A rectangular swimming pool is 8 m wide and 20 m long. Its bottom is a sloping plane, the depth increasing from 1 m at the shallow end to 3 m at the deep end. Water is draining out of the pool at a rate of 1 m3/min. How fast is the surface of the water falling when the depth of the water at the deep end is (a) 2.5 m? (b) 1 m?
When the water is 2.5m deep, the section draining is just a rectangular prism with area 160m^2 and height 1/2 m.
So, draining at 1m^3/min, the water level is dropping at a rate of 1/120 m/min
When the water is y m deep (y<2), the side cross-section of the pool is a right triangle with legs y and 10y.
So, the volume of the water is
v = (1/2)(y)(10y)(8) = 40y^2 m^3
dv/dt = 80y dy/dt
so, at y=1,
-1 = 80*1 dy/dt
dy/dt = -1/80 ft/min
typo alert
dy/dt at y=2.5 is -1/160, not -1/120
To find the rate at which the surface of the water is falling, we can use the related rates formula:
dV/dt = (dA/dt) * dh/dA
Where:
- dV/dt is the rate of change of volume of the water (given as 1 m^3/min)
- dA/dt is the rate of change of the surface area of the water
- dh/dA is the rate of change of depth with respect to the surface area
Step 1: Find the equation relating the depth (h) and the surface area (A) of the water.
The shape of the pool is rectangular, so the cross-sectional area at any given depth can be calculated using the formula:
A = w * h
Where:
- A is the cross-sectional area
- w is the width of the pool
- h is the depth of the water
Step 2: Find the rate of change of the surface area with respect to time (dA/dt).
We already have the values for the width (w) and the rate of change of volume (dV/dt).
dA/dt = dV/dt / dh/dA
Step 3: Find the value of dh/dA.
dh/dA is the reciprocal of the slope of the depth versus area graph at any given depth.
Step 4: Substitute the values into the formula to find the rate at which the surface of the water is falling (dA/dt).
(a) When the depth at the deep end is 2.5 m:
- Calculate the value of A at this depth using A = w * h.
- Calculate the value of dh/dA by taking the reciprocal of the slope of the depth versus area graph at this depth.
- Substitute the values into the formula to find dA/dt.
(b) When the depth at the deep end is 1 m:
- Follow the same steps as in part (a) to find dA/dt.
To find the rate at which the surface of the water is falling, we need to determine the rate at which the depth of the water is decreasing. We can use the chain rule to calculate this.
Let's start by labeling the dimensions of the pool:
Width of the pool (w) = 8 m
Length of the pool (l) = 20 m
We are given that the depth at the shallow end is 1 m, and at the deep end, it is 3 m. We need to find the rate at which the surface is falling when the depth at the deep end is (a) 2.5 m and (b) 1 m.
Using similar triangles, we can see that the depth at any point x along the length of the pool can be expressed as:
H(x) = 1 + (3 - 1) * (x/l)
Where:
H(x) is the depth at position x along the length of the pool (in meters)
x is the distance from the shallow end to the point we are interested in (in meters)
l is the length of the pool (20 m in this case)
Now we need to find the rate at which the depth is changing with respect to time (dh/dt) using the chain rule.
For part (a), where the depth at the deep end is 2.5 m:
We want to find dh/dt when H(x) = 2.5 m. We can solve the equation for x and substitute it into the equation to get dh/dt.
2.5 = 1 + (3 - 1) * (x/l)
1.5 = 2x/20
1.5 * 20 = 2x
x = 15
Substituting x = 15 into H(x) = 1 + (3 - 1) * (x/l), we obtain H(15) = 2.5.
Now let's differentiate H(x) with respect to time (t), using the chain rule:
dh/dt = dH/dt = dH/dx * dx/dt
To find dx/dt, we know that the water is draining out of the pool at a rate of 1 m^3/min. The volume of the pool can be calculated as the area of the pool's base (A) multiplied by the depth (H(x)). So, we have:
V = A * H(x)
To find dx/dt, we differentiate this with respect to time:
dV/dt = d(A * H(x))/dt
1 m^3/min = (dA/dt) * H(x) + A * (dH(x)/dt)
Now, the area A is constant since it's based on the dimensions of the pool, so dA/dt = 0. We can solve for dH/dt:
1 m^3/min = A * (dH(x)/dt)
dH(x)/dt = 1 / A
In this case, A is equal to the width of the pool:
A = w = 8 m
So, dH(x)/dt = 1 / 8 m/min.
Therefore, when H(x) = 2.5 m, dh/dt = 1 / 8 m/min.
For part (b), when the depth at the deep end is 1 m:
Following the same steps as above, we can solve for x:
1 = 1 + (3 - 1) * (x/l)
0 = 2x/20
2x = 0
x = 0
Since x = 0, we are at the shallow end of the pool. The rate at which the surface of the water is falling is 0 m/min since the depth is not changing at the shallow end.
In summary:
(a) When the depth at the deep end is 2.5 m, the surface of the water is falling at a rate of 1/8 m/min.
(b) When the depth at the deep end is 1 m, the surface of the water is not falling (0 m/min).