A spherical ball of salt is dissolving in water in such a way that the rate of decrease in volume at any instant is proportional to the surface. Prove that the radius is decreasing at a constant rate.
My Approach:
$$\dfrac {dV}{dt}\propto Surface (S)$$ $$\dfrac {dV}{dt}=k.S$$ where $k$ is a proportionality constant. $$\dfrac {dV}{dt}=k.4\pi r^2$$
How do I proceed?
On
The volume is a function of the radius, which itself is a function of time. So really, your equation is $$ \frac{d}{dt}V(r(t)) = K\cdot (r(t))^2 $$ (where I set the constant $K$ to be $4\pi k$). Now use the chain rule, and what you know about $V(r)$.
On
Area is the volume growth rate with respect to radius:
$$\dfrac{dV}{dr}=\dfrac {d( \dfrac43 \pi r^3)}{dr}= 4 \pi r^2 = A; $$
Take time rates
$$ \dfrac{\dfrac {dV}{dt}}{\dfrac {dr}{dt}}= A; $$
Interchange cross multiplication constant product
$$ \quad \dfrac{\dfrac {dV}{dt}}{A}={\dfrac {dr}{dt}}$$
Since the left hand side is given as some constant $k$ , the right hand side should also equal the same constant $k$.
You have $V=\frac{4}{3} \pi r^3$ so: $$\frac{d V}{dt}=\frac{d}{dt}\left(\frac{4}{3} \pi r^3 \right)=\frac{4}{3} \pi \times 3 \frac{dr}{dt} r^2 $$ so the equation is: $$4 \pi \frac{dr}{dt} r^2 = k 4 \pi r^2$$ i.e (as long as $r\neq 0$): $$\frac{dr}{dt}=k$$.