Q. 57 - Calculus

Question

Riley is holding an ice cream cone on a hot summer day. As usual, the cone has a small hole at the bottom, and ice cream is melting and dripping through the hole at a rate of half a cubic inch per minute. The cone has a radius of 2 inches and a height of 5 inches. How fast is the height of the ice cream changing when the height of the ice cream in the cone is 3 inches?

Step-by-Step Solution

Answer

The rate of change is $-\frac{25}{72\pi }\mathrm{in}$

Step 1. Given information

Given the cone has a radius of $2$ inches and a height of $5$ inches

Step 2: Calculate volume in terms of height

Calculating, we get

$$\begin{gathered} V=\frac{1}{3}\pi r^{2}h \\ V=\frac{1}{3}\pi {\left(\frac{2}{5}h\right)}^{2}h \\ V=\frac{4}{75}\pi h^3 \end{gathered}$$

Step 3: Differentiating the function

Differentiating, we get

$$\begin{gathered} \frac{dV}{dt}=\frac{4}{75}\pi \left(3h^2\right)\times \frac{dh}{dt} \\ -\frac{1}{2}=\frac{4}{75}\pi \left(3(3{)}^2\right)\times \frac{dh}{dt} \\ \frac{dh}{dt}=-\frac{25}{72\pi }\mathrm{in} \end{gathered}$$