1. ## Related rate (right answer, wrong method)

Question = "If a snowball melts so that its surface area decreases at the rate of $1 \frac{cm^2}{min}$, find the rate at which the diamter decreases when the diameter is 10cm."

My math went like this: $\frac{da}{dt} = 1 \frac{cm^2}{min}$; Surface area formula is $S = 4*PI*r^2$, and I want to the derivative of the rate with respect to the area (I think). $\frac{dr}{da}$.

So I kind of mapped out the chain rule $\frac{da}{dr} = \frac{da}{dt}*\frac{dt}{da}$, and I did this $\frac{da}{dt} = 8PIr \frac{dt}{da}$, and subsituted r = 5cm (the radius) to end up getting $\frac{1}{40pi}$ which was wrong. So, I check out the answer sheet and saw that it was $\frac{1}{20PI}$. Seeing that I was close to the right answer I tweaked my math somewhat like this: $2 = 8PIr \frac {dt}{da}$, and then solved for $\frac{dt}{da}$.

My issue is that I have no idea what I'm really doing with all these Leibniz notations and I won't have the luxury of an answer sheet on an exam. So, I'm asking if anyone has a chance to simplify and show how the chain rule would work here. More specifically, what notations should I be using, area, time, and the rate? It seems to be a theme in these kinds of problems (that I've been doing anyway) to focus on those 3 measurements.

Thanks.

2. ## Re: Related rate (right answer, wrong method)

Originally Posted by AZach
Question = "If a snowball melts so that its surface area decreases at the rate of $1 \frac{cm^2}{min}$, find the rate at which the diamter decreases when the diameter is 10cm."

$\frac{dS}{dt} = -1 \, cm/sec^2$

$\frac{d}{dt} \left(S = 4\pi r^2\right)$

$\frac{dS}{dt} = 8\pi r \cdot \frac{dr}{dt}$

$-1 = 8\pi \cdot 5 \cdot \frac{dr}{dt}$

$\frac{dr}{dt} = -\frac{1}{40\pi} \, cm/sec$

$\frac{d}{dt} \left(D = 2r\right)$

$\frac{dD}{dt} = 2\frac{dr}{dt} = -\frac{1}{20\pi} \, cm/sec$

3. ## Re: Related rate (right answer, wrong method)

I think I see it now. I haven't really done too many related rates problems as of yet. I guess it's really just a lot of chain rule and implicit differentiation most of the time.

Thanks again.