Suppose we have the equation $V = \frac{1}{3}\pi r^2h$. Find $\frac{dr}{dh}$.

[Chain Rule]

We have $\frac{dV}{dh}=\frac{dV}{dr}\cdot\frac{dr}{dh}$. $$ \begin{cases} \frac{dV}{dh}=\frac{1}{3}\pi r^2\\ \frac{dV}{dr}=\frac{2}{3}\pi rh \end{cases} \implies \frac{1}{3}\pi r^2 = \frac{2}{3}\pi rh\cdot\frac{dr}{dh} \implies r=2h\cdot\frac{dr}{dh} \implies \frac{dr}{dh}=\frac{r}{2h} $$

[Implicit Differentiation]

$$ \begin{align*} \frac{d}{dh}V&=\frac{\pi}{3}\frac{d}{dh}\big(r^2 h\big)\\ 0&=\frac{\pi}{3}\big(r^2\cdot\frac{d}{dh}h + h\frac{d}{dh}r^2\big)\\ 0&=r^2+2\cdot h\cdot r\frac{dr}{dh}\\ \frac{dr}{dh}&=-\frac{r^2}{2\cdot h\cdot r}=-\frac{r}{2h} \end{align*} $$

Why does the solution using the chain rule method have a different sign compared to the one using implicit differentiation? Did I make any mistake?


There's a few things going on here. It's important to distinguish between $V(r,h)$ the function and $V$ the constant.

First, your chain rule is wrong (for what you are asking). It should be: $$\dfrac{d V}{d h} = \dfrac{\partial V(r,h)}{\partial r} \dfrac{d r}{d h} + \dfrac{\partial V(r,h)}{\partial h} \dfrac{d h}{d h}$$ Notice that the left hand side is $V$ the constant, which when you did implicit differentiation you set equal to zero. Then we have \begin{align} \dfrac{d V}{d h} &= \dfrac{\partial V(r,h)}{\partial r} \dfrac{d r}{d h} + \dfrac{\partial V(r,h)}{\partial h} \dfrac{d h}{d h}\\ 0 &= (\dfrac{2}{3} \pi r h) \dfrac{d r}{d h} + \dfrac{1}{3} \pi r^2 (1)\\ \implies \dfrac{d r}{d h} &= -\dfrac{r}{2h} \end{align}

Now chain rule gives the same answer as implicit differentiation.

To give a bit more intuiton.

  • Your chain rule calculation that you originally had answered said "How much do I have to change $r$ by to get the same volume change if I changed $h$?"
  • Your implicit differentiation question said "How much do I have to change $r$ by to keep volume the same if I changed $h$?"

You are solving two different problems.

In your first part, you have V a function of h and r where $r$ and $h$ are both independent variables and V is the dependent variable. That means your $$ \frac {dr}{dh} =0$$ and $$ \frac {dh}{dr} =0$$

In your second part, you keep $V$ constant in which case you have $\frac {dV}{dh}=0.$ and $\frac {dV}{dr}=0.$ but you can solve for $\frac {dh}{dr}$ or $\frac {dr}{dh}$

  • $\begingroup$ But, this does not explain why the signs of the two solutions are different. $\endgroup$ – Ku Zijie Oct 5 '18 at 19:32
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    $\begingroup$ The negative sign appears when you use implicit differentiation to find $dy/d{x}$ from $ F(x,y)=C $ $\endgroup$ – Mohammad Riazi-Kermani Oct 5 '18 at 20:24
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    $\begingroup$ @KuZijie The OP is solving two different problems and coming up with two different solutions. Please check my edited answer. $\endgroup$ – Mohammad Riazi-Kermani Oct 5 '18 at 20:35

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