Suppose oil spills from a ruptured tanker and spreads in a circular pattern. If the radius of the oil spill increases at a constant rate of 1m/s, how fast is the area of the spill increasing when the radius is 30m?

Area = πr^2

d(area)/dt = 2πr dr/dt
but dr/dt = 1 m/s

d(area)/dt = 2π(30)(1) = 60π m^2/s

Well, it seems like you're in quite a slippery situation with that oil spill! Let's see if we can make some laughs while solving this problem.

To find the rate at which the area of the spill is increasing, we can use the formula for the area of a circle, which is A = πr^2. Now, let me put on my thinking nose and solve this for you!

First, we need to find the rate at which the radius is changing. Given that the radius is increasing at a constant rate of 1m/s, we know dr/dt = 1 m/s.

Next, we take the derivative of the area formula with respect to time (t). So, dA/dt = d(πr^2)/dt.

Applying the chain rule, we get dA/dt = 2πr(dr/dt).

Now we can plug in the values. When the radius is 30m, the area of the spill is A = π(30)^2. Simplifying that gives A = 900π square meters.

So, dA/dt = 2π(30)(1) = 60π square meters per second.

Voila! The area of the spill is increasing at a whopping rate of 60π square meters per second when the radius is 30m. That's quite a gooey mess, huh?

To find the rate at which the area of the oil spill is increasing, we can use the formula for the area of a circle:

A = πr^2

Where A is the area and r is the radius of the oil spill.

We are given that the radius of the oil spill is increasing at a constant rate of 1m/s. Let's denote the rate of change of the radius as dr/dt.

Given that dr/dt = 1m/s, we want to find dA/dt, the rate at which the area is changing with respect to time.

To do this, we'll need to differentiate the area function with respect to time. Differentiating both sides of the equation A = πr^2 with respect to time gives us:

dA/dt = d/dt (πr^2)

Using the power rule, the derivative of r^2 with respect to t is 2r(dr/dt), since r is changing with respect to time.

dA/dt = 2πr(dr/dt)

Now we can substitute the given values into the equation. When the radius is 30m, r = 30m and dr/dt = 1m/s.

dA/dt = 2π(30)(1)

dA/dt = 60π m^2/s

Therefore, when the radius is 30m, the area of the oil spill is increasing at a rate of 60π square meters per second.

To find how fast the area of the oil spill is increasing, we can use the formula for the area of a circle:

A = πr^2

Where A is the area of the circle and r is the radius. Since the radius is increasing at a constant rate of 1m/s, we need to find the rate of change of the area with respect to time.

We can differentiate both sides of the equation with respect to time (t) using implicit differentiation:

dA/dt = d/dt (πr^2)

Using the chain rule, the derivative of r^2 with respect to t is 2r(dr/dt):

dA/dt = 2πr(dr/dt)

Now we need to find the value of dr/dt when the radius is 30m, which is given as 1m/s.

Plugging in the values we know:

dr/dt = 1m/s
r = 30m

dA/dt = 2π(30)(1)

Simplifying the equation:

dA/dt = 60π

Therefore, the area of the oil spill is increasing at a rate of 60π square meters per second when the radius is 30m.