How fast is the water level rising when the water is 6 inches deep?
A trough is 14 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 filled with water at a rate of 14 ft^3 / min, how fast is the water level rising when the water is 6 inches deep?
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Filed under: Water Filter System
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Related Rates
GIVEN dV/dt = 14
t=4 and a=1 and h=1/2 (6in=.5ft same units)
LOOKING for dh/dt
V = 14 (1/2 base * height ) = 7 base * height.
Set up similar triangles to get in terms of same variabble by the simliar triangles we see that 1:4 is h:b ===> b = 4h
now V = 7(4h)(h) = 28 h^2
dV/dt = 56 h (dh/dt)…plug in h=1/2 and dV/dt = 14 to solve for dh/dt
14 = 56 * 1/2 * dh/dt
dh/dt = 14/28 = 1/2
Water is rising at a rate of 1/2 foot per minute
A related rates problem, not much different from one last night.
Your trough has a volume of 0.5 base x height x length. The trick is express base in terms of height, which, from your statement, is 4h = b. Then we have,
V = 28 h^2. Then, we can determine dV/dh as
dV/dh = 56h. We know dV/dt=14, and we can convert dV/dh to dV/dt x dt/dh= 56h.
Then 14 dt/dh = 56h, or dt/dh = 56/14 h= 4h.
We really want dh/dt, which is the reciprocal, or
dh/dt = 1/(4h)
For h of 0.5 ft, we have 0.5 ft/minute
Let
h = height water in trough
V = volume water in trough
Given
dV/dt = 14 ft³ / min
Find
dh/dt when h = 1/2 ft
We have
V = 14*(1/2)(4h)h = 28h²
dV/dh = 56h
dh/dt = (dV/dt) / (dV/dh) = 14 / (56h) = 1 / (4h)
dh/dt = 1/[4*(1/2)] = 1/2 ft/min