Question

A circular oil slick is expanding at a rate of 3m/min. How fast is the area of the oil slick changing when the diameter is 48 meters?

Answer 1

The area of the oil slick changes at a rate of 144π square meters per minute when the diameter is 48 meters, which is determined by differentiating the area of a circle to time using the given radius expansion rate.

Step-by-step explanation:

The question involves the application of derivatives for time, more specifically related rates of change, which is typically taught in high school calculus.

To find out how fast the area of the oil slick is changing when the diameter is 48 meters, we can set up a relationship between the radius r of the oil slick and its area A. The area of a circle is given by A = πr². Since the diameter is 48 meters, the radius is half of that, which is 24 meters.

Assuming that the radius is changing at a constant rate of 3 meters per minute (expansion rate), we can denote the rate of change of the radius for time as dr/dt = 3 m/min. We can then differentiate the area formula concerning time to find dA/dt, the rate of change of the area. Using the chain rule, d(πr²)/dt = 2πr dr/dt. Plugging in the radius (24m) and dr/dt (3m/min), we get:

dA/dt = 2π(24m)(3m/min) = 144π m²/min.

Therefore, the area is changing at a rate of 144π square meters per minute when the diameter is 48 meters.

Answer 2

`48\pi \:m/min`

Step-by-step explanation:

Diameter = 48 m

Therefore radius r = 48/2 = 24 m

`Area \: of \: circular \: oil \: slick \\ A = \pi {r}^(2) \\ differentiating \: w \: r \: to \: t \: on \: \\ both \: sides \\ \\ (dA)/(dt) = (d)/(dt) (\pi {r}^(2) )\\ \\ (dA)/(dt) =\pi (d)/(dt) {r}^(2) \\ \\ (dA)/(dt) =\pi * 2{r}\\ \\ (dA)/(dt) =2\pi {r} \\ \\ \bigg((dA)/(dt) \bigg)_(r=24) =2\pi * {24} \\ \\ \bigg((dA)/(dt) \bigg)_(r=24) =48\pi \: m \: per \: min`