Chapter 3.6, Problem 9MRPS

Solution

To calculate: $A\left(r\left(t\right)\right)$ and evaluate $A\left(r\left(2\right)\right)$ if a pebble is dropped into a calm pond, causing ripples of concentric circles.

  $A\left(r\left(t\right)\right)$ is $36\pi t^2$ and $A\left(r\left(2\right)\right)$ is equal to $144\pi$ . $A\left(r\left(2\right)\right)$ represents the area of outer ripple formed after 2 seconds.

Given information:

The given area of the outer ripple is $A\left(r\right)=\pi r^2$ and its radius is $r\left(t\right)=6t$ .

The radius r (in feet) of the outer ripple is given by $r\left(t\right)=6t$ where t is the time (in seconds) after the pebble hits the water. The area A (in square feet) of the outer ripple is given by $A\left(r\right)=\pi r^2$ . Also explain what $A\left(r\left(2\right)\right)$ represent.

Explanation:

Write the given equation:

  $A\left(r\right)=\pi r^2$ .

Substitute $r\left(t\right)$ for r $A\left(r\right)$ to get $A\left(r\left(t\right)\right)$ ,

  $A\left(r\left(t\right)\right)=A\left(6t\right)$ /

Now find $A\left(6t\right)$ .

  $\begin{array}{c}A\left(6t\right)=\pi {\left(6t\right)}^2\\ =\pi \left(36t^2\right)\\ =36\pi t^2\end{array}$

Therefore, $A\left(r\left(t\right)\right)$ is $36\pi t^2$ .

Now substitute 2 for t in $36\pi t^2$ to get $A\left(r\left(2\right)\right)$ .

  $\begin{array}{c}36\pi t^2=36\pi {\left(2\right)}^2\\ =36\pi {\left(4\right)}^2\\ =144\pi \end{array}$

Therefore, $A\left(r\left(2\right)\right)$ is equal to $144\pi$ .

It is given that $r\left(t\right)$ represents the radius of the ripple after t seconds and $A\left(r\right)$ represents the area of the outer ripple.

Thus, $r\left(t\right)$ represents the radius of the ripple after 2 seconds, and $A\left(r\left(2\right)\right)$ represents the area of the outer ripple formed after 2 seconds.