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?
Please show and explain ALL work. Thank you.
so, let the width of the room be x feet. So, at time t,
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.
To solve this problem, we can use similar triangles and apply the chain rule of differentiation.
Let's first set up a diagram to visualize the situation:
```
_______________________
| |
| |
B | |
| |
|-------- ----------|
A Justin C
```
A and C represent the two closing walls, while B represents the water level. Justin's height is 6 feet, and the room is 8 feet deep. The distance between A and C is decreasing at a rate of 1 foot per minute.
To find the rate at which the water level is rising when it reaches the top of Justin's head, we need to find the rate at which the distance BC is changing with respect to time.
Let's assign variables to the quantities in the problem:
- Let x be the distance BC (the depth of the water).
- Let t be the time in minutes since the walls start closing.
- Let dA/dt be the rate at which the distance between the walls is changing (given as -1 ft/min).
- Let dx/dt be the rate at which the depth of the water is changing (what we want to find).
From the given information, we have:
- x = 2 feet (when the water is 2 feet deep).
- dA/dt = -1 ft/min.
Using similar triangles, we can relate the height of Justin's head to the depth of the water:
Justin's height/depth of water = 6 feet/x.
Taking the derivative of both sides with respect to time (t), we get:
(d/dt)(Justin's height/depth of water) = (d/dt)(6 feet/x).
To find (d/dt)(6 feet/x), we need to apply the quotient rule of differentiation:
(d/dt)(6 feet/x) = (6 * (d/dt)(1/x) - (1/x^2) * (d/dt)(x)).
Now, let's calculate each term separately:
1) (d/dt)(1/x):
Using the chain rule, we have:
(d/dt)(1/x) = -(1/x^2) * (dx/dt).
2) (d/dt)(x):
Given that x = 2 feet (constant), the derivative is zero:
(d/dt)(x) = 0.
Plugging these values back into the equation, we have:
(d/dt)(6 feet/x) = (6 * (-(1/x^2) * (dx/dt)) - (1/x^2) * 0).
Simplifying, we get:
(d/dt)(6 feet/x) = -(6/x^2) * (dx/dt).
Now, let's substitute the known values:
Justin's height/depth of water = 6 feet/x,
(dx/dt) = the rate at which the depth of the water is changing (what we want to find), and
(dA/dt) = -1 ft/min.
Plugging these values back into the equation, we get:
(d/dt)(6 feet/x) = -(6/x^2) * (dx/dt).
-1 = -(6/x^2) * (dx/dt).
Solving for (dx/dt), we get:
(dx/dt) = (-1 * x^2) / 6.
Since we want to find the rate at which the water level is rising when it reaches the top of Justin's head, we need to find (dx/dt) when x = 6 feet (depth of the water at Justin's height).
Substituting x = 6 into the equation, we have:
(dx/dt) = (-1 * 6^2) / 6.
(dx/dt) = -6.
Therefore, the rate at which the water level is rising when it reaches the top of Justin's head, dx/dt, is -6 feet per minute.
Note: The negative sign indicates that the water level is actually decreasing as it reaches the top of Justin's head.