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anonymous

  • 5 years ago

If Theta is the acute angle between al line through the origin and the positive x-axis,the area in cm^2, of part of a rose petal is A=(9/6).(4Theta-sin(4Theta)) If the angle Theta is increasing at a rate of 0.2 radians per minute, at what rate is the area changing when Theta=pi/4 ?

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  1. amistre64
    • 5 years ago
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    polar coordinates eh...

  2. anonymous
    • 5 years ago
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    let me rewrite the question

  3. anonymous
    • 5 years ago
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    If \[\theta\] is the acute angle between al line through the origin and the positive x-axis,the area in \[cm^{2}\], of part of a rose petal is \[A=(9/16).(4\theta-\sin(4\theta))\] If the angle \[\theta\] is increasing at a rate of 0.2 radians per minute, at what rate is the area changing when \[\theta=\pi/4\] ?

  4. amistre64
    • 5 years ago
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    the normal integral for polar regions is: (1/2) [S] [f(t)]^2 dt ; [a,b]

  5. amistre64
    • 5 years ago
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    but you seems to want derivatives..... so thats no good to you lol

  6. anonymous
    • 5 years ago
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    this answer involve topic rate of change under diiferentation

  7. amistre64
    • 5 years ago
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    A=(9/6).(4t-sin(4t) A' = 0(4t-...) + (9/6) D(4t-sin(4t))

  8. amistre64
    • 5 years ago
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    A' = (9/6).(-4cos(4t)) if i see it right

  9. amistre64
    • 5 years ago
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    A' = (9/6).(0t' -4cos(4t)) A' = (9/6).(-4cos(4t))

  10. amistre64
    • 5 years ago
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    that dt=.2 is throwing me; not sure how it applies in here...

  11. amistre64
    • 5 years ago
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    maybe its my choice of t as a substitute for theta.... A' = (9/6).(-4cos(4t)) (.2) is what i wanna say is correct.... do you have an answer sheet to go by?

  12. anonymous
    • 5 years ago
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    sadly no.. anyway thanks

  13. amistre64
    • 5 years ago
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    i wanna say A' = 1.2 rads/min

  14. amistre64
    • 5 years ago
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    at t=pi/4

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