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Math Help - Conceptual Question on Solving Related Rates Problems

  1. #1
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    Conceptual Question on Solving Related Rates Problems

    Let's use this problem as an example:

    Air is being pumped into spherical balloon so that its volume is increasing at a rate of 100 cm^3/s. How fast is the radius of the balloon increasing when the diameter is 50 cm?

    This problem is an example in our textbook, so I have the entire solution. I'm not looking for an answer to the text question. My question is more about why they do what they do.

    The basic formula is the volume of a sphere V = \frac{4}{3}\pi r^3. They differentiate each side of the equation as follows:

    \frac{dV}{dt} = \frac{dV}{dr} \frac{dr}{dt} = 4\pi r^2\frac{dr}{dt}

    What confuses me is that it looks like the chain rule is being applied differently on the right and left sides of the equation. On the right side there is a r^2 \frac{dr}{dt} but the left side only winds up with \frac{dV}{dt}. What happened to the V? (i.e. why isn't the left side V \frac{dV}{dt}?).

    I get that the answer is that the derivative of V is 1, but if that is true, then why isn't the derivative of r also 1? If we are differentiating with respect to t, how can the derivative be 1? Is is simply because the problem sets this rate up as a constant or should I be able to see that some other way?

    I hope that this makes sense. Thanks.
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  2. #2
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    Quote Originally Posted by joatmon View Post

    I get that the answer is that the derivative of V is 1, but if that is true, then why isn't the derivative of r also 1?
    The derivative of r is 1, but you are taking the derivative of r^3. So, you apply the Power Rule to differentiate r^3, resulting in 3r^2. Because r is assumed to be a function of t, apply the Chain Rule (multiply by the derivative of the inside function, where the inside function of r^3 is r):

    3r^2 \times \dfrac{dr}{dt}
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