The radius of the base of a cylinder is decreasing at a rate of 12 kilometers per second. The height of the cylinder is fixed at 2.5 kilometers. At a certain instant, the radius is 40 kilometers. What is the rate of change of the volume of the cylinder at that instant (in cubic kilometers per second)? Choose 1 answer: (A) -2400 pi (B) -4000 pi (C) -1600 pi (D) -360 pi The volume of a cylinder with radius r and height h is pir^(2)h.

Question

The radius of the base of a cylinder is decreasing at a rate of 12 kilometers per second. The height of the cylinder is fixed at 2.5 kilometers. At a certain instant, the radius is 40 kilometers. What is the rate of change of the volume of the cylinder at that instant (in cubic kilometers per second)? Choose 1 answer: (A)  -2400 pi (B)  -4000 pi (C)  -1600 pi (D)  -360 pi The volume of a cylinder with radius  r and height  h is  pir^(2)h.

Bytelearn · Accepted Answer

Differentiate Volume Formula: First, let's differentiate the volume formula with respect to time t to find dV/dt. dV/dt = d(πr^2h)/dt Since h is constant, we can take it outside the differentiation. dV/dt = πh * d(r^2)/dtDifferentiate r^2: Now, we differentiate r^2 with respect to time t. d(r^2)/dt = 2r * dr/dtCalculate dV/dt: We plug in the values for dr/dt, r, and h into the differentiated volume formula. dV/dt = πh * (2r * dr/dt) dV/dt = π * 2.5 * (2 * 40 * -12)Calculate final value: Now, we calculate the value. dV/dt = π * 2.5 * (2 * 40 * -12) dV/dt = π * 2.5 * (-960) dV/dt = -2400π km^3/s

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