assignment 16 solution - weebly · assignment 16 solution please do not copy and paste my answer....
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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.
10
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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