Rebecca, Justin, and Robert are trapped in a rectangular room 8 feet deep and 10 feet tall. Two opposing walls are closing in at a rate of 1 foot per minute. If the water in the room is 2 feet deep when the moving walls are 12 feet apart, how fast is the water level rising when it reaches the top of Justin’s head, if Justin is 6 feet tall?

Question

I have all of this: x = 12-2t
If the height of the water is h, then the volume of water is
v = 8xh
dv/dt = 8h dx/dt + 8x dh/dt = 0
So now plug in your numbers to find dh/dt when h=6

I still don’t understand what numbers to plug in and how to get the final answer. PLEASE explain the WHOLE problem
Thank you!

Answer 1

when h=6, 192 = 8x*6, so x=4
8*6(-2) + 8*4 dh/dt = 0
dh/dt = 96/32 = 3 ft/min

If the width of the room is x feet, then since the two side walls are closing in at 1 ft/min each, x = 12-2t. As it turns out, you don't really need that. You just need to know that the width is shrinking by 2 ft/min, so dx/dt = -2

The volume of water does not change, but stays constant at 102 ft^3
v = length * width * height.

Answer 2

actually, the problem as stated is a bit dodgy, since you do not know the volume of the people in the room. That will affect the rate of water's rising.

Just another of those simplifying assumptions...

Answer 3

Thank you!!