Problema Solution
A trough is 16 ft long and its ends have the shape of isosceles triangles that are 4 ft across at the top and have a height of 1 ft. If the trough is being filled with water at a rate of 9 ft3/min, how fast is the water level rising when the water is 9 inches deep?
Answer provided by our tutors
We are given the rate of change of the volume is:
dV/dt = 9 ft^3/min
We need to find the rate of change of the water level: dh/dt, for h = 9 in = 9/12 ft = 3/4 ft
The volume is V = (1/2)bh16 = 8bh, where b is the length of the base of the triangle
Using similar triangles we get:
b : h = 4 : 1
b = 4h
No we can substitute this into our volume equation and implicitly differentiate:
V = 8bh
V = 8(4h)h
V = 32h^2
dV/dt = 32h dh/dt
32h dh/dt = 9
dh/dt = 9/(32h)
dh/dt = 9/(32*(3/4))
dh/dt = 0.375 ft/min
When the water level is h = 9 inches, the water is rising at the rate of 0.375 ft/min.
We are given the rate of change of the volume is:
dV/dt = 9 ft^3/min
We need to find the rate of change of the water level: dh/dt, for h = 9 in = 9/12 ft = 3/4 ft
The volume is V = (1/2)bh16 = 8bh, where b is the length of the base of the triangle
Using similar triangles we get:
b : h = 4 : 1
b = 4h
No we can substitute this into our volume equation and implicitly differentiate:
V = 8bh
V = 8(4h)h
V = 32h^2
dV/dt = 32*2h dh/dt
64h dh/dt = 9
dh/dt = 9/(64h)
dh/dt = 9/(64*(3/4))
dh/dt = 0.1875 ft/min
When the water level is h = 9 inches, the water is rising at the rate of 0.1875 ft/min.