a horizontal trough is 16 meters long and its ends are isosceles trapezoids with an altitude of 4 meters, an upper base of 6 meters and lower base of 4 meters. water is being poured in the trough at a rate of 10 cubic meters per minute.
a) how fast is the water level rising when the water is 2 meters deep?
b) if the water level is decreasing at a rate of 25 centimeters per minute when the water is 3 meters deep, at what rate is water being drawn from the trough?
When the height of the water is h, the surface of the water is 4 + h/2 wide
That means the cross-section of the water is a trapezoid with area 4(4 + h/2)/2 * h = 8h+h^2
Thus the volume of the water is
v = 16(8h+h^2)
dv/dt = 16(8+2h) dh/dt = 32(h+4) dh/dt
Now plug in your numbers to find dh/dt or dv/dt
a) Well, let's dive into this problem! First, we need to find the area of the trapezoidal cross-section of the trough at a water depth of 2 meters.
The formula for the area of a trapezoid is:
Area = (1/2) × (upper base + lower base) × altitude
So, plugging in the values, we get:
Area = (1/2) × (6 + 4) × 4 = 20 square meters
The trough is 16 meters long, so the volume of water at a depth of 2 meters is:
Volume = area × length = 20 × 16 = 320 cubic meters
Now, since water is being poured into the trough at a rate of 10 cubic meters per minute, we can find how fast the water level is rising when the water is 2 meters deep by finding the derivative of the volume with respect to time:
dVolume/dt = 10
But we want to find dH/dt, where H is the water depth. So we can use a similar triangle relationship:
dH/dt = (dH/dV) × (dV/dt)
We know dV/dt = 10, and since H = 2 at this point:
dH/dV = 1/(dV/dH) = 1/(area × length) = 1/(20 × 16) = 1/320
Substituting these values into the equation, we get:
dH/dt = (1/320) × 10 = 0.03125 meters per minute
So, the water level is rising at a rate of approximately 0.03125 meters per minute when the water is 2 meters deep.
b) Alright, let's get to the second part of the riddle! If the water level is decreasing at a rate of 25 centimeters per minute when the water is 3 meters deep, we need to find the rate at which the water is being drawn from the trough.
Since H = 3 at this point, let's first convert the given rate of 25 centimeters per minute to meters per minute:
25 centimeters = 0.25 meters
Now, similarly to part a), we know that:
dH/dt = (dH/dV) × (dV/dt)
We are given that dH/dt = -0.25, and we need to find dV/dt.
Using the same formula as before, we know dH/dV = 1/(20 × 16) = 1/320
Rearranging the equation, we get:
dV/dt = (dH/dt) × (dV/dH) = (-0.25) × 320 = -80 cubic meters per minute
Therefore, the water is being drawn from the trough at a rate of 80 cubic meters per minute. Quite a thirsty trough, I must say!
To solve this problem, we'll need to use similar triangles and the formula for the volume of a trapezoidal prism. Let's break it down step by step.
Step 1: Find the width of the trough at a given water level.
We'll use the similar triangles formed by the water level and the trapezoidal ends of the trough. Since the altitude of the trapezoids is 4 meters, and the distance from the water level to the top base is a constant 6 - 4 = 2 meters, we can set up the following proportion:
(Width of trough)/(Distance from water level to top base) = (Width of trapezoid)/(Altitude of trapezoid)
Let's represent the width of the trough at a given water level as x meters. Plugging in the values, we have:
x / 2 = 6 / 4
Simplifying, we find:
x = 3
Step 2: Find the area of the trapezoid at a given water level.
Since the bases of the trapezoid are 4 meters and 6 meters, and the altitude is 4 meters, we can use the formula for the area of a trapezoid:
Area = ((b1 + b2) * h) / 2 = ((4 + 6) * 4) / 2 = 20 square meters
Step 3: Find the volume of the water in the trough at the given water level.
The volume of a trapezoidal prism is equal to the area of the base times the length. In this case, the length is 16 meters. So, the volume of the water in the trough at the given water level is:
Volume = Area * Length = 20 * 16 = 320 cubic meters
Now, let's solve the questions.
a) How fast is the water level rising when the water is 2 meters deep?
To find the rate at which the water level is rising, we need to find the derivative of the volume with respect to time. Let's call the height of the water level h(t), where t represents time. We are given that dh/dt (the rate at which the water level is rising) is what we need to find when h(t) = 2 meters.
To find dh/dt at h(t) = 2 meters, we need to find dh/dt as a function of h. We can do this by differentiating the volume equation with respect to time:
dh/dt = (dV/dt) / (dV/dh)
The rate at which the volume is changing is given as 10 cubic meters per minute, so dV/dt = 10.
