Solutions to Exercises, Section 6 - Department of …park/Fall2014/precalculus/6.6sol.pdf ·...

Instructor’s Solutions Manual, Section 6.6 Exercise 1

Solutions to Exercises, Section 6.6

In Exercises 1–12, convert the polar coordinates given for each point torectangular coordinates in the xy-plane.

1. r = √19, θ = 5π

solution We have

x = √19 cos(5π) and y = √19 sin(5π).

Subtracting even multiples of π does not change the value of cosineand sine. Because 5π − 4π = π , we have cos(5π) = cosπ = −1 andsin(5π) = sinπ = 0. Thus the point in question has rectangularcoordinates (−√19,0).


2. r = 3, θ = 21000π

solution We have

x = 3 cos(21000π) and y = 3 sin(21000π).

Because 21000 is an even multiple of π , we have cos(21000π) = 1 andsin(21000π) = 0. Thus the point in question has rectangularcoordinates (3,0).


3. r = 4, θ = π2

solution We have

x = 4 cos π2 and y = 4 sin π2 .

Because cos π2 = 0 and sin π2 = 1, the point in question has rectangular

coordinates (0,4).


4. r = 5, θ = −π2solution We have

x = 5 cos(−π2 ) and y = 5 sin

(−π2 ).Because cos

(−π2 ) = 0 and sin(−π2 ) = −1, the point in question has

rectangular coordinates (0,−5).





(−π4 ) = √2

2 and sin(−π4 ) = −√2

2 , the point in question hasrectangular coordinates (3

√2,−3

√2).


6. r = 7, θ = π4

solution We have


Because cos π4 =√

22 and sin π

4 =√

22 , the point in question has

rectangular coordinates(7√

22 , 7

√2

2

).


7. r = 8, θ = π3

solution We have


Because cos π3 = 12 and sin π

3 =√


rectangular coordinates (4,4√

3).





(−π3 ) = 12 and sin

(−π3 ) = −√32 , the point in question has

rectangular coordinates(9

2 ,−9√

32

).


9. r = 10, θ = π6

solution We have

x = 10 cos π6 and y = 10 sin π6 .

Because cos π6 =√

32 and sin π

6 = 12 , the point in question has

rectangular coordinates (5√

3,5).



x = 11 cos(−π6 ) and y = 11 sin


(−π6 ) = √3

2 and sin(−π6 ) = −1


rectangular coordinates(11

√3

2 ,−112

).


11. r = 12, θ = 11π4

solution We have

x = 12 cos 11π4 and y = 12 sin 11π

4 .

Because cos 11π4 = −

√2

2 and sin 11π4 =

√2

2 , the point in question hasrectangular coordinates (−6

√2,6

√2).


12. r = 13, θ = 8π3

solution We have

x = 13 cos 8π3 and y = 13 sin 8π

3 .

Because cos 8π3 = −1

2 and sin 8π3 =

√3


rectangular coordinates(−13

2 ,13√

32

).


In Exercises 13–28, convert the rectangular coordinates given for eachpoint to polar coordinates r and θ. Use radians, and always choose theangle to be in the interval (−π, π].13. (2,0)

solution The point (2,0) is on the positive x-axis, 2 units from theorigin. Thus we have r = 2, θ = 0.


14. (−√3,0)

solution The point (−√3,0) is on the negative x-axis,√

3 units fromthe origin. Thus we have r = √3, θ = π .


15. (0,−π)solution The point (0,−π) is on the negative y-axis, π units fromthe origin. Thus we have r = π , θ = −π2 .


16. (0,2π)

solution The point (0,2π) is on the positive y-axis, 2π units fromthe origin. Thus we have r = 2π , θ = π

2 .


17. (3,3)

solution We have

r =√

32 + 32 =√

32 · 2 =√

32√

2 = 3√

2.

The point (3,3) is on the portion of the line y = x that makes a 45◦

angle with the positive x-axis. Thus θ = π4 .


