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So I was reading Polya's book and in it, there was a problem involving finding the rate of change of depth of water in a cone.

At some point, we come to the conclusion that V = $\pi a^2 y^3/(3b^2)$ where $V$ = volume of water, $a$= radius of base of cone, $y$ = depth of water and $b$ = height of cone.

This is where I get lost...They then come to the conclusion that since $V$ increases along with $y$, then upon differentiation, the result should be $dV/dt = (\pi a^2 y^2 / b^2 )dy/dt$.

How come they're multiplying with $dy/dt$? whats the rule ? I know this must be extremely basic but it's been a while since I touched maths and wanted to get back into it.

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  • $\begingroup$ Chain rule: $\frac{dV}{dt} = \frac{dV}{dy}\frac{dy}{dt}$ $\endgroup$ – Ben Grossmann Aug 6 '15 at 15:54
  • $\begingroup$ $y$ is a function of $t$, so they apply the chain rule. $\endgroup$ – BadAtMaths Aug 6 '15 at 15:58
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The answer is easy: $$ \frac{dV}{dt} = \frac{d}{dt} \frac{\pi a^2}{3b^2} y^3 = \frac{\pi a^2}{3b^2} \cdot 3 y^2 \frac{dy}{dt} $$ by the chain rule applied to $t \mapsto y(t)^3$.

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  • $\begingroup$ Wow that was fast. Thanks!! $\endgroup$ – Newbie Aug 6 '15 at 16:04
  • $\begingroup$ So...would it be safe to say that d/dt $x^4$ = 4$x^3$dx/dt ? $\endgroup$ – Newbie Aug 6 '15 at 17:41
  • $\begingroup$ Yes, it is even correct when $x$ is independent of $t$. $\endgroup$ – Siminore Aug 6 '15 at 18:44

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