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If a snowball melts so that its surface area decreases at a rate of $1\ cm^2/min$, find the rate at which the diameter decreases when the diameter is 10 cm.

So I know that the surface area of a sphere is $A = 4\pi r^2$, $ds / dt = -1$, and we are trying to find $dd/dt$

I tried the following but got a wrong answer. Not sure where I made my mistake. Can someone clarify where I made a mistake? $$ ds/dt = 4 \pi r^2 (dd/dt) \\ -1 = 8 \pi r^2 (dd/dt) \\ -1/8 \pi(10/2) = dd/dt \\ = -0.00795 $$

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it seems that actually you are finding the rate of change of the radius. We can rewrite the equation of the surface area as follows:

$A = 4\pi (d/2)^{2}$

If we differentiate implicitly with respect to t, we will then get that:

$\displaystyle\frac{dA}{dt} = 2\pi d \displaystyle\frac{dd}{dt}$

Clearing out the terms gives that the answer is $\approx -0.0159$.

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  • $\begingroup$ why is it $2 \pi d$ and not $4 \pi d$ $\endgroup$
    – Mr. 1.0
    Mar 3 '14 at 0:58
  • $\begingroup$ Clear out the terms. We will get by differentiating implicitly the terms $4\pi$, $2$, $1/2$ and $d/2$ (which comes out from the exponent). If you cancel out the terms, at the end you will get a $2$. Hope this helps. $\endgroup$
    – arcbloom
    Mar 3 '14 at 1:12

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