Q 38. - Calculus

Question

A rock dropped into a pond causes a circular wave of ripples whose radius increases at $6$ inches per second. How fast is the area of the circle of ripples expanding at the instant that the area of the circle is $100$ square inches? $200$ square inches? $1000$ square inches? Explain why it makes sense that the rate of change of the area increases as the area increases 

Step-by-Step Solution

Answer

For $Area=100 i{n}^2$: The rate of the area of the circle of ripples expanding at the instant  is $212.63 i{n}^2/sec$

For $Area=200 i{n}^2:$ The rate of the area of the circle of ripples expanding at the instant  is $300.839 i{n}^2/sec$

For $Area=1000 i{n}^2:$ The rate of the area of the circle of ripples expanding at the instant  is $672.552 i{n}^2/sec$

The rate of change of the area increases as the value of radius increases because circle is expanding.

Step 1. Given Information

It is given that

$\frac{dr}{dt}=6 in/sec$

Areas are 

$Area=100 i{n}^2,Area=200 i{n}^2 , Area = 1000 i{n}^2$

Step 2. Calculation of radius for different areas

(I)

$A=100 i{n}^2$

Apply area of circle formula

$A=\pi r^2$

Plug values

$$\begin{gathered} 100=\pi {\left(r\right)}^2{r}^2=\frac{100}{\pi } \\ r=\sqrt{\frac{100}{\pi }} \\ r=5.64 in \end{gathered}$$

(II)

$A=200 i{n}^2$

Apply area of circle formula

$A=\pi r^2$

Plug values

$$\begin{gathered} 200=\pi {\left(r\right)}^2 \\ {r}^2=\frac{200}{\pi } \\ r=\sqrt{\frac{200}{\pi }} \\ r=7.98 in \end{gathered}$$

(III)

$A=1000 i{n}^2$

Apply area of a circle formula

$A=\pi r^2$

Plug values

$$\begin{gathered} 1000=\pi {\left(r\right)}^2 \\ {r}^2=\frac{1000}{\pi } \\ r=\sqrt{\frac{1000}{\pi }} \\ r=17.84 in \end{gathered}$$

Step 3. Setting up the rate of change of the area equation

The formula for the area of a circle is 

$A=\pi r^2$

Find derivative with respect to $t$ using the chain rule

$$\begin{gathered} \frac{dA}{dt}=\pi \left(2r\right)\frac{dr}{dt} \\ \frac{dA}{dt}=2\pi r\frac{dr}{dt} \end{gathered}$$

Step 4. Finding the rate of change of area value

(I)

It is given that 

$\frac{dr}{dt}=6 in/sec , r=5.64 in$

Plug these values into rate of change of area formula

$$\begin{gathered} \frac{dA}{dt}=2\pi (5.64)\left(6\right) \\ \frac{dA}{dt}= 212.623 i{n}^2/sec \end{gathered}$$

(II)

It is given that

$\frac{dr}{dt}=6 in/sec, r=7.98 in$

Plug these values into rate of change of area formula

$$\begin{gathered} \frac{dA}{dt}=2\pi (7.98)\left(6\right) \\ \frac{dA}{dt}=300.839 i{n}^2/sec \end{gathered}$$

(III)

It is given that

$\frac{dr}{dt}=6in/sec , r=17.84 in$

Plug these values into rate of change of area formula

$$\begin{gathered} \frac{dA}{dt}=2\pi (17.84)\left(6\right) \\ \frac{dA}{dt}=672.552 i{n}^2/sec \end{gathered}$$

The rate of change of area is increasing because as radius increases ripple area will increase.