Assignment 16 Solution

Please do not copy and paste my answer. You will get similar questions but with different numbers!

This question tests you:

Given a polar coordinate (r,θ), the Cartesian coordinate is (x, y) where x = r cosθ, y = r sinθ.

If r > 0, the point (r,θ) lies in the same quadrant as θ; If r < 0, the point (r,θ) lies in the opposite quadrant as θ.

(−r,θ) represents the same point as (r,θ+π)

(r,θ) can be also represented by (r,θ+2nπ) and (−r,θ+ (2n +1)π), n = 0,±1,±2....

r > 0 : I am using n=1 (you can use n=2,3,....), (r,θ+2π) = (5, 5π3 +2π) = (5, 11π

3 )

r < 0 : I am using n=0 (you can use n=1,2,....), (−r,θ+ (2×0+1)π) = (−5, 5π3 +π) = (−5, 8π

3 )

Cartesian: x = r cosθ = 5cos( 5π3 ) = 5× 1

2 = 2.5, y = r sinθ = 5sin( 5π3 ) =−5×

p3

2 .

1

So I am going to choose the upper right graph.

r > 0 : I am using n=0 (you can use n=1,2,....), (−r,θ+ (2×0+1)π) = (6, π2 +π) = (6, 3π2 )

r < 0 : I am using n=1 (you can use n=2,3,....), (r,θ+2×1π) = (−6, π2 +2π) = (−5, 5π2 )

Cartesian: x = r cosθ =−6cos(π2 ) = 0, y = r sinθ =−6sin(π2 ) =−6.

So I am going to choose the lower right graph.

2

r > 0 : I am using n=1 (you can use n=2,3,....), (r,θ+2π) = (5,−1+2π)

r < 0 : I am using n=0 (you can use n=1,2,....), (−r,θ+ (2×0+1)π) = (−5,−1+π)

Cartesian: x = r cosθ = 5cos(−1) =≈ 2.7, y = r sinθ = 5sin(−1) ≈−4.2.

So I am going to choose the upper right graph.

3

This question tests you:

Given you a Cartesian coordinates, you can find polar coordinates by r =√

x2 + y2 and tanθ = yx .

(r,θ) can be also represented by (r,θ+2nπ) and (−r,θ+ (2n +1)π), n = 0,±1,±2....

(a) then r =√

(2p

3)2 +22 = 4 and tanθ = 22p

3= 1p

3⇒ θ = π

6

(2p

(3),2) is in the first quadrant. If r > 0, then r = 4 and θ = π6 .

If r < 0, then r =−4, θ = π6 + (2×0+1)π= 7π

6 .

(b) r =√

22 + (−5)2 =p29 and tanθ = −5

2 ⇒ θ = ar ct an(−52 ).

Since (2,−5) is in the fourth quadrant, if r > 0, r =p29 and θ = ar ct an(−5

2 )+2π

If r < 0, then r =−p29, θ = ar ct an(−52 )+ (2×0+1)π= ar ct an(−5

2 )+π.

4

r ≥ 2 means we consider the part outside of the circle x2 + y2 = 22. You can eliminate upper right and lower right plots.

π≤ θ ≤ 2π means we only want the lower half (upper half means 0 ≤ θ ≤π). Hence I will choose lower left plot.

5

(a) I think cartesian is easier because the center of the circle is not the origin. You can use circle equation with radius r and center

(a,b): (x −a)2 + (y −b)2 = r 2

(x −1)2 + (y −1)2 = 32.

(b) Center is origin, so the polar equation is easier. You can use this r = r , 0 ≤ θ ≤ 2π.

r = 1, 0 ≤ θ ≤ 2π.

6

This question tests you:

Given a polar coordinate (r,θ), the Cartesian coordinate is (x, y) where x = r cosθ, y = r sinθ.

Choose some points such as (a) r = θ = π2 , then x = π

2 cos π2 = 0 and y = π

2 sin π2 = π

2 . So you can eliminate upper right, lower left.

(b) r = θ =π, then x =πcosπ=−π and y =πsinπ= 0. You can eliminate upper left.

Hence I will choose lower right.

7

This question tests you:

Given a polar coordinate (r,θ), the Cartesian coordinate is (x, y) where x = r cosθ, y = r sinθ.

