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?
Best Answer
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 $\frac {dr}{dh}$