A tank of water in the shape of a cone is being filled with water at a rate of 12 m^3/sec. The base radius of the tank is 26 meters and the height of the tank is 8 meters At what rate is the depth of the water in the tank changing when the radius of the top of the water is 10 meters?

Answer 1

r = 8/26 h
so when r=10, h=65/2
v = 1/3 πr^2 h = 1/3 π (8/26 h)^2 h = 16π/507 h^3
dv/dt = 16π/169 h^2 dh/dt
now plug in your numbers and find dh/dt