SOLUTION: A stone dropped into a still pond sends out a circular ripple whose radius increases at a constant rate of 2ft/s. How rapidly is the area enclosed by the ripple increasing at the e

Question 1060474: A stone dropped into a still pond sends out a circular ripple whose radius increases at a constant rate of 2ft/s. How rapidly is the area enclosed by the ripple increasing at the end of 13s?
Answer by rothauserc(4718) (Show Source): You can put this solution on YOUR website!
we use the formula for the area of a circle and use the implicit derivative
:
A = pi * r^2, where A is the area, r is the radius and pi = 22/7
:
we now take the implicit derivative with respect to time
:
dA/dt = 2 * pi * r * (dr/dt)
:
we are given that (dr/dt) = 2 and asked to evaluate the Area's rate of change after 13 seconds
:
we use the circumference of the circle to get r after 13 seconds
:
C = 2 * pi * r, where C is the circumference, then
:
dC/dt = 2 * pi * (dr/dt) = 2 * pi * 2 = 4 * pi per second
:
after 13 seconds C = 4 * pi * 13 = 52 * pi
:
52 * pi = 2 * pi * r
:
r = 26
:
*******************************************************
dA/dt = 2 * pi * 26 * 2 = 326.7256 square feet / second
*******************************************************
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