0
$\begingroup$

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?

$\endgroup$
1
$\begingroup$

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 $h$?"
$\endgroup$
2
$\begingroup$

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 $\frac {dr}{dh}$

$\endgroup$
  • $\begingroup$ But, this does not explain why the signs of the two solutions are different. $\endgroup$ – Ku Zijie Oct 5 '18 at 19:32
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ @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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.