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 3m/s, how fast is the area of the spill increasing when the radius is 15m?
a = oi r^2
da/dt = 2pi r dr/dt
plug and chug.
To solve this problem, we can use the formulas for the area of a circle and its rate of change. The area of a circle is given by the formula A = πr^2, where A is the area and r is the radius. We are given that the radius of the oil spill is increasing at a constant rate of 3m/s.
To find the rate at which the area is increasing, we can take the derivative of the area formula:
dA/dt = d/dt(πr^2)
Using the chain rule, we can rewrite this as:
dA/dt = 2πr(dr/dt)
We know that dr/dt, the rate at which the radius is increasing, is 3m/s. Substituting the values, we get:
dA/dt = 2π(15)(3)
Simplifying this expression, we find:
dA/dt = 90π m^2/s
Therefore, when the radius is 15m, the area of the spill is increasing at a rate of approximately 90π m^2/s.