SOLUTION: A trough is 75cm long and its ends in the form of an isosceles triangle have an altitude of 20cm and a base of 30cm. Water is flowing into the trough at the rate of 100cm3/sec. Fin

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Question 1193214: A trough is 75cm long and its ends in the form of an isosceles triangle have an altitude of 20cm and a base of 30cm. Water is flowing into the trough at the rate of 100cm3/sec. Find the rate at which water level is rising when the water is 10cm deep?
Answer by greenestamps(13198) About Me  (Show Source):
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The trough can be seen as a triangular prism with a cross section of the trough as the base and the length of the trough as the height. The volume of the triangular prism is base times height.

The height (length of the trough) is fixed at 75.

As water is poured into the trough, the cross section of the water is always an isosceles triangle whose base and height are always in the same ratio as the cross section of the whole trough -- 30:20 = 3:2.

We are given dV/dt; we are to determine dh/dt (where h is the height of the isosceles triangular cross section of the water -- not the "height" (length) of the trough.)

V=%2875%29%28%281%2F2%29%28base%29%28height%29%29

The base of the cross section is always 3/2 of the height:

V=%2875%29%28%281%2F2%29%28%283%2F2%29h%29%28h%29%29=%28225%2F4%29h%5E2

dV%2Fdh=%28225%2F2%29h

dV%2Fdt=%28dV%2Fdh%29%28dh%2Fdt%29
100=%28%28225%2F2%29h%29%28dh%2Fdt%29
dh%2Fdt=200%2F%28225h%29=8%2F%289h%29

When the depth (height) of the water is 10cm,

dh%2Fdt=8%2F90=4%2F45

ANSWER: When the depth of the water in the trough is 10cm, the water level is rising at the rate of 4%2F45 cm/sec