Anonymous
Anonymous asked in Science & MathematicsMathematics · 9 years ago

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

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  • 9 years ago
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    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:

    http://i50.tinypic.com/30w5mh4.jpg

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