An open hemispherical storage tank has radius 13 m. Oil flows into the tank such that the depth ' h ' of oil in the tank changes at the rate of 3 m / hr. When the depth h = 1 m, the rate of change of the area of the top surface of the oil is

Step-by-step solution

Let R be the radius of the hemispherical tank, so R = 13 m. Let r be the radius of the top surface of the oil at depth h . The distance from the center of the hemisphere to the oil surface is R - h . Using the Pythagorean theorem in the cross-section of the hemisphere: r^2 + (R - h)^2 = R^2 r^2 = R^2 - (R^2 - 2Rh + h^2) = 2Rh - h^2 The area of the top surface of the oil is A = r^2 . A = (2Rh - h^2) Differentiating with respect to time t : dA dt = (2R - 2h) dh dt Given R = 13 m, h = 1 m, and dh dt = 3 m/hr, substitu