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Math Help - Cylinder in related rates

  1. #1
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    Cylinder in related rates

    A cylinder is of 0.08 foot thickness and when the radius is 150 feet, the radius is increasing at the rate of 0.5 foot per minute.It's asking for the rate in cubic feet per minute.

    I decided to derive the formula for a cylinder, which is pi*r^2*h = y and got;
    dy/dt = 2*pi*r*dr/dt*dh/dt
    dy/dt = 2*pi*150*0.5*0.08
    dy/dt = 12pi ... which isn't the answer. (~37.7 ft^3/min)

    What did I do wrong?
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  2. #2
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    To quote an educator of my acquaintance, "That's the product rule, Children!"

    Assuming r(t) and h(t), we have

    \frac{d}{dt} \pi r^{2}h\;=\;\pi (r^{2}\frac{dh}{dt} + h\cdot 2r \frac{dr}{dt})
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  3. #3
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    Quote Originally Posted by TKHunny View Post
    To quote an educator of my acquaintance, "That's the product rule, Children!"

    Assuming r(t) and h(t), we have

    \frac{d}{dt} \pi r^{2}h\;=\;\pi (r^{2}\frac{dh}{dt} + h\cdot 2r \frac{dr}{dt})
    Thanks broseidon. h and dh/dt are the same in this case, right?
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  4. #4
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    I think not, but I really can't understand the question Are you sure this is the exact wording of the question? The more I look at it, the less it makes sense.

    My impression is that dh/dt = 0, but I'm willing to be wrong.
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