I just wanted to see if I did this correctly. Only asking for B.
So the question ask : The area of a circle increases at a rate of $1cm^2/s$.
a. How fast is the radius changing when the radius is $2cm$?
B. How fast is the radius changing when the circumference is $2cm$?
my solution : So I took the circumference equation $C=2{\pi}r$ and Isolated and derived for the $dr/dt$.
I got $(dc/dt)/2{\pi}r= dr/dt$. The answer in the back of my book is $1/2$ and when radius is $2cm$ I do get that answer since $dc/dt= 2\pi$. However, the radius was stated in a. and not necessarily the general question. So I am not sure if I can use $2cm$ as my radius.
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$\begingroup$We have the area of the circle as $$A=\pi r^2.$$ Treating $A$ and $r$ as implicitly differentiable functions of $t$, we get $$\frac{dA}{dt}=2\pi r\frac{dr}{dt}.$$
We are given that $$c=2\pi r=2 \Rightarrow r=\frac{1}{\pi}.$$ We are also given that, $$\frac{dA}{dt}=1.$$ Putting this together, $$1=2\pi \frac{1}{\pi}\frac{dr}{dt}\Rightarrow 1=2\frac{dr}{dt}\Rightarrow \frac{dr}{dt}=\frac{1}{2}cm/sec.$$
$\endgroup$ 1 $\begingroup$The related rate you are given is that of the area $A$ with respect to time $t$. It is $$\frac{dA}{dt}=1 \text { (cm/s)}$$ We also know that area is related to radius by $$A=\pi r^2$$ We differentiate to find that $$\frac{dA}{dr}=2 \pi r$$ or, alternatively, $$dA=2 \pi r dr$$ and so $$\frac{2 \pi r dr}{dt}=1$$ Re-arranging, we get $$\frac{dr}{dt}=\frac{1}{2 \pi r}$$ So when the circumference is a certain value ($2$), $r$ is a certain value. Figure that out and plug it in the get $f'(2)$ where $f$ is a function of $r$ with respect to $t$.
$\endgroup$ $\begingroup$$$c=2\pi r , A = \pi r^2 $$
$$ \dot A = 2 \pi r \dot r = c \dot r = 1 $$
$$ \dot r = \frac12 \frac{cm}{sec} $$
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