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Math Help - Calc III: Partial Derivatives

  1. #1
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    Calc III: Partial Derivatives

    The radius of a right circular cone is increasing at a rate of 5 inches per second and its height is decreasing at a rate of 3 inches per second. At what rate is the volume of the cone changing when the radius is 50 inches and the height is 40 inches?

    I went to get help with this problem and I was advised to differentiate the Volume equation in terms of r and h and plug in the numbers. But I didn't get the right answer. I did this:

    V=\pi r^2(t)h(t)
    dV/dt = 2\pi r(t)r'(t)h(t)+h'(t)\pi r^2(t)
    Plugged in the numbers and got 12500\pi, which isn't right!

    What am I doing wrong?
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  2. #2
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    Hello, saxyliz!

    Did you forget the denominator?


    The radius of a right circular cone is increasing at a rate of 5 in/sec
    and its height is decreasing at a rate of 3 in/sec.
    At what rate is the volume of the cone changing
    when the radius is 50 inches and the height is 40 inches?
    Your derivative is correct . . .

    We have: . V \;=\;\frac{\pi}{3}r^2h

    Then: . \frac{dV}{dt} \;=\;\frac{2\pi}{3}rh\!\cdot\!\frac{dh}{dt} + \frac{\pi}{3}r^2\!\cdot\!\frac{dh}{dt}


    We are given: . r = 50,;\;h=40,\;\;\frac{dr}{ht} = 5,\;\;\frac{dh}{dt} = \text{-}3


    Therefore: . \frac{dV}{dt} \;=\;\frac{2\pi}{3}(50)(40)(5) + \frac{\pi}{3}(50^2)(-3) \;=\;\frac{12,500}{3}\pi in³/sec.

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  3. #3
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    Oh my goodness! It's a cone!! Wow. Neither me nor the person who helped me caught that, we read it as a cylinder for some reason... I feel dumb now.

    Thank you so much!
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