Question

The radius of the circle is decreasing at a rate of 1 meter per day and the sides of the square are decreasing at a rate of 4 meters per day.

Answer 1

SOLUTION:

Step 1:

In this question, we have the following:

Step 2:

$\begin{gathered} \frac{dr}{\differentialDt t}\text{ = - 1m/day } \\ \frac{dl}{\differentialDt t}=\text{ -4 m / day} \end{gathered}$$\begin{gathered} \text{ A = l}^2-\pir^2 \\ \frac{dA}{\differentialDt t}\text{ =2l}\frac{dl}{\differentialDt t}\text{ -2}\pi r\frac{dr}{\differentialDt t} \\ =\text{ 2l ( - 4 ) - 2}\pi r(-1) \\ =\text{ -8l +2}\pi r \\ =\text{ -8(15) + 2}\pi(4) \\ =\text{ -120 + 8}\pi \end{gathered}$$\begin{gathered} \text{Assume }\pi\text{ = 3.142} \\ =\text{ -120 + 8 ( 3. 142)} \\ =\text{ -120 + 25.136} \\ =\text{ -94.864 square meters per day} \end{gathered}$