Derivatives to Physical Systems:

A stone is dropped into a quiet lake, & waves move within circles outward from the location of the splash at a constant velocity of 0.5 feet per second.  Determine the rate at that the area of the circle is increasing when the radius is 4 feet.

Solution:

Using the formula for the area of a circle,

A = πr²

obtain the derivative of both sides of this equation along with respect to time t.

dA/dt = 2πr (dr/dt)

But, dr/dt is the velocity of the circle moving outward which equals  0.5 ft/s and dA /dt  is the  rate  at  which  the  area  is increasing,  that  is the  quantity  to  be determined.   Set r equal to 4 feet, substitute the known values into the equation, and solve for dA /dt.

dA/dt = 2πr(dr/dt)

dA/dt = (2)(3.1416)(4 ft)(0.5 ft/s)

dA/dt = 12.6 ft²/s

Therefore, at a radius of 4 feet, the area is raising at a rate of 12.6 square feet per second.