A cup has the shape of a right circular cone. The height of the cup is 12 cm, and the radius of the opening is 3 cm. Water is poured into the cup at a constant rate of 2 cm^3/sec. What is the rate at which the water level is rising when the depth of the water in the cup is 5 cm? (The volume of a cone of height h and radius r is given by V = 1/3 pi r^2 h)

(A) 32/25 pi cm/sec

(B) 96/125 pi cm/sec

(C) 2/3 pi cm/sec

(D) 2/9 pi cm/sec

(E) 1/200 pi cm/sec

(A) 32 / (25π) cm/sec

Step-by-step explanation:

Let's say h is the depth of the water and r is the radius of the water at that depth.

Using similar triangles, we can say:

r / h = 3 / 12

r / h = ¼

r = ¼ h

The volume of the water is:

V = ⅓ π r² h

Substituting:

V = ⅓ π (¼ h) ² h

V = π h³ / 48

Take derivative with respect to time:

dV/dt = 3π h² / 48 dh/dt

dV/dt = π h² / 16 dh/dt

Solve for dh/dt:

dh/dt = 16 dV/dt / (π h²)

dV/dt = 2, and h = 5, so:

dh/dt = 16 (2) / (π 5²)

dh/dt = 32 / (25π)

The water rises at at rate of 32 / (25π) cm/sec.