Question

Oil spreads on a frying pan so that its radius is proportional to t^(1/2), where t represents the time from the moment when the oil is poured. Find the rate of change dT/dt of the thickness T of the oil.

Answer 1

First, here is the answer from the solutions:

Answer 2

And now, I'll post my work. Unfortunately I get a different answer. Hopefully someone can tell me where I am going astray.

First, here is what I get for dr/dt. I'll use that for substitution later.
\[r = t^{\frac{1}{2}}\\r' = \frac{1}{2t^{\frac{1}{2}}} \\\\ \]
Now a couple of definitons for volume, V. I'll use the 2nd definition for subs. later: \[
V = \pi r^2T \\ T = \frac{V}{\pi r^2}\\ r^2T = \frac{V}{\pi}\\ (r^2T)' = 2rr'T + r^2T' = 0\\\] So far, my answer corresponds with Jerison's solution key.

Here I substitue in for T\[ \frac{2rr'V}{\pi r^2} + r^2T' = 0 \\\] Divide by r^2 \[ \frac{2r'V}{\pi r^3}+T' = 0\\ \] Substitute for r' \[ T' = \frac{-2V}{2\pi t^\frac{1}{2}r^3}\\\\\] Subs. for r \[T' = \frac{-V}{\pi t^2}
\]

Answer 3

\[T' = \frac{-V}{\pi t^2} = \frac{-T}{t} \]... So, the only difference between my answer and Jerison's is a factor of T.

Answer 4

In the step following "Divide by r^2, you removed the r from the numerator, but the denominator remained r^2.

Answer 5

Thank you for checking my work. The denominator went from r^2 to r^3 because I actually did two things when under "divide by r^2". I divided everything by r^2, and then I simplified the fraction on the left removing one r from the denominator and numerator.

So, breaking it down would look something like this:

\[\frac{2rr'V}{\pi r^4}+T' = 0 \\\frac{2r'V}{\pi r^3}+T' = 0 \] Part of me feels it must be a typo in the solutions key, but I'm not confident...

Answer 6

Here is the line where Jerison get's rid of the T (the one that I never get rid of). Is this a mistake?