Here are my answers to the questions vs the expected solutions:

A trough is 10 ft long with ends the shape of isosceles triangles that are 3 feet across the top and have a height of 1 ft. If the trough is being filled with water at a rate of 12 , how fast is the water level rising in when the water is 6 inches deep? Give an exact answer with no decimals.

My answer = 6/5
Their answer = 4/5

Relevant equations: V = (1/2) bh*L
L = 10, dV/dt = 12, 6 " = 1/2'


b = h / 1'
V = (h^2/1') * 10
dV/dt = 10/1 (2h) (dh/dt)

We sub 6" for h below because it is the height at which we want to know the rate of water level increase. This is 1/2'

12 = (10/1) (2 * 1/2') (dh/dt)

12/10 = dh/dt
6/5 = dh/dt

Apparently the correct answer is 4/5. Any thoughts on this would be GREATLY appreciated!!

As the trough fills up, the cross-section of the water is a similar (same-proportions) triangle to the cross-section of the trough. So the width of the water, w, is related to the height of the water in the V-shaped trough like this:

w = (3 feet/1 foot) h = 3 h

And the volume of the rising water, V, is the length of the trough, L, times the cross-sectional area of the water, A, which is the area of an (inverted) triange with base w and height h. But we can simplify this since we have a formula for w.

V = L × A = L × ½ ( h × w ) = (3/2)×L×h²

Now from the chain rule, we know the rate of change of the volume of water with respect to time, dV/dt, is the rate of change of Volume as a function of the height of the water, dV/dh, times the rate of change of the height of the water with respect to time, dh/dt.

But since we have V = (3/2)×L×h², then we know what dV/dh is at any height of water.

dV/dt =(dV/dh)×(dh/dt) = 3Lh dh/dt

So we can solve for dh/dt.

dh/dt = 1/(3Lh) dV/dt

And plug in the variables from the values given in the problem. (Note that the rate is not specified as cubit-feet per second, or per minute or per hour, so I have a place-holder for the time units.)

L = 10 feet
h = 6 inches × (1 foot/12 inches) = ½ foot
dV/dt = 12 feet³/TIME

dh/dt = 1/(3 × 10 feet × ½ foot) 12 feet³/TIME= (12/15) feet/TIME= 4/5 feet/TIME

It looks like you failed to correctly setup the volume of the water in the trough.
You write "b = h / 1'" which makes no sense from the problem or even in the units of b (my w) and h. b = (3'/1') h is correct in units and for this problem.

You write "V = (h^2/1') * 10" bu this does not follow from the formula: "V = (1/2) bh*L" -- you left out the (1/2) and your formula for b is wrong and the units have been left off of the 10 feet. V should be a number of cubic feet, but is not as you wrote it.

Thank you so much,

Knowing this I was able to re-create the problem and the work checked out!