Q. 54 - Calculus

Question

In Exercises 52-55, Linda is bored and decides to pour an entire container of salt into a pile on the kitchen floor. She pours $3$ cubic inches of salt per second into a conical pile whose height is always two-thirds of its radius.

How fast is the height of the conical salt pile changing when the radius of the pile is $2$ inches?

Step-by-Step Solution

Answer

The rate of change is $\frac{3}{4\pi }\frac{\text{ inches }}{\text{ sec }}$

Step 1. Given information

Given the radius of pile is $2$ inches

Step 2: Calculate the height and the rate of change

Calculating, we get

$$\begin{gathered} h=\frac{2}{3}r \\ h=\frac{2}{3}\times 2 \\ h=\frac{4}{3} \\ V=\frac{1}{3}\pi r^{2}h \\ V=\frac{1}{3}\pi {\left(\frac{3}{2}h\right)}^{2}h \\ V=\frac{9}{12}\pi h^3 \end{gathered}$$

Step 3: Differentiate and calculate the rate of change

Calculating, we get

$$\begin{gathered} \frac{dV}{dt}=\frac{9}{12}\pi \times 3h^2\frac{dh}{dt} \\ 3=\frac{9}{12}\pi \times 3{\left(\frac{4}{3}\right)}^2\frac{dh}{dt} \\ \frac{dh}{dt}=\frac{3}{4\pi }\frac{\text{ inches }}{\sec } \end{gathered}$$