# Related rate (right answer, wrong method)

• Oct 25th 2012, 01:45 PM
AZach
Related rate (right answer, wrong method)
Question = "If a snowball melts so that its surface area decreases at the rate of $\displaystyle 1 \frac{cm^2}{min}$, find the rate at which the diamter decreases when the diameter is 10cm."

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

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

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.
• Oct 25th 2012, 02:45 PM
skeeter
Re: Related rate (right answer, wrong method)
Quote:

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

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

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

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

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

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

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

$\displaystyle \frac{dD}{dt} = 2\frac{dr}{dt} = -\frac{1}{20\pi} \, cm/sec$
• Oct 25th 2012, 06:24 PM
AZach
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.