A trough is 12 ft long and has ends that are isosceles triangles that are 1 ft high and 1.5 ft wide. If the trough is being filled at a rate of 11 cubic feet per minute, how fast is the height of the water increasing when the height is 7 inches?
Check some of the Related Questions below. This trapezoid trough has come up several times.
DUCK
To find the rate at which the height of the water is increasing, we can apply related rates, which involves differentiating the given information with respect to time.
Let's first define the variables:
- Let h(t) be the height of the water in the trough at time t.
- Let A(t) be the cross-sectional area of the water in the trough at time t.
We need to find dh/dt, the rate at which the height of the water is increasing.
Given information:
- Length of the trough (base of the triangle) = 12 ft
- Height of the triangle = 1 ft
- Width of the triangle = 1.5 ft
- The trough is being filled at a rate of 11 cubic feet per minute.
To find the cross-sectional area A(t) of the water in the trough at time t, we need to consider the shape of the triangle formed by the water.
The cross-sectional area A(t) of the water in the trough is equal to the area of an isosceles triangle. The formula for the area of a triangle is A = (1/2) * base * height.
The base of the triangle at any given time t is equal to (12 - 2h(t)) ft, where 2h(t) accounts for the length of each triangular end.
Thus, the formula for the cross-sectional area A(t) is:
A(t) = (1/2) * (12 - 2h(t)) * h(t)
To find dh/dt, we differentiate A(t) with respect to t and apply the chain rule:
dA/dt = (1/2) * (12 - 2h(t)) * dh/dt + (1/2) * (-2h(t)) * dh/dt
Given that dA/dt = 11 and h(t) = 7 inches (0.5833 ft), we can substitute these values into the equation:
11 = (1/2) * (12 - 2(0.5833)) * dh/dt + (1/2) * (-2(0.5833)) * dh/dt
Simplifying the equation, we have:
11 = (1/2) * (11.834) * dh/dt - (0.5833) * dh/dt
Now, let's solve for dh/dt:
11 = (1/2) * (11.834 - 0.5833) * dh/dt
11 = 11.251 * dh/dt
dh/dt = 11 / 11.251
Finally, we can calculate the value of dh/dt:
dh/dt ≈ 0.977 ft/min
Therefore, the height of the water is increasing at a rate of approximately 0.977 feet per minute when the height is 7 inches.