assume that oil spilled from a ruptured tanker spreads in a circular pattern whose radius increases at a constant rate of 2 feet per second. how fast is the area of spill increasing when the radius of the spill is 60 feet? Note: you must state and use the linking function !
a = π r^2
differentiating ... da/dt = 2 π r dr/dt
To solve this problem, we need to apply the concept of derivatives. Let's denote the radius of the spill as r and the area of the spill as A. We are given that the radius is increasing at a constant rate of 2 feet per second.
The formula for the area of a circle is A = πr^2. Differentiating both sides of the equation with respect to time t, we get:
dA/dt = d/dt (πr^2)
Using the chain rule, we'll differentiate both terms separately:
dA/dt = 2πr(dr/dt)
Now, we are given that the rate of change of the radius (dr/dt) is constant at 2 feet per second. Thus, we can substitute this value into the equation:
dA/dt = 2πr(2)
Simplifying the equation:
dA/dt = 4πr
Now we need to find the value of dA/dt when the radius (r) is 60 feet. Substituting r = 60 into the equation, we have:
dA/dt = 4π(60)
Calculating this:
dA/dt = 240π
So, the area of the spill is increasing at a rate of 240π square feet per second when the radius of the spill is 60 feet.
To summarize, we have used the linking function dA/dt = 4πr, which relates the rate of change of the area of the spill to the radius.