The rate at which the area changes with respect to the height can be found by differentiating the area equation with respect to height:
dA/dh = (d/dh)(((b1 + b2) * h) / 2) = (1/2)(b1 + b2) = (1/2)(4 + 6) = 5
(Note: We can treat b1 and b2 as constants since they remain constant as the water level rises)
Now, let's plug in the values into the equation for dh/dt:
dh/dt = (10) / (5) = 2 meters per minute
Therefore, the water level is rising at a rate of 2 meters per minute when the water is 2 meters deep.
b) If the water level is decreasing at a rate of 25 centimeters per minute when the water is 3 meters deep, at what rate is water being drawn from the trough?
Since the height is given in centimeters, we need to convert it to meters: 25 centimeters = 0.25 meters.
We need to find dV/dt (the rate at which water is being drawn from the trough) when h(t) = 3 meters.
Following the same approach as above, we have:
dh/dt = (dV/dt) / (dV/dh)
We are given that dh/dt = -0.25 (since the water level is decreasing), and we need to find dV/dt when h(t) = 3.
Again, dV/dt = 10 (since water is being poured into the trough at a constant rate of 10 cubic meters per minute).
We already determined dV/dh in the previous question, which is 5 (since the area changes with respect to height at a constant rate of 5 square meters per meter).
Now, let's plug in the values into the equation for dV/dt:
-0.25 = (10) / (5)
-0.25 = 2 cubic meters per minute
Therefore, water is being drawn from the trough at a rate of 2 cubic meters per minute when the water level is 3 meters deep.
To solve this problem, we need to apply the concept of related rates. Let's start by understanding the geometric properties of the trough.
First, let's draw a diagram of the trough. We can represent the trough as a rectangular-shaped box with isosceles trapezoidal ends.
The trough's length is 16 meters. This means that the width of the rectangular base is also 16 meters.
The ends of the trough are isosceles trapezoids with an altitude of 4 meters. The upper base of each trapezoid is 6 meters, and the lower base is 4 meters.
Now let's move on to solving the problems.
a) To find how fast the water level is rising when it is 2 meters deep, we need to find the rate of change of the water level with respect to time.
Let's denote the depth of the water as "h" (in meters) and the time as "t" (in minutes).
We are given that the water is poured into the trough at a rate of 10 cubic meters per minute. This means that the rate at which the volume of water in the trough is changing with respect to time is 10 cubic meters per minute.
The volume of the water in the trough is equal to the cross-sectional area of the trough multiplied by the depth of the water. Since the cross-section of the trough is a trapezoid, we can find its area using the formula for the area of a trapezoid.
The formula for the area of a trapezoid is: Area = (base1 + base2) * height / 2
In this case, base1 is the upper base of the trapezoid (6 meters), base2 is the lower base of the trapezoid (4 meters), and the height is the depth of the water (h meters).
So the cross-sectional area of the trough is: Area = (6 + 4) * h / 2 = 5h.
Now we can express the volume of the water in the trough as a function of the depth of the water: Volume = 16 * 5h = 80h.
Since we know that the volume of water is changing at a rate of 10 cubic meters per minute, we can write: dV/dt = 10.
Differentiating the volume equation with respect to time gives us: dV/dt = d(80h)/dt = 80 * dh/dt.
Substituting the given value for dV/dt, we get: 80 * dh/dt = 10.
Now we can solve for dh/dt, which is the rate at which the water level is rising when the water is 2 meters deep.
Given: h = 2. To find dh/dt, we rearrange the equation: dh/dt = 10 / 80 = 1/8 meters per minute.
Therefore, the water level is rising at a rate of 1/8 meters per minute when the water is 2 meters deep.
b) To find the rate at which water is being drawn from the trough when the water level is 3 meters deep, we need to find the rate of change of the volume of water with respect to time.
We are given that the water level is decreasing at a rate of 25 centimeters per minute when the water is 3 meters deep. The rate of change of the water level with respect to time is given as -25 centimeters per minute (negative because the water level is decreasing).
Using the same reasoning as in part (a), the volume of water in the trough is given by the equation: Volume = 80h.
Differentiating the volume equation with respect to time gives us: dV/dt = d(80h)/dt = 80 * dh/dt.
Substituting the given value for dh/dt, we get: 80 * (-0.25) = dV/dt.
Therefore, the rate at which water is being drawn from the trough is -20 cubic meters per minute.
Note: The negative sign indicates that water is being drawn or removed from the trough.
In summary:
a) The water level is rising at a rate of 1/8 meters per minute when the water is 2 meters deep.
b) Water is being drawn from the trough at a rate of 20 cubic meters per minute when the water level is 3 meters deep.