This question has several parts that must be completed sequentially. If you skip

Question

This question has several parts that must be completed sequentially. If you skip a part of the question, you will not receive any points for the skipped part, and you will not be able to come back to the skipped part. Tutorial Exercise A trough is 8 ft long and its ends have the shape of isosceles triangles that are 3 ft across at the top and have a height of 1 ft. If the trough is being filled with water at a rate of 12 ft3/min, how fast is the water level rising when the water is 8 inches deep?

Explanation / Answer

Solution:

We know that V=(1/2)bhl

where b is the base and represents the top of the water.

h is height... and l is length = 8 feet

we also know that there is a ratio of 1.5/1 = x/y

where x = 0.5b and y = h

because if we draw a picture of the end of the trough...

we see this relationship through the right triangles that we have

So 1.5/1 = 0.5b/h (or 3/1 = b/h)

0.5b = 1.5h

b = 3h which makes perfect sense if the height is 1' deep... then the trough is 3 feet wide... at the top...

now once again...

V = (1/2)bhl

V = 0.5(3h)h(8)

V = 12h2

differentiate both sides with respect to time...

dV/dt = 24h dh/dt

dV/dt is the rate of water coming into the tank = 12 ft3/min

dh/dt is the rate that the water is rising = ?

h is the height = 8 inches = 0.5 ft.

So we get...

12 = 24(0.5) dh/dt

12/12 = dh/dt

dh/dt = 1 ft/min.

Thus the water level is rising at the rate of 4/5 ft/min when the water is 8 inches deep

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