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# Can someone find theta when r is maximized?

make x = theta below

r = 4sinx

r = 3+2cos(x)

r = 10(1-sinx)

r 6+12sinx

r = 4cos3x

r = 5sin2x

...and also the zeros of r

### 1 Answer

- JánošíkLv 79 years agoFavorite Answer
1)

r = 4∙sin(θ)

r = 0 for

4∙sin(θ) = 0

sin(θ) = 0

θ = 0 or θ = π

r' = 4∙cos(θ)

r' = 0 for

4∙cos(θ) = 0

cos(θ) = 0

θ = π/2 or θ = 3π/2

4∙sin(π/2) = 4∙1 = 4

4∙sin(3π/2) = 4∙(-1) = -4

so r-max at θ = π/2

Graph:

http://i48.tinypic.com/2ccmbrd.jpg

2)

r = 3 + 2∙cos(θ)

r = 0 for

3 + 2∙cos(θ) = 0

2∙cos(θ) = -3

cos(θ) = -3/2 <-- not possible (cos can not be less than -1), so r=0 never happens

r' = -2∙sin(θ)

r' = 0 for

-2∙sin(θ) = 0

sin(θ) = 0

θ = 0 or θ = π

3 + 2∙cos(0) = 5

3 + 2∙cos(π) = -5

so r-max at θ = 0

Graph:

http://i48.tinypic.com/doauci.jpg

3)

r = 10∙(1 - sin(θ))

r = 10 - 10∙sin(θ)

r = 0 for

10 - 10∙sin(θ) = 0

-10∙sin(θ) = -10

sin(θ) = 1

θ = π/2

r' = -10∙cos(θ)

r' = 0 for

-10∙cos(θ) = 0

cos(θ) = 0

θ = π/2 or θ = 3π/2

r = 10 - 10sin(π/2) = 0

r = 10 - 10sin(3π/2) = 20

so r-max at θ = 3π/2

Graph:

http://i46.tinypic.com/2qt9zdl.jpg

4)

r = 6 + 12∙sin(θ)

r = 0 for

6 + 12∙sin(θ) = 0

12∙sin(θ) = -6

sin(θ) = -1/2

θ = 5π/4 or θ = 7π/4

r' = 12∙cos(θ)

r' = 0 for

12∙cos(θ) = 0

cos(θ) = 0

θ = π/2 or θ = 3π/2

r = 6 + 12∙sin(π/2) = 18

r = 6 + 12∙sin(3π/2) = -6

so r-max at θ = π/2

Graph:

http://i49.tinypic.com/2aep1eo.jpg

5)

r = 4cos(3θ)

r = 0 for

4∙cos(3θ) = 0

cos(3θ) = 0

3θ = π/2 or 3θ = 3π/2

θ = π/6 or θ = π/2 or θ = 5π/6 or θ = 7π/6 or θ = 3π/2 or θ = 11π/6

r' = -12∙sin(3θ)

r' = 0 for

-12∙sin(3θ) = 0

sin(3θ) = 0

3θ = 0 or 3θ = π

θ = 0 or θ = π/3 or θ = 2π/3 or θ = π or θ = 4π/3 or θ = 5π/3

r = 4cos(3∙0) = 4

r = 4cos(3∙π/3) = -4

r = 4cos(3∙2π/3) = 4

r = 4cos(3∙π) = -4

r = 4cos(3∙4π/3) = 4

r = 4cos(3∙5π/3) = -4

so r-max at θ = 0 or θ = 2π/3 or θ = 4π/3

Graph:

http://i48.tinypic.com/2my116g.jpg

6)

r = 5∙sin(2θ)

r = 0 for

5∙sin(2θ) = 0

sin(2θ) = 0

2θ = 0 or 2θ = π

θ = 0 or θ = π/2 or θ = π or θ = 3π/2

r' = 10∙cos(2θ)

r' = 0 for

10∙cos(2θ) = 0

cos(2θ) = 0

2θ = π/2 or 2θ = 3π/2

θ = π/4 or θ = 3π/4 or θ = 5π/4 or θ = 7π/4 or

r = 5∙sin(2π/4) = 5

r = 5∙sin(2∙3π/4) = -5

r = 5∙sin(2∙5π/4) = 5

r = 5∙sin(2∙7π/4) = -5

so r-max at θ = π/4 or θ = 5π/4

Graph: