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A ball of radius $24cm$ has its volume decreasing at a rate of three times its surface area per second. What is the time taken for the volume to be reduced to one-eighth of the initial size?

Initial volume is $18432π$ and one with of this is $2304π$. I tried to integrate as follows

$$\int _{2304\pi }^{14832\pi }\:\frac{dV}{dt}$$ Where $\frac{dV}{dt}=4\pi r^2\cdot\frac{dr}{dt}$

Also from the given statement $\frac{dV}{dt}=-3\:\cdot4\pi r^2$ implying that $\frac{dr}{dt}=-3$

I couldn't solve it this way so I assume it is wrong. How do I do it?

Etemon
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user71207
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2 Answers2

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Nice job obtaining $\dfrac {dr}{dt} = -3$. This means that the radius is decreasing at a constant rate of $3$ cm per second.

Initial radius of the ball is $24$cm. At one-eighth of its volume, the radius would be halved, i.e. $12$cm. This takes:

$$\frac {24-12}{3} = 4 \text{ seconds}$$

player3236
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One observation. The model does not make much phyisical sense, as it is written.

The OP and the suggested solution use this :

$\frac{dV}{dt}=-3A$

But this dimensionally does not make sense. As a check, a change of units $r \rightarrow \lambda r$, would lead to:

$\frac{dV}{dt} \rightarrow \lambda^3\frac{dV}{dt} $

and

$-3A \rightarrow -3\lambda^2A $

Clearly the equation is than not a physical one because it cannot be valid for every unit system.

Thomas
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