we compare with the graph of x = y2. Fromhere, the vertex of the given parabola is (1,−5), the directrix is x = 1− 1/4 = 3/4 and thefocus is at (1 + 1/4,−5) = (5/4,−5). Here is the graph

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Problem 14/ page 696. Find the vertices and foci of the ellipse 4x2 + 25y2 = 25 andsketch its graph.Solution. The ellipse has a center at (0, 0). If x = 0 we obtain y = ±1. If y = 0 we obtainx = ±5/2 = ±2.5. Hence, the vertices of the ellipse are (−2.5, 0) and (2.5, 0). On the otherhand c =

√(5/2)2 − 12 =

√25/4− 1 =

√21/4 and from here the foci of the ellipse are at

(−√

21/2, 0) and (√

21/2, 0). Here is the graph

Problem 20/ page 696. Find the vertices, foci and asymptotes of the hyperbola y2/16−x2/36 = 1 and sketch its graph.Solution. The center of the hyperbola is (0, 0). If x = 0 we obtain y = ±4 and from herethe vertices of the hyperbola are (0,−4) and (0, 4). If we divide the given equation by x2

and multiply by 16 we obtainy2

x2=

1636

+16x2

and when x→∞, then 16/x2 → 0 and the two asymptotes are determined by

y2

x2=

1636

⇒ y

x= ±2

3⇒ y = ±2

3x.

We compute c =√

16 + 36 =√

52 = 2√

13 to conclude that (0,−2√

13) and (0, 2√

13) arethe foci of the hyperbola. Here is the graph of the hyperbola given by a Cartesian equationy2/16− x2/36 = 1.

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Problem 28/ page 696. Identify the type of conic section with Cartesian equation y2 −8y = 6x− 16 and find its vertices and foci.Solution. The given Cartesian equation can be written in the following form

(y − 4)2 − 16 = 6x− 16⇒ (y − 4)2 = 6x.

Consider a conic with a cartesian equation y2 = 6x. Its graph is the graph of the given conicbut moved 4 units below. This is a parabola so, the given cartesian equation represents aparabola. In addition 4p = 6 and from here p = 3/2 hence, the directrix of the parabolay2 = 6x is x = −3/2, its focus is at (3/2, 0) and its vertex is at (0, 0).Going back to the given conic we are to move all above data with 4 units above. From here,the directrix of the given conic is x = −3/2, its vertex is at (0, 4) and its focus is at (3/2, 4).Here is the graph of the conic given by the Cartesian equation y2 − 8y = 6x− 16.

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Problem 46/ page 697. Find a Cartesian equation for the hyperbola with vertices(−1, 2), (7, 2) and foci (−2, 2) and (8, 2).Solution. The center of the hyperbola with the given data is at (3, 2). In order to obtaindata with a center at the origin we move the data 3 units to the left and 2 units below tolook for a hyperbola with vertices (−4, 0), (4, 0) and foci (−5, 0) and (5, 0). Such a hyperbolahas a Cartesian equation

x2

16− y2

b2= 1.

with c = 5 and c2 = 16 + b2 ⇒ 25 = 16 + b2 ⇒ b = 3 and from here we have

x2

16− y2

9= 1.

Now, moving back 3 units to the right and 2 units above we obtain a Cartesian equationfor the hyperbola satisfying the given conditions

(x− 3)2

16− (y − 2)2

9= 1.

This hyperbola has asymptotesy − 2x− 3

= ±34⇒ y = ±3

4(x− 3) + 2⇒ y =

3x− 14

, y =−3x+ 17

4.

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Here is the graph of the hyperbola conic having vertices at (−1, 2), (7, 2) and foci at (−2, 2)and (8, 2).

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