A pebble is dropped into a pond, causing ripples in the form of concentric circles. The radius r of the outer ripple is increasing at a constant rate of 2 feet per second. When the radius is 10 feet, at what rate is the total area A of the disturbed water changing?
a = pi r^2
da/dt = 2pi r dr/dt
when r = 10,
da/dt = 2pi * 10 ft * 2 ft/s = 40pi ft^2/s
To find the rate at which the total area of the disturbed water is changing, we need to differentiate the formula for the area of a circle with respect to time.
The formula for the area of a circle is A = πr^2, where A is the area and r is the radius.
We are given that the radius is increasing at a constant rate of 2 feet per second. This means that dr/dt = 2 ft/s.
To find how the area is changing with respect to time, we need to differentiate the area formula with respect to time. Using the chain rule, we have:
dA/dt = dA/dr * dr/dt
Let's differentiate the area formula:
dA/dt = d/dt(πr^2)
Applying the power rule, we get:
dA/dt = 2πr(dr/dt)
Now we can substitute the given values:
r = 10 ft
dr/dt = 2 ft/s
Substituting these values into the equation, we have:
dA/dt = 2π(10)(2)
Simplifying, we get:
dA/dt = 40π
So, when the radius is 10 feet, the total area of the disturbed water is changing at a rate of 40π square feet per second.