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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?

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    $\begingroup$ Trick question... the NaCl crystals are a cubic shape. So there cannot be a truly spherical ball of salt, only a quantized approximation. $\endgroup$
    – TrinitronX
    Apr 25, 2018 at 15:52
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    $\begingroup$ And even if you had a polished ball of salt, there would probably be a difference in dissolution rate between the crystal face, edge, and vertex. $\endgroup$
    – Nick T
    Apr 25, 2018 at 16:37
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    $\begingroup$ @TrinitronX As opposed to other types of matter, which totally can be idealized geometrical spheres an not approximations at all $\endgroup$
    – wedstrom
    Apr 25, 2018 at 17:37

4 Answers 4

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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 $r\neq 0$): $$\frac{dr}{dt}=k$$.

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    $\begingroup$ How is this a "hint"? You're giving the entire solution! $\endgroup$
    – Arthur
    Apr 25, 2018 at 10:58
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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 $k$.

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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 $V(r)$.

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Volume is $c\cdot r^3$. So $r^2$ can be culled from both sides.

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  • $\begingroup$ Your first sentence provides only a small portion of the work, and not enough to justify your second sentence. Additionally, your answer is later than the other answers. $\endgroup$
    – Acccumulation
    Apr 25, 2018 at 15:27
  • $\begingroup$ why would my second sentence need special justification? and i answered second, so... what's your point, exactly? $\endgroup$
    – bukwyrm
    Apr 25, 2018 at 19:23
  • $\begingroup$ The question asks you to prove it. Justification is exactly what is requested. For your answer to mistake, someone would have to take your cr^3 that you give, take the derivative, substitute it into the equation, and then recognize that your second sentence applies to that equation. You're asking the reader to do most of the work. If there were no answers before yours, then posting yours would be only barely informative, but as it is, it adds nothing. $\endgroup$
    – Acccumulation
    Apr 25, 2018 at 20:44
  • $\begingroup$ I answered while the first answer was written, without seeing it. The asker WAS ASKED to prove it, but ASKED how to proceed. I gave a hint how to proceed. You need to read the questions more carefully. $\endgroup$
    – bukwyrm
    Apr 26, 2018 at 5:31

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