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?
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!
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.
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...
Thank you!!
To solve this problem, we can use related rates, which relates the rates at which two quantities change with respect to time. Here's a step-by-step explanation:
1. Let's first define the variables:
- x represents the distance between the two moving walls.
- h represents the height of the water in the room.
2. According to the problem, x = 12 - 2t, where t represents the time in minutes. This equation represents the distance between the two walls as a function of time.
3. Now let's determine the volume of the water in the room. The volume, V, can be calculated by multiplying the length, width, and height:
V = 8 * x * h
Since the width and length of the room are constant at 8 feet and the height is h, we only need to consider the variables x and h.
4. We will differentiate both sides of the equation with respect to time (t):
dV/dt = d(8xh)/dt
5. Using the product rule of differentiation, we can find the derivative of V with respect to t:
dV/dt = 8h * dx/dt + 8x * dh/dt
6. According to the problem statement, the walls are closing in at a rate of 1 foot per minute. Therefore, dx/dt is -1 since x decreases over time:
dx/dt = -1
7. We want to find dh/dt, the rate at which the water level is rising when it reaches the top of Justin's head. At that point, h = 6.
dh/dt = ?
8. We can substitute the known values into our equation:
0 = 8h * (-1) + 8x * dh/dt
9. Since we want to find dh/dt, we can solve the equation for it:
8h = -8x * dh/dt
dh/dt = -8h / (8x)
10. We know that h = 6 and x = 12 - 2t, so we can substitute these values into the equation:
dh/dt = -8(6) / [8(12 - 2t)]
11. Simplifying the equation further:
dh/dt = -48 / (96 - 16t)
dh/dt = -1 / (2 - t/6)
That's the final answer! The rate at which the water level is rising when it reaches the top of Justin's head is given by the expression dh/dt = -1 / (2 - t/6), where t represents the time in minutes.