The radius of a circle is increasing at a rate of 3 ft/min. If the radius is 0 when the time is 0, how fast is the circle's area changing when time is at 10 minutes?
dr/dt = 3
r = 3t + c
since r(0) = 0,
r = 3t
a = πr^2
da/dt = 2πr dr/dt
...
To find how fast the circle's area is changing, we can use the formula for the area of a circle, which is A = πr^2, where A is the area and r is the radius.
First, let's find the equation for the radius as a function of time. We know that the radius is increasing at a rate of 3 ft/min, so we can write r(t) = 3t, where t is the time in minutes.
Now, let's differentiate the area function with respect to time. We have:
dA/dt = d(πr^2)/dt
Using the chain rule, we can differentiate this equation as follows:
dA/dt = 2πr * dr/dt
Substituting the expression for the radius, we have:
dA/dt = 2π(3t) * 3
Simplifying, we get:
dA/dt = 18πt
Now, we can find the rate of change of the circle's area when time is 10 minutes by substituting t = 10 into the equation:
dA/dt = 18π * 10
Calculating this value, we get:
dA/dt = 180π ft^2/min
Therefore, the circle's area is changing at a rate of 180π square feet per minute when time is at 10 minutes.