Question

A trough has ends shaped like isosceles triangles, with a length of 9 m, width of 3 m, and height of 4 m. Water is being pumped into the trough at a rate of 3 m³/min. At what rate does the height of the water change when the water is 2 m deep?

(A question about the rate of change and related rates in calculus would need to be provided for options)
A) The correct rate of change
B) An incorrect rate of change
C) Not enough information to determine
D) None of the above

Answer 1

The rate at which the height of the water in the trough is changing when it is 2 m deep is 1 m/min.

Step-by-step explanation:

To find the rate at which the height of the water in the trough is changing, we can use related rates. Let h be the height of the water and t be the time in minutes. We know that dh/dt = 3 m³/min, which represents the rate at which the volume of water in the trough is changing. We want to find dH/dt, the rate at which the height of the water is changing when it is 2 m deep.

First, we can find the formula for the volume of the trough. The trough can be divided into two congruent right triangular prisms, so the total volume of the trough is V = 2(1/2 * bh * h) = bh * h, where b is the base of the triangular end. We are given that b = 3 m, so the volume is V = 3h.

Since the volume is changing with respect to time, we can take the derivative of both sides of the equation to find an expression for dh/dt in terms of dH/dt: d(V)/dt = d(3h)/dt. This gives us 3(dh/dt) = dV/dt = 3. We can solve this equation for dh/dt: dh/dt = 1.

Now, we can find dH/dt when h = 2 by substituting the values into the equation we found: dH/dt = dh/dt = 1 m/min.