18. (4,−4)

solution We have

r =√

42 + (−4)2 =√

42 · 2 =√

42√

2 = 4√

2.

The point (4,−4) is on the portion of the line y = −x that makes a−45◦ angle with the positive x-axis. Thus θ = −π4 .


19. (−5,5)

solution We have

r =√

52 + (−5)2 =√

52 · 2 =√

52√

2 = 5√

2.

The point (−5,5) is on the portion of the line y = −x that makes a135◦ angle with the positive x-axis. Thus θ = 3π

4 .


20. (−6,−6)

solution We have

r =√(−6)2 + (−6)2 =

√62 · 2 =

√62√

2 = 6√

2.

The point (−6,−6) is on the portion of the line y = x that makes a−135◦ angle with the positive x-axis. Thus θ = −3π

4 .


21. (3,2)

solution We have

r =√

32 + 22 = √13 ≈ 3.61.

Because both coordinates of (3,2) are positive, we have

θ = tan−1 23 ≈ 0.588 radians.


22. (4,7)

solution We have

r =√

42 + 72 = √65 ≈ 8.06.

Because both coordinates of (4,7) are positive, we have

θ = tan−1 74 ≈ 1.052 radians.


23. (3,−7)

solution We have

r =√

32 + (−7)2 = √58 ≈ 7.62.

Because the first coordinate of (3,−7) is positive, we have

θ = tan−1(−73

) ≈ −1.166 radians.


24. (6,−5)

solution We have

r =√

62 + (−5)2 = √61 ≈ 7.81.

Because the first coordinate of (6,−5) is positive, we have

θ = tan−1(−56

) ≈ −0.695 radians.


25. (−4,1)

solution We have

r =√(−4)2 + 12 = √17 ≈ 4.12.

Because the first coordinate of (−4,1) is negative and the secondcoordinate is positive, we have

θ = tan−1( 1−4

)+π = tan−1(−14

)+π≈ 2.897 radians.


26. (−2,5)

solution We have

r =√(−2)2 + 52 = √29 ≈ 5.39.

Because the first coordinate of (−2,5) is negative and the secondcoordinate is positive, we have

θ = tan−1( 5−2

)+π = tan−1(−52

)+π≈ 1.951 radians.


27. (−5,−2)

solution We have

r =√(−5)2 + (−2)2 = √29 ≈ 5.39.

Because both coordinates of (−5,−2) are negative, we have

θ = tan−1(−2−5

)−π = tan−1 25 −π

≈ −2.761 radians.


28. (−3,−6)

solution We have

r =√(−3)2 + (−6)2 = √45 ≈ 6.71.

Because both coordinates of (−3,−6) are negative, we have

θ = tan−1(−6−3

)−π = tan−1 2−π≈ −2.034 radians.


29. Find the center and radius of the circle whose equation in polarcoordinates is r = 3 cosθ.

solution Multiply both sides of the equation r = 3 cosθ by r ,obtaining

r 2 = 3r cosθ.

Now convert the equation above to rectangular coordinates in thexy-plane, getting

x2 +y2 = 3x.

Subtract 3x from both sides, getting

x2 − 3x +y2 = 0.

Completing the square, we can rewrite this equation as

(x − 32)

2 +y2 = 94 .

Thus we see that the polar equation r = 3 cosθ describes a circlecentered at (3

2 ,0) with radius 32 .


30. Find the center and radius of the circle whose equation in polarcoordinates is r = 10 sinθ.

solution Multiply both sides of the equation r = 10 sinθ by r ,obtaining

r 2 = 10r sinθ.

Now convert the equation above to rectangular coordinates in thexy-plane, getting

x2 +y2 = 10y.

Subtract 10y from both sides, getting

x2 +y2 − 10y = 0.

Completing the square, we can rewrite this equation as

x2 + (y − 5)2 = 25.

Thus we see that the polar equation r = 10 sinθ describes a circlecentered at (0,5) with radius 5.

