Water leaks from the bottom of an inverted cone of semi-vertical angle 45 degrees at a rate of 2 m³/s. When the height of water in the cone is 2 m, find the rate d (in m/s) at which the circumference of the water surface is changing. What is |5d|?

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5. Since semi-vertical angle = 45 deg, r = h always. Volume V = (1/3) pi h³. Differentiating: dV/dt = pi h² * dh/dt. Water leaks at -2 m³/s: -2 = pi * 4 * dh/dt → dh/dt = -1/(2*pi). Circumference C = 2*pi*h, so dC/dt = 2*pi * dh/dt = 2*pi * (-1/(2*pi)) = -1 m/s. So d = -1, and |5d| = 5.

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