Question

If by dropping a stone in a quiet lake a wave moves in a circle at a speed of $3.5cm/sec$, then the rate of increase of the enclosed region when the radius of the circular wave is $10cm$, is $(\pi =\frac{22}{7})$

• A. $220sq.cm/sec$
• B. $110sq.cm/sec$
• C. $35sq.cm/sec$
• D. $350sq.cm/sec$

Answer 1

The correct option is A

$220sq.cm/sec$

Step 1: Given Data

1. The radius of the circular wave is$r$=$10cm$.
2. The speed of the wave moving at a circular wave is $v=3.5cm/sec$

Step 2: Formula Used

The area($A$) of the circular wave in terms of the radius($r$) is,

$A=\pi r^2$

When a stone is dropped into a lake, waves move in a circle of the speed($v$) of $3.5cm/sec$, i.e. Radius of the circle increases at a rate of $3.5cm/sec$,

$v=$$\frac{dr}{dt}$

$t$ is the time taken.

The rate of increase in the area concerning time is

$\frac{dA}{dt}$

Step 3: Calculate the rate of increasing area

The radius of the circle increases at a rate of $3.5cm/sec$,

We know, that the area ($A$) of the circular wave in terms of the radius ($r$) is,

$A=\pi r^2$

Now,

$$\begin{gathered} \frac{dA}{dt}=\frac{d\left(\pi r^2\right)}{dt} \\ \frac{dA}{dt}=\pi \frac{d\left(r^2\right)}{dt} \\ \frac{dA}{dt}=\pi \frac{d\left(r^2\right)}{dt}\times \frac{dr}{dr} \\ \frac{dA}{dt}=\pi \frac{d\left(r^2\right)}{dr}\times \frac{dr}{dt} \\ \frac{dA}{dt}=\pi \times 2r\times \frac{dr}{dt} \\ \frac{dA}{dt}=\pi \times 2r\times 3.5 \\ \frac{dA}{dt}=7\pi r \end{gathered}$$ we know,$\frac{dr}{dt}$=$3.5$

When $r=10cm$

$$\begin{gathered} {\left|\frac{dA}{dt}\right|}_{r=10}=7\times \pi \times 10 \\ {\left|\frac{dA}{dt}\right|}_{r=10}=70\pi \end{gathered}$$

${\left|\frac{dA}{dt}\right|}_{r=10}=70\times \frac{22}{7}c{m}^2/sec$$=220sq.cm/sec$

Therefore, Area is increasing at the rate $220sq.cm/sec$ when $r=10cm$.

Hence, option A is the correct option.