Question

A stone is dropped into a quiet lake and waves move in circles at the speed of $5\raisebox{1ex}{$cm$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$. At that instant, when the radius of the circular wave is $8cm$, how far is the enclosed area increasing?

• A. $6\pi \raisebox{1ex}{$c{m}^2$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$
• B. $8\pi \raisebox{1ex}{$c{m}^2$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$
• C. $\raisebox{1ex}{$8$}\!\left/ \!\raisebox{-1ex}{$3$}\right.\pi \raisebox{1ex}{$c{m}^2$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$
• D. $80\pi \raisebox{1ex}{$c{m}^2$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$

Answer 1

The correct option is D

$80\pi \raisebox{1ex}{$c{m}^2$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$

Step 1: Given data and assumptions.

The radius of the wave, $r=8cm$

Change in radius with respect to time, $\frac{dr}{dt}=5\raisebox{1ex}{$cm$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$

Step 2: Finding the enclosed area increasing.

Since, the enclosed area of the wave, $A=\pi r^2$

Therefore, the rate of change of area with respect to time,

$\frac{dA}{dt}=\frac{d\left(\pi r^2\right)}{dt}$

$\frac{dA}{dt}=2\pi r\frac{dr}{dt}$

$\frac{dA}{dt}=2\pi \times 8\times 5$

$\frac{dA}{dt}=80\pi \raisebox{1ex}{$c{m}^2$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$

Hence, option D is correct. the enclosed area increasing at the rate of $80\pi \raisebox{1ex}{$c{m}^2$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$.