Hi Joyce,
Picture the inverted cone with the radius, the height, and the slant height forming a right triangle. The surface of the gasoline has a radius, a height, and a slant height each smaller than the cone which also form a right triangle similar to the one in the cone. Since the triangles are similar, a proportion can be set up with x(radius of the cone) and y(height of the cone) to the corresponding sides in the smaller right triangle resulting in x/y = 4/5. It may not be apparent why this is needed at this time, but it is necessary. To begin solving the problem we know that dV/dt = 8 m3/sec and the level of the gasoline is at 4 meters.We want to find dy/dt or how fast the y-value or height of the gasoline is rising when y = 4 meters.
since we are given that the change in the volume is dV/dt = 8, we must use the formula for the volume of a cone
V = (1/3)πr2h or in terms of x and y V = (1/3)πx2y Since we do not know a value for x, this is where the similar triangles are needed.We can use the proportion x/y = 4/5 and solve for x to get x = 4y/5. Now
V = (1/3)π(4y/5)2 y = (16πy3/75 Differentiating with respect to t we have
dV/dt = (16π/75) 3y2 (dy/dt) = (48π/75) y2 (dy/dt) Since dV/dt = 8 and y = 4, substitute these values into the equation above and get
8 = (48π/75)(42) (dy/dt) Solve for dy/dt
dy/dt = 8(75/48•16π) = 0.25 m/sec