Instructor’s Solutions Manual, Section 6.6 Problem 31

Solutions to Problems, Section 6.6

31. Use the law of cosines to find a formula for the distance (in the usualrectangular coordinate plane) between the point with polar coordinatesr1 and θ1 and the point with polar coordinates r2 and θ2.

solution The distance between the point with polar coordinates r1

and θ1 and the point with polar coordinates r2 and θ2 is the length cshown below:

r1

Θ1

�

c

r2

Θ2�Θ1

�

By the law of cosines, we have

c2 = r12 + r2

2 − 2r1r2 cos(θ2 − θ1).

Thus the length between the two points equals√r1

2 + r22 − 2r1r2 cos(θ2 − θ1).


32. Describe the set of points whose polar coordinates are equal to theirrectangular coordinates.

solution Consider a typical point that has polar coordinates r and θand rectangular coordinates (x,y). This problem asks us to describethe set of points such that r = x and θ = y .

If we assume that r = x and θ = y , then we have√x2 +y2 = r = x.

Squaring both sides gives

x2 +y2 = x2,

which implies that y = 0. The equation θ = y then implies that θ = 0.And indeed, if r ≥ 0 then the point with polar coordinates r and 0 hasrectangular coordinates (r ,0).

Thus the set of points whose polar coordinates are equal to theirrectangular coordinates is the nonnegative horizontal axis (or thepositive horizontal axis if we want to exclude the origin because it canbe represented in infinitely many ways in polar coordinates).


33. What is the relationship between the point with polar coordinatesr = 5, θ = 0.2 and the point with polar coordinates r = 5, θ = −0.2?

solution The point with polar coordinates r = 5, θ = −0.2 is thereflection of the point with polar coordinates r = 5, θ = 0.2 through thehorizontal axis, as can be seen in the figure below:

�

�


34. What is the relationship between the point with polar coordinatesr = 5, θ = 0.2 and the point with polar coordinates r = 5, θ = 0.2+π?

solution The point with polar coordinates r = 5, θ = 0.2+π is thereflection of the point with polar coordinates r = 5, θ = 0.2 through theorigin, as can be seen in the figure below:

�

�


35. Explain why the polar coordinate θ corresponding to a point withrectangular coordinates (x,y) can be chosen as follows:

• If x > 0, then θ = tan−1 yx .

• If x < 0, then θ = tan−1 yx +π .

• If x = 0 and y ≥ 0, then θ = π2 .

• If x = 0 and y < 0, then θ = −π2 .

Furthermore, explain why the formula above always leads to a choice ofθ in the interval [−π2 , 3π

2 ).

solution Suppose we want to choose the polar coordinate θcorresponding to a point with rectangular coordinates (x,y). If x �= 0,then we must choose θ to satisfy the equation tanθ = y

x and such thatcosθ has the same sign as x and sinθ has the same sign as y .

First consider the case where x > 0 and y ≥ 0. Taking θ = tan−1 yx

certainly implies that tanθ = yx . Because y

x ≥ 0, we conclude that θ isin the interval [0, π2 ), which then implies that cosθ > 0 and sinθ ≥ 0,showing that cosθ has the same sign as x and sinθ has the same signas y . Thus this choice of θ works in this case.

Next consider the case where x > 0 and y < 0. Taking θ = tan−1 yx


x < 0, we conclude that θ isin the interval (−π2 ,0), which then implies that cosθ > 0 and sinθ < 0,showing that cosθ has the same sign as x and sinθ has the same signas y . Thus this choice of θ works in this case.


Next consider the case where x < 0 and y ≥ 0. Taking θ = tan−1 yx +π

implies that θ −π = tan−1 yx , which implies that tan(θ −π) = y

x . Thus

tanθ = tan(θ −π) = yx .

