The area of a circular oil slick is increasing at a constant rate of 2 square miles per day. How fast is the radius increasing when the area is 4(pi) square miles (approximately 12.57 square miles)?

So far I have that the area increases at the rate of x+(2mi^2/d) but I don't think that's right

a = πr^2

da/dt = 2πr dr/dt

So, plug in your numbers. When a = 4π, r=2, so

2 = 2π*2 dr/dt
dr/dt = 1/(2π) miles/day

To find how fast the radius is increasing when the area is 4π square miles, you can use the formula for the area of a circle:

A = πr^2

To differentiate with respect to time (t), you will need to use implicit differentiation:

dA/dt = 2πr(dr/dt)

Given that dA/dt = 2 square miles per day and the area A = 4π square miles, we can substitute the values into the equation:

2 = 2πr(dr/dt)

Simplifying the equation, we get:

r(dr/dt) = 1

Now we need to solve for (dr/dt) when the area is 4π square miles. Plugging in A = 4π, we have:

r(dr/dt) = 1
4π(dr/dt) = 1

Now we can solve for (dr/dt):

(dr/dt) = 1/(4π)
(dr/dt) ≈ 0.0796 miles per day

Therefore, the radius is increasing at a rate of approximately 0.0796 miles per day when the area is 4π square miles.

To find the rate at which the radius is increasing, you can use the relationship between the area and the radius of a circle. The formula for the area of a circle is A = πr^2, where A is the area and r is the radius.

Given that the area of the circular oil slick is increasing at a constant rate of 2 square miles per day, you can express this rate as dA/dt = 2 mi^2/day.

Now, you want to find how fast the radius (dr/dt) is changing when the area is 4π square miles.

To solve this problem, let's differentiate both sides of the area formula with respect to time:

d/dt(A) = d/dt(πr^2)

Using the chain rule, we have:

dA/dt = 2πr(dr/dt)

From the problem statement, we know that dA/dt = 2 mi^2/day and we want to find dr/dt when A = 4π mi^2.

Substituting these values into the equation, we have:

2 = 2πr(dr/dt)

Simplifying the equation, we get:

dr/dt = 1 / (πr)

Now we can find the rate at which the radius is increasing by substituting the area A = 4π mi^2 into the equation:

dr/dt = 1 / (π * r)

dr/dt = 1 / (π * 2 * √(A / π))

dr/dt = 1 / (2√(A / π))

dr/dt = 1 / (2 * √(4π / π))

dr/dt = 1 / (2 * √4)

dr/dt = 1 / 4

Therefore, when the area is 4π square miles, the rate at which the radius is increasing is 1/4 miles per day.