NOTE: THIS QUESTION HAS ALREADY BEEN ANSWERED AND CAN BE FOUND AT THIS LINK (Related rates circle problem)
Question: Two circles $A$ and $B$ have the same center. The radius of the inner circle $A$ is increasing at a rate of $1$ unit/sec, and the radius of the larger circle $B$ is also increasing such that the area between the two circles is always $10\pi$. When the radius of $A$ is 5, how fast is the radius of $B$ increasing?
So I know that this question has been answered already in the past, but I'm wondering if there is another solution. Could this problem perhaps be solved with differentiation? All I'm asking is if there is another way to solve the problem (another solution perhaps). My thinking was rather than substitute variables for others like the previous solution did (smaller radius is R and 2R and so forth), we could solve it directly. Of course, this may not be possible, and I'm asking the most brilliant minds in STEM to help me out :D If a solution does arrive, I'd love to see it! Alternate explanations are always possible in math. Note: Could it stick within the realm of Calculus only? Number theory and other advanced topics like that aren't necessary.
It can be solved without differentiation although the logic of solving it is based on calculus.
When $r = 5$ and the area between R and r is $10\pi$, then $R = \sqrt35$
For an infinitesimally small change in radius dr, the area of the smaller circle increases by $2\pi r dr$. To maintain the same area of $10\pi$, the larger circle must increase in area by the same amount.
Hence $2\pi r\ dr = 2\pi R\ dR$
$10\pi\ dr = 2\sqrt{35}\pi\ dR$
$5\ dr = \sqrt{35}\ dR$
$dR = \frac{5}{\sqrt{35}}dr$
Therefore $$\frac{dR}{dt} = \frac{5}{\sqrt{35}} \frac{dr}{dt}$$
and $$\frac{dR}{dt} = \frac{5}{\sqrt{35}}\cdot 1$$ $$=\frac{\sqrt{35}}{7} \text{unit/sec}$$