AP Calc AB questions

utche11
Wed, 15 Dec 2010 19:10:48 +0000

A cube is expanding so that its surface area is increasing at a constant rate of 4 square inches per second. How fast is the volume increasing at the instant when the surface area is 24 square inches?
answer is 2cubic in/sec


Chain rule:

dV/dt = dV/dA * dA/dt

dV/dt = ?
dA/dt = 4 in^2/second

to find dV/dA, we need volume in terms of surface area

V = a^3
A = 6a^2 —> a = sqrt(A/6)

so,
V = (A/6)^(3/2)

dV/dA = 1/6 * 3/2 * (A/6)^(1/2)

when surface area = 24 inches, dV/dA = 1/2

dV/dt= dV/dA * dA/dt = 1/2 * 4 = 2 in^3/second