Question

CALCULUS 1A stone is thrown into a pond, creating a circular wave whoseradius increases at the rate of 1 foot per second. In squarefeet per second, how fast is the area of the circular rippleincreasing 3 seconds after the stone hits the water?

Answer 1

6π

Explanation

The area of the circular ripple is calculated as follows:

`A=\pi r^2`

where r is the radius of the circle.

To find how fast is the area of the circular ripple increasing, we need to compute the derivative of the area with respect to time as follows:

`\begin{gathered} (dA)/(dt)=(d)/(dt)(\pi r^2) \\ (dA)/(dt)=\pi(d)/(dt)(r^2) \\ (dA)/(dt)=\pi\cdot2r\cdot(dr)/(dt)\text{ \lparen r is also a function of time\rparen} \end{gathered}`

Given that the radius of the circle increases at the rate of 1 foot per second, then:

`(dr)/(dt)=1(ft)/(s)`

Considering the rate of 1 foot per second, then after 3 seconds the radius will be:

`r=3\text{ ft}`

Substituting these values into the formula of dA/dt, we get:

`\begin{gathered} (dA)/(dt)=\pi\cdot2\cdot3\cdot1 \\ (dA)/(dt)=6\pi(ft^2)/(s) \end{gathered}`