# Differentiating both sides with respect to time.

So I have this problem:

An active volcanic mountain grows in the shape of a cone while maintaining its base diameter equal to its height. The volume of the mountain increases at a rate of 10,000,000 cubic feet per year.

At what rate is the height of the mountain rising when its height is 2,000 feet?

Express your answer to the nearest tenth (i.e., 0.1) of a foot per year.

[Note: The volume of a cone is $V=\frac{π}{3}r^{2}h$, where r is the radius of the base and h is the height of the cone.]

Step 1

This is a function in two variables (i.e. V(r,h)) and there is an advantage if we remove one of the variables. Since it is given that the diameter of the base is equal to the height of the cone, we can remove r from the equation as follows.

$r = h/2$

$V = \frac{π}{3}(\frac{h}{2})^{2}(h)$

$V = \frac{1}{12}(π)(h^{3})$

Step 2

Now we take the new equation and differentiate both sides with respect to time.

$\frac{d}{dt}(V)= \frac{d}{dt}\frac{1}{12}(π)(h^{3}))$

This becomes

$\frac{dV}{dt}=\frac{1}{4}(π)(h^{2})\frac{dh}{dt}$

So my question is, where did dh/dt come from? I thought when you took the derivative of something you subtract the exponent by 1 put whatever you subtracted in the front..

• This is a key aspect of implicit differentiation, so you could study that. It comes from the chain rule. And it is where you use your 0.01 ft/sec value. – turkeyhundt Jan 27 '15 at 0:35

$$\frac{dV}{dt} = \frac{dV}{dh}\cdot \frac{dh}{dt}$$
The expression $\displaystyle \frac{1}{4}\pi h^2$ represents $\displaystyle \frac{dV}{dh}$.
• You have a function $V(h)$ which gives $V$ in terms of $h$. You should know how to work out $\frac{dV}{dh}$. But here you're differentiating wrt time $t$. So the answer will be different and has to be computed by using chain rule. When the derivatives are expressed in Leibniz notation ("fractions"), like in my answer, it's often easier for a novice to see what's going on. Multiplying derivatives seems to behave much like multiplying fractions, with like terms in the denominator and numerator cancelling. That's why $\frac{dV}{dh} \cdot \frac{dh}{dt} = \frac{dV}{dt}$. – Deepak Jan 27 '15 at 1:53
• Of course, the above is also equivalent to saying $V'(t) = V'(h)\cdot h'(t)$, which is exactly the same thing (chain rule) but written up using different notation (Lagrange's). The mechanics of the rule are not as easy to see here, but maybe you might be more comfortable with this form. – Deepak Jan 27 '15 at 1:54