Because yx ≤ 0, the equation θ −π = tan−1 y

x implies that θ −π is inthe interval (−π2 ,0], which implies that θ is in the interval (π2 , π],which then implies that cosθ < 0 and sinθ ≥ 0, showing that cosθ hasthe same sign as x and sinθ has the same sign as y . Thus this choiceof θ works in this case.

Next consider the case where x < 0 and y < 0. Taking θ = tan−1 yx +π


x . Thus


Because yx > 0, the equation θ −π = tan−1 y

x implies that θ −π is in

the interval (0, π2 ), which implies that θ is in the interval (π, 3π2 ), which

then implies that cosθ < 0 and sinθ < 0, showing that cosθ has thesame sign as x and sinθ has the same sign as y . Thus this choice of θworks in this case.

Next consider the case where x = 0 and y ≥ 0. Then the point (x,y) ison the nonnegative vertical axis, which is described by the equationθ = π

2 .

Finally, consider the case where x = 0 and y < 0. Then the point (x,y)is on the negative vertical axis, which is described by the equationθ = −π2 .


In all the cases considered above, our choice of θ was in the interval[−π2 , 3π

2 ).


36. Give a formula for the polar coordinate θ corresponding to a point withrectangular coordinates (x,y), similar in nature to the formula in theprevious problem, that always leads to a choice of θ in the interval[0,2π).

solution The following formula always produces a polar coordinateθ in the interval [0,2π):

• If x > 0 and y ≥ 0, then θ = tan−1 yx .

• If x > 0 and y < 0, then θ = tan−1 yx + 2π .

• If x < 0, then θ = tan−1 yx +π .

• If x = 0 and y ≥ 0, then θ = π2 .

• If x = 0 and y < 0, then θ = 3π2 .

Here is the detailed justification that the polar coordinate θ as definedabove is in the interval [0,2π):

First consider the case where x > 0 and y ≥ 0. Taking θ = tan−1 yx


x ≥ 0, we conclude that θ isin the interval [0, π2 ), which then implies that cosθ > 0 and sinθ ≥ 0,showing that cosθ has the same sign as x and sinθ has the same signas y . Thus this choice of θ works in this case.

Next consider the case where x > 0 and y < 0. Takingθ = tan−1 y

x + 2π implies that θ − 2π = tan−1 yx , which implies that

tan(θ − 2π) = yx . Thus

tanθ = tan(θ − 2π) = yx .


Because yx < 0, the equation θ − 2π = tan−1 y

x implies that θ − 2π is in

the interval (−π2 ,0), which implies that θ is in the interval (3π2 ,2π),

which then implies then implies that cosθ > 0 and sinθ < 0, showingthat cosθ has the same sign as x and sinθ has the same sign as y .Thus this choice of θ works in this case.

Next consider the case where x < 0 and y ≥ 0. Taking θ = tan−1 yx +π


x . Thus


Because yx ≤ 0, the equation θ −π = tan−1 y

x implies that θ −π is inthe interval (−π2 ,0], which implies that θ is in the interval (π2 , π],which then implies that cosθ < 0 and sinθ ≥ 0, showing that cosθ hasthe same sign as x and sinθ has the same sign as y . Thus this choiceof θ works in this case.

Next consider the case where x < 0 and y < 0. Taking θ = tan−1 yx +π


x . Thus


Because yx > 0, the equation θ −π = tan−1 y

x implies that θ −π is in

the interval (0, π2 ), which implies that θ is in the interval (π, 3π2 ), which

then implies that cosθ < 0 and sinθ < 0, showing that cosθ has thesame sign as x and sinθ has the same sign as y . Thus this choice of θworks in this case.


Next consider the case where x = 0 and y ≥ 0. Then the point (x,y) ison the nonnegative vertical axis, which is described by the equationθ = π

2 .

Finally, consider the case where x = 0 and y < 0. Then the point (x,y)is on the negative vertical axis, which is described by the equationθ = 3π

2 .

In all the cases considered above, our choice of θ was in the interval[0,2π).

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