Q. 55 - 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 radius of the conical salt pile changing when the height of the pile is 4 inches?

Step-by-Step Solution

Answer

The rate of change is $\frac{dr}{dt}=\frac{1}{8\pi }\frac{\text{ inches }}{\text{ sec }}$

Step 1. Given information

Given the height of pile is four inches

Step 2: Calculate the height and the rate of change

Calculating, we get

$$\begin{gathered} h=\frac{2}{3}r \\ 4=\frac{2}{3}\times r \\ r=6 \\ V=\frac{1}{3}\pi r^{2}h \\ V=\frac{1}{3}\pi r^2\left(\frac{2}{3}r\right) \\ V=\frac{{\displaystyle 2}}{{\displaystyle 9}}\pi r^3 \end{gathered}$$

Step 3: Differentiate and calculate rate of change

Calculating, we get

$$\begin{gathered} \frac{dV}{dt}=\frac{2}{9}\pi \times 3r^2\frac{dr}{dt} \\ 3=\frac{2}{9}\pi \times 3(6{)}^2\frac{dr}{dt} \\ \frac{dr}{dt}=\frac{1}{8\pi }\frac{\text{ inches }}{\sec } \end{gathered}$$