Choose (a) θ = π2 , r = ln(π2 ) and x = r cosθ = 0 and y = r sinθ = r = ln(π2 ) ≈ 0.45. You can eliminate upper right and lower right.

(b) θ =π, then r = lnπ, x = lnπcosπ=− lnπ≈−1.14 and = lnπsinπ= 0. You can eliminate upper left.

So I will choose lower left.

8

This question tests you:

Given a polar coordinate (r,θ), the Cartesian coordinate is (x, y) where x = r cosθ, y = r sinθ.

Here r = 3sin(2θ), so x = r cosθ = 3sin(2θ)cosθ and y = 3sin(2θ)sinθ.

θ x y eliminate

π4

3p

22 ≈ 2.12 3

p2

2 ≈ 2.12 upper left, upper right, lower right

Only lower left is correct.

9

This question tests you:

Given a polar coordinate (r,θ), the Cartesian coordinate is (x, y) where x = r cosθ, y = r sinθ.

Here r = cos(4θ), so x = r cosθ = cos(4θ)cosθ and y = cos(4θ)sinθ.

θ x y eliminate

π4 −

p2

2 ≈−0.7 −p

22 ≈−0.7 lower left, upper right, lower right

Only upper left is correct.

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Question 9: see handwriting version. Too difficult to type.

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This question tests you:

slope = d yd x =

drdθ sinθ+r cosθdrdθ cosθ−r sinθ

d y

d x=

drdθ sinθ+ r cosθdrdθ cosθ− r sinθ

= 3cosθ sinθ+ r cosθ

3cosθcosθ− r sinθ

When θ = π6 , sinθ = 1

2 and cosθ =p

32 and r = 3sinθ = 3

2

d y

d x= 3cosθ sinθ+ r cosθ

3cosθcosθ− r sinθ

= 3p

32

12 + 3

2

p3

2

3p

32

p3

2 − 32

12

=p3

12

This question tests you:

slope = d yd x =

drdθ sinθ+r cosθdrdθ cosθ−r sinθ

d y

d x=

drdθ sinθ+ r cosθdrdθ cosθ− r sinθ

=− 5θ2 sinθ+ r cosθ

− 5θ2 cosθ− r sinθ

When θ =π, sinθ = 0 and cosθ =−1 and r = 5/θ = 5π

d y

d x=

− 5θ2 sinθ+ r cosθ

− 5θ2 cosθ− r sinθ

=− 5π2 0− 5

π

5π2 −0

=−π

13

This question tests you:

slope = d yd x =

drdθ sinθ+r cosθdrdθ cosθ−r sinθ

To get horizontal tangent, let drdθ sinθ+ r cosθ = 0. That is −5sinθ sinθ+ 5cosθcosθ = 5(cos2θ− sin2θ) = 0. Then we have

sinθ =±cosθ. It means that θ = π4 or θ = 3π

4 . (We want 0 ≤ θ <π ).

If θ = π4 , r = 5cos(π4 ) = 5

p2

2 .

If θ = 3π4 , r = 5cos( 3π

4 ) =− 5p

22 .

To get vertial tangent, let drdθ cosθ− r sinθ = 0. That is −5sinθcosθ−5cosθ sinθ =−10sinθcosθ = 0. Here sinθ = 0 or cosθ = 0.

We have θ = 0, or θ = π2 .

If θ = 0, r = 5cos(0) = 5

If θ = π2 , r = 5cos π

2 = 0.

14

x = r cosθ = cosθ−cos2θ and y = r sinθ = sinθ−cosθ sinθ.

θ x y eliminate

π2 0 1 upper left, lower left

π −2 0 lower right

So I will choose upper right.

15

This question tests you:

slope = d yd x =

drdθ sinθ+r cosθdrdθ cosθ−r sinθ

d y

d x=

drdθ sinθ+ r cosθdrdθ cosθ− r sinθ

= − 13 sin( θ3 )sinθ+ r cosθ

− 13 sin( θ3 )cosθ− r sinθ

When θ =π, sinθ = 0 and cosθ =−1 and r = cos(π3 ) = 12 and sin(π3 ) =

p3

2

d y

d x= − 1

3 sin( θ3 )sinθ+ r cosθ

− 13 sin( θ3 )cosθ− r sinθ

= 0− 12

13

p3

2 −0=−p3

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