Suppose a mothball loses volume at a rate proportional to its SURFACE AREA.
If the radius of the ball decreases from 2cm to 1cm in 2 months, how long will it take till the radius is 1 mm?

Question

dumbcow · Answer

$$\frac{dV}{dt} = k(4 \pi r^{2})$$
$$\frac{dV}{dt} = \frac{dV}{dr}*\frac{dr}{dt}$$
$$\frac{dV}{dr} = 4 \pi r^{2}$$
$$\rightarrow (4 \pi r^{2})\frac{dr}{dt} = k(4 \pi r^{2})$$
$$\frac{dr}{dt} = k \rightarrow dr =k*dt $$
$$\int\limits_{}^{}dr = \int\limits_{}^{} k dt$$
$$r = kt + C$$
$$r(0) = 2 \rightarrow C = 2$$
$$r(2) = 1$$
$$1 = 2k +2$$
$$\rightarrow k = -\frac{1}{2}$$
$$r = -\frac{1}{2}t +2 = \frac{1}{10}$$
$$t = \frac{19}{5} = 3.8 months$$