How fast is the water level rising when the water is 6 inches deep?

Question

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?

Answer 1

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

Answer 2

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

Answer 3

Firstly, find the volume as a function of the height (x) of the water level; drawing a diagram makes it easy to deduce: V = 10 * 1/2 * x * 2x = 10x² ft³ You know that dV/dt=13 ft³/s, and V=10x². So: dV/dx = 20x And you ultimately want to find dx/dt = dx/dV * dV/dt (by the chain rule). So: dx/dt = 1/(20x) * 13 = 13/(20x) So when x=6 in=0.5 ft: dx/dt = 13/(20 * 0.5) = 1.3 ft/s

Answer 4

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