Question 1193214
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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.<br>
The height (length of the trough) is fixed at 75.<br>
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.<br>
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.)<br>
{{{V=(75)((1/2)(base)(height))}}}<br>
The base of the cross section is always 3/2 of the height:<br>
{{{V=(75)((1/2)((3/2)h)(h))=(225/4)h^2}}}<br>
{{{dV/dh=(225/2)h}}}<br>
{{{dV/dt=(dV/dh)(dh/dt)}}}
{{{100=((225/2)h)(dh/dt)}}}
{{{dh/dt=200/(225h)=8/(9h)}}}<br>
When the depth (height) of the water is 10cm,<br>
{{{dh/dt=8/90=4/45}}}<br>
ANSWER: When the depth of the water in the trough is 10cm, the water level is rising at the rate of {{{4/45}}} cm/sec<br>