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 = πr^{2}
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 ft^{2}/s
Therefore, at a radius of 4 feet, the area is raising at a rate of 12.6 square feet per second.