a swimming pool is 5m wide, 10m long, 1m deep at the shallow end, and 3 m deep at its deepest point. a cross section is shown in the figure. if the pool is being filled at a rate of 0.1 m3/min, how fast is the water level rising when the depth at the deepest point is 1m?
Well, I'm not a swimming expert, but I'll give it a splash! Let's see if I can dive into this problem.
First, let's find the volume of the swimming pool. We can use the formula for the volume of a rectangular prism: length × width × height. So, the volume is 5m × 10m × (3m - 1m) = 50m³.
Now, we need to find the rate at which the water level is rising when the depth at the deepest point is 1m. To do this, we can use the related rates formula: rate of change of volume = area of cross-section × rate of change of depth.
The cross-sectional area can be found by multiplying the width and the average depth. The average depth at a depth of 1m is (1m + 3m)/2 = 2m.
So, the cross-sectional area is 5m × 2m = 10m².
Now, let's plug in the values we have:
0.1m³/min = 10m² × rate of change of depth.
Next, let's solve for the rate of change of depth:
rate of change of depth = 0.1m³/min ÷ 10m² = 0.01m/min.
So, the water level is rising at a rate of 0.01m/min when the depth at the deepest point is 1m.
And there you have it! The water level is making quite a splash with a rise of 0.01m/min. I hope that sinks in!
To find the rate at which the water level is rising, we need to use related rates. Let's consider the volume of the pool as a function of time.
The volume of a rectangular swimming pool can be calculated using the formula:
Volume = length × width × depth
For the given pool, the width is 5m, the length is 10m, and the depth varies linearly from 1m at the shallow end to 3m at the deepest point.
Since the depth is a linear function, we can write it as:
depth = mx + b,
where x represents the distance from the shallow end and b is the initial depth (1m in this case). To find the value of m, we can use the information given.
At the deepest point (x = 10), the depth is 3, so we have:
3 = 10m + 1.
Solving this equation, we get:
m = (3 - 1) / 10 = 2/10 = 0.2.
Therefore, the equation for the depth as a function of x is:
depth = 0.2x + 1.
Now, we can express the volume of the pool as a function of x:
Volume = length × width × depth = 10 × 5 × (0.2x + 1).
Differentiating both sides of this equation with respect to time (t) gives:
dV/dt = 10 × 5 × d/dt (0.2x + 1).
However, we are given that the pool is being filled at a rate of 0.1 m^3/min, so dV/dt = 0.1.
Now, we need to find d/dt (0.2x + 1). Since the only variable is x, we need to express x in terms of time.
Using the relation between x and the depth (0.2x + 1), we can solve for x:
0.2x + 1 = 1.
Solving this equation, we get: x = 0.
Therefore, when the depth at the deepest point is 1m, x = 0. Plugging this into the equation d/dt (0.2x + 1), we find:
d/dt (0.2x + 1) = d/dt (0.2 × 0 + 1) = d/dt (1) = 0.
Hence, the rate at which the water level is rising when the depth at the deepest point is 1m is 0 m/min.
To find how fast the water level is rising when the depth at the deepest point is 1m, we can use the concept of related rates.
The first step is to relate the variables involved in the problem. We need to find a relationship between the volume of the pool and the rate at which it is being filled.
The volume of the pool can be calculated by multiplying the width, length, and average depth:
Volume = Width × Length × Average Depth
The average depth can be calculated by finding the average of the shallow end depth and the deepest point depth:
Average Depth = (1m + 3m) / 2 = 2m
Substituting the given values, we have:
Volume = 5m × 10m × 2m = 100m³
Now, let's differentiate both sides of the equation with respect to time (t) to find the rate at which the volume is changing:
dV/dt = d/dt (Width × Length × Average Depth)
The rate at which the volume is changing (dV/dt) is given as 0.1 m³/min.
0.1 m³/min = d/dt (5m × 10m × 2m)
Now, let's solve for d/dt, the rate at which the water level is rising when the depth at the deepest point is 1m:
d/dt = 0.1 m³/min ÷ (5m × 10m × 2m)
d/dt = 0.1 m³/min ÷ (100m²)
d/dt = 0.001 m/min
Therefore, the water level is rising at a rate of 0.001 m/min when the depth at the deepest point is 1m.
When the water is y meters deep, the surface has a length of 5y and a width of 5. So, its volume is
v(y) = 25/2 y^2
dv/dt = 25y dy/dt
Now plug in y=1 and dv/dt=0.1 to find dy/dt