A trough is 6 feet long and has ends that are isosceles triangles that are 1 foot high and 3.5 feet wide. If the trough is being filled at a rate of 9 cubic feet per minute, how fast is the height of the water increaseing when the height is 5 inches?
using similar triangles, we see that when the water is of height h, the width of the cross-section is
w = 3.5h
So, the volume of water at height h is
v = 1/2 6*h(3.5h) = 10.5h^2
so, we have
dv/dt = 21h dh/dt
Now just plug in your numbers and solve for dh/dt
To find the rate at which the height of the water in the trough is increasing, we can use related rates. Let's denote the height of the water in the trough as "h" (in feet) at any given time "t" (in minutes).
We are given that the trough is being filled at a rate of 9 cubic feet per minute, which means the volume of water in the trough is increasing at a rate of 9 cubic feet per minute. Since the trough has the shape of an isosceles triangle, we can use the formula for the volume of a triangular prism:
V = (1/2) * b * h * L
Where:
V = volume in cubic feet
b = base width of the triangle (3.5 feet)
h = height of the triangle (1 foot)
L = length of the trough (6 feet)
We can express the volume "V" as a function of time "t":
V(t) = (1/2) * b * h(t) * L
To find dh/dt, the rate at which the height of the water in the trough is increasing with respect to time, we differentiate both sides of the equation with respect to time:
dV/dt = (1/2) * b * d(h(t))/dt * L
Given that dV/dt = 9 cubic feet per minute and the values of b and L are constant, we can solve for d(h(t))/dt:
9 = (1/2) * 3.5 * d(h(t))/dt * 6
Simplifying the equation:
9 = 10.5 * d(h(t))/dt
Now, we can solve for d(h(t))/dt, the rate at which the height of the water in the trough is increasing:
d(h(t))/dt = 9 / 10.5
d(h(t))/dt = 0.8571 feet per minute
Since we are interested in the rate of change of the height in inches per minute when the height is 5 inches, we need to convert the units:
0.8571 feet per minute * 12 inches per foot = 10.2857 inches per minute
Therefore, when the height of the water is 5 inches, the rate at which the height is increasing is approximately 10.2857 inches per minute.