At what rate is the level of the shaving cream changing when it is still 3 centimeters from the top of the cone?

Question

Problem

Sarah is filling ice cream cones with shaving cream to hand out to trick or treaters. The cones have a largest radius of $3$ centimeters and a depth of 13 centimeters. She fills them at a rate of $5$ cubic centimeters per second. (She works carefully so that shaving cream is compact and even) At what rate is the level of the shaving cream changing when it is still $3$ centimeters from the top of the cone?

What I know

1. Radius is $3$ cm and it is a constant
2. Height is $13$ cm
3. Rate of change of the volume is $\dfrac {dV}{dt} = 5\text{cm}^3/\text{sec}$
4. Trying to find $\dfrac{dh}{dt}$ by using the formula $V=\left(\dfrac 13\right)\pi r^2h$
5. Find the derivative of $V=\left(\dfrac 13\right)\pi r^2h$

Answer 1

We know that the volume of the cone is given by:

$$V=\frac 13 \pi r^2h$$

From the information that the greatest radius is $3 \text{cm}$, and height is $13\text{cm}$, we can deduce the following relationship between $r$ and $h$:

$$\frac rh = \frac3{13} \rightarrow3h=13r\rightarrow r=\frac3{13}h$$

Then plug that back into $V$ to get it in terms of $h$ purely. Differentiating with respect to time, we get:

$$V=\frac 13 \pi \left({3\over 13}h\right)^2h \rightarrow\frac{dV}{dt} = \frac 9{169}\pi h^2\frac{dh}{dt}$$

The goal is to find $\frac{dh}{dt}$. Since you know $\frac{dV}{dt}$ and the current $h$, you can solve for $\frac{dh}{dt}$.

Answer 2

Hint: the rate of change of height at some fixed point in time doesn't care what the shape of the container is (i.e. that it is a cone). It only cares about the cross section at that specific height (a circle of some radius). So you will get the same answer for your cone as you get for a cylinder of the right radius.