So I've got a question where I'm given the area of a circle: $$a = 12m^2$$ as well as the rate of change of the decrease in the circle's area w.r.t. time: $$ \frac {da}{dt} = -0.5m^2s^{-1}$$ and I'm being asked to calculate the R.o.C. of the radius w.r.t. time.
This is my working and answer:
$$ r = \sqrt {\frac {a}{\pi}} = 1.9544… $$
$$ \frac {dr}{dt}= \frac {dr}{da} * \frac {da}{dt} $$
$$ a = \pi r^2$$
$$ \frac {da}{dr} = 2 \pi r $$
$$ \frac {dr}{da} = \frac {1}{2 \pi r} $$
$$ \frac {dr}{dt} = \frac {1}{2 \pi r} * -0.5 = -0.0407 (3s.f.) $$
So the radius is decreasing at a rate of: $$ 0.0407ms^{-1} $$ to 3 significant figures when the area is: $$ 12m^2 $$
Have I worked this out correctly?
Best Answer
To cross-check the answer, we can do a rough estimation about how the area should have changed given a small $\Delta t$, let's say $\Delta t = 1$ s. Then $A_t = A_0 + \frac{dA}{dt} \Delta t = 12 + (-0.5)(1) = 11.5$ m$^2$ s$^{-1}$, and the corresponding radius would be $r_t = \sqrt{\frac{A_t}{\pi}} = 1.9133$ m.
This leads to an estimation of $\frac{dr}{dt}\approx \frac{\Delta r}{\Delta t }= \frac{1.9544-1.9133}{1} \approx 0.0411$ m s$^{-1}$, which is very close to what you get.