Thread: [SOLVED] Related rates: trapezoidal water trough

1. [SOLVED] Related rates: trapezoidal water trough

A water trough is 2 ft deep and 10 ft long. It has a trapezoidal cross section with base lengths 2 and 5 feet.
• Find a relationship between the volume of water in the tank at any given time and the depth of the water at that time.

Where V= volume, L = length, D = depth of water
V = L [(b1 + b2)/2] D
V = 35D

• If water enters the trough at a rate of 10 ft^3/min, how fast is the water level rising (to the nearest 0.5 in./min) when the water is 1 ft deep.

V' = 35

Obviously, that doesn't work. Any advice?

2. We want $\frac{dh}{dt}$ when h=1, given $\frac{dV}{dt}=10$.

Let b=length of top of cross-section of the water.

h=depth of water.

By the area of a trapezoid, we have the volume of the trough:

V=h((b+2)/2)(10)

By similar triangles, think of the two triangles on the ends as one isosceles triangle.

(b-2)/h=3/1

b=3h+2

V=5(3h^2+4h)

Now, can you finish up by differentiating impicitly wrt time?.

Enter in your givens, h, dV/dt, and solve for dh/dt.

3. Originally Posted by sinewave85
A water trough is 2 ft deep and 10 ft long. It has a trapezoidal cross section with base lengths 2 and 5 feet.
• Find a relationship between the volume of water in the tank at any given time and the depth of the water at that time.

Where V= volume, L = length, D = depth of water
V = L [(b1 + b2)/2] D
V = 35D
• If water enters the trough at a rate of 10 ft^3/min, how fast is the water level rising (to the nearest 0.5 in./min) when the water is 1 ft deep.

V' = 35
Obviously, that doesn't work. Any advice?
Similar triangles

EDIT: oops, missed Galactus's post

Sorry

4. Originally Posted by galactus
We want $\frac{dh}{dt}$ when h=1, given $\frac{dV}{dt}=10$.

Let b=length of top of cross-section of the water.

h=depth of water.

By the area of a trapezoid, we have the volume of the trough:

V=h((b+2)/2)(10)

By similar triangles, think of the two triangles on the ends as one isosceles triangle.

(b-2)/h=3/1

b=3h+2

V=5(3h^2+4h)

Now, can you finish up by differentiating impicitly wrt time?.

Enter in your givens, h, dV/dt, and solve for dh/dt.

Do'h -- I was using top of the trough instead of the top of the water for the second b! No wonder I was a variable short of an equation, so to speak. Thanks so much for the help!

5. i have a question in galactus post, in similar triangles, is it really really (b-2)/h=3/1 or should it be (b-2)/h=3/2 ?? please rep.. im really confused... tnx

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the ends of a horizontal water trough

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