Multiple choice

The radius of a right circular cylinder increases at the rate of $0.1 cm/min$, and the height decreases at the rate of $0.2 cm/min$. The rate of change of the volume of the cylinder, in $cm^3/min.$, when the radius is $2 cm$ and the height is $3 cm$ is

  1. $-2p$
  2. $-\displaystyle \frac{8\pi}{5}$
  3. $-\displaystyle \frac{3\pi}{5}$
  4. $\displaystyle \frac{2\pi}{5}$
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D Correct answer
Explanation

V = pi * r^2 * h. dV/dt = pi * (2rh * dr/dt + r^2 * dh/dt). Given r=2, h=3, dr/dt=0.1, dh/dt=-0.2. dV/dt = pi * (2*2*3*0.1 + 2^2*-0.2) = pi * (1.2 - 0.8) = 0.4pi = 2pi/5.

AI explanation

The volume of a cylinder is given by the formula V = pi * r squared * h. Differentiating this with respect to time gives dV/dt = pi * (2r * h * dr/dt + r squared * dh/dt). Substituting r = 2, h = 3, dr/dt = 0.1, and dh/dt = -0.2 results in dV/dt = pi * (2 * 2 * 3 * 0.1 + 4 * -0.2). This simplifies to pi * (1.2 - 0.8), making the rate of change 0.4 * pi, which is equivalent to 2pi/5.

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