Question

The radius of a circular oil slick expands at a rate of 4 m/min. (a) how fast is the area of the oil slick increasing when the radius is 21 m? (b) if the radius is 0 at time t=0 , how fast is the area increasing after 3 mins?

Answer 1

The area of the oil slick is increasing at a rate of 168π m^2/min when the radius is 21 m. After 3 minutes, the area of the oil slick is increasing at a rate of 96π m^2/min.

Step-by-step explanation:

**(a) How fast is the area of the oil slick increasing when the radius is 21 m?**

To solve this problem, we can use the formula for the area of a circle, \(A = \pi r^2\), where \(r\) is the radius. We are given that the radius is increasing at a rate of
`\(4 \, \text{m/min}\)`, which is the rate of change of \(r\) with respect to time \(t\).

We can express this relationship using the chain rule: \(\frac{dA}{dt} = 2\pi r
`(dr)/(dt)\`). Here,
`\((dA)/(dt)\)`is the rate of change of the area, \(r\) is the radius, and \(\frac{dr}{dt}\) is the rate of change of the radius.

Now, substitute the given values into the formula: \[\frac{dA}{dt} = 2\pi \times 21 \times 4\]

Calculate this expression to find the rate at which the area is increasing when the radius is
`\(21 \, \text{m}\)`.

**(b) If the radius is 0 at time \(t=0\), how fast is the area increasing after 3 mins?**

For this part, we continue to use the formula \(\frac{dA}{dt} = 2\pi r \frac{dr}{dt}\). However, since we are given that the radius is 0 at \(t=0\), we need to determine the radius at \(t=3\) minutes.

Using the initial condition \(r(0) = 0\) and the fact that the radius is increasing at \(4 \, \text{m/min}\), we find that \(r(3) = 4 \times 3\).

Now, substitute \(r = 4 \times 3\) and \(\frac{dr}{dt} = 4\) into the formula to calculate
`\((dA)/(dt)\) at \(t=3\)`minutes.

Ensure that the units are consistent throughout the calculations to obtain the correct rate of change.