The height of a cone is 6 inches, and, at it's opening it has a 3 inch diameter. Find a formula for the height as a function of the volume. Compute $h'(V)$
So if the volume, $V$ of a cone is $V = \pi r^2 {h\over 3}$
then my function $h(V)$ can be defined as: $3v\over \pi r^2$
So now I have $$h(V) = {3v \over \pi r^2}$$
to find $h'(V)$ I'm rewriting the function as $h(V) = 3v * (\pi r^2)^{-1} $
this now requires the product rule, chain rule, and implicit differentiation correct? If so, this is where I'm stuck as I'm uncertain about with respect to which variable I should be doing the implicit differentiation.
Would $$3 *(\pi r^2)^{-1} + 3v*-(\pi r^2)^{-2} * (2 \pi {dr \over dv})$$ be correct?
If you draw the cone as in the diagram, then you can determine a relationship between $h$ and $r$ and use that to eliminate the variable $r$:
Using similar triangles, we have the relationship: $$ \frac{r}{1.5} = \frac{h}{6} $$ And so: $$ r = \frac{h}{4} $$
We already know that $V = \frac{\pi r^2h}{3}$ for a cone, so substituting $r$ from the above equation, we get: $$ V = \frac{\pi h^3}{48} $$ We can now rearrange to get the required relationship: $$ h = \left( {\frac{48V}{\pi}} \right)^{\frac{1}{3}} $$ Now that we have $h(V)$, we can easily differentiate with respect to V to find the result.
PS: Better to rewrite as $$ h = \left( {\frac{48}{\pi}} \right)^{\frac{1}{3}} V^{\frac{1}{3}} $$ to differentiate easily.