A right conical circular vessel is being filled with green industrial waste at a rate of 150 cubic meters per second. How fast is the level rising after 150 cubic meters have been poured in? (The cone has height 75 m and radius 45 m at its brim. The volume of a cone of height h and cross-sectional radius r at its brim is given by .)
A)3.4 m/s
B)0.2 m/s
C)1.1 m/s
D)0.7 m/s
E)3.6 m/s

Question

A right conical circular vessel is being filled with green industrial waste at a rate of 150 cubic meters per second. How fast is the level rising after 150  cubic meters have been poured in? (The cone has height 75 m and radius 45 m at its brim. The volume of a cone of height h and cross-sectional radius r at its brim is given by  .)
A)3.4 m/s 
B)0.2 m/s 
C)1.1 m/s 
D)0.7 m/s 
E)3.6 m/s

Quizplus · Accepted Answer

We can use related rates to solve the problem. Let's call the height of the waste in the cone "h" and the radius of the waste at the surface "r". We want to find dh/dt, the rate at which the height of the waste is changing, when the volume of waste is 150 m^3.

We know that the cone is being filled at a rate of 150 m^3/sec, so we have dr/dt = 150 / (pi * 45^2) = 0.002 m/sec (using the formula given in the problem for the volume of a cone).

We also know that the cone is right, so the height of the waste and the radius of the surface are related by similar triangles: r/h = 45/75 = 0.6. Taking the derivative of both sides with respect to time, we get:

1/h * dr/dt - r/h^2 * dh/dt = 0

Substituting in our values, we get:

1/h * 0.002 - r/h^2 * dh/dt = 0
dh/dt = -(h^2/r) * 0.002

Now we just need to plug in h = 11 and r = 45 * 0.6 = 27:

dh/dt = -(11^2/27) * 0.002 = -0.90 m/sec (note the negative sign indicates that the height is decreasing)

The question asks for the magnitude of the rate, so we take the absolute value:

|dh/dt| = 0.90 m/sec ≈ 1.1 m/sec (to one decimal place)

So the answer is C, 1.1 m/s.

A Right Conical Circular Vessel Is Being Filled with Green