# Solving Rate of Change when variable isn't given

I'm completely stuck in solving questions with rates of change when no variable is given in the formula.

Example: The volume, $V\text{ cm}^3$, of water in a container is given by $V=(1/3)\pi h^3$ where $h$ cm is the depth of water in the container at time $t$ minutes. Water is draining from the container at a constant rate of $300\text{ cm}^3/\text{min}$. The rate of decrease of $h$, in cm/min, when $h = 5$ is?

Cheers

Here, when they say "$h$ cm is the depth of water in the container at time $t$ minutes," they are saying three things:

1. The value of $h$ is in centimeters, and gives a depth of water in the container.
2. The value of $h$ varies with the time $t$--that is, $h=h(t).$
3. The value of $t$ is in minutes.

So, your independent variable is the time, $t.$ Applying the chain rule to the time-derivative of the formula $$V=\frac13\pi h^3$$ gives us $$\frac{dV}{dt}=\frac13\pi\cdot 3h^2\cdot\frac{dh}{dt}=\pi h^2\cdot\frac{dh}{dt}.$$

Now, you know that $\frac{dV}{dt}=-300,$ and you must assume that $h=5,$ which yields $$-300=25\pi\cdot\frac{dh}{dt},$$ or equivalently, $$-\frac{12}\pi=\frac{dh}{dt}.$$

The above gives us the rate of change of $h$ with respect to $t$--denoted $\frac{dh}{dt}$--when $h=5.$ However, the rate of decrease of $h$ with respect to $t$ is $-\frac{dh}{dt},$ so multiplying both sides by $-1$ gives us our answer.

I leave it to you to find the units.

$$V= \frac 1 3 \pi h^3$$ $$\frac{dV}{dt} = \pi h^2 \frac{dh}{dt}$$ $$-300 = \pi\cdot5^2\cdot\left. \frac{dh}{dt} \right|_{h=5}.$$ $$\text{So}\quad\left. \frac{dh}{dt} \right|_{h=5} = \text{ ?}$$