Question

Water is poured into a conical tank 6m across the top and 8m deep at the rate of 10 m^2/min. how fast is the water level rising when the tank is 5m deep.

Answer 1

To solve the student's problem, we use the volume formula for a cone and apply the chain rule to relate the different variables. Given the rate at which volume changes, we can find the rate at which the water level, represented by the height, changes.

Step-by-step explanation:

The student is presenting a calculus problem involving related rates. Specifically, the student is asking how fast the water level is rising in a conical tank as water is added at a known rate. To find this, we can use the formula for the volume of a cone V = 1/3πr^2h and apply the chain rule to relate the variables.

First, since the tank is 6m across, the radius r is 3m. When the water is 5m deep, using similar triangles, we find the radius of the water's surface at that depth is (5/8) r. Then, differentiating both sides of the volume formula with respect to time t gives us dV/dt = (1/3)π(2rdr/dt)h + (1/3)πr^2dh/dt. Since the problem provides dV/dt and requires dh/dt, we can solve for dh/dt once dr/dt is found using similar triangles.

Using the given rate of change of the volume 10 m^3/min, we can find dh/dt. It's a matter of substituting known values and solving for the unknown dh/dt to find the answer.