A water trough is 7 m long and has a cross-section in the shape of an isosceles trapezoid that is 40 cm wide at the bottom, 100 cm wide at the top, and has height 60 cm. If the trough is being filled with water at the rate of 0.3 m3/min how fast is the water level rising when the water is 40 cm deep?
Draw a diagram. When the water has depth y, its surface has width 40+(y/60)(100-40) = y+40
So, the volume of water when depth is y is
v = y(40 + y+40)/2 * 7 = 280y + 7/2 y^2
so,
dv/dt = 280 + 7y dy/dt
Now just plug in your numbers
Yeah, I know the units for v are in m-cm^2, but the factor of 100 is just a linear scale. You can include it if you want.
To find the rate at which the water level is rising, we will differentiate the volume of the water with respect to time.
Let's start by finding the volume of water in the trough as a function of its height.
1. Let's denote the height of the water by 'h' (in meters) and the length of the trough by 'x' (in meters).
2. Since we have an isosceles trapezoid shape, we can see that the area of the cross-section at height 'h' is:
A = (1/2)(b1 + b2)(h)
Where b1 is the bottom width of the trapezoid, b2 is the top width of the trapezoid, and h is the height.
Substituting the given values:
A = (1/2)(0.4 + 1)(h) = (0.7)(h)
3. The volume of water in the trough at height 'h' is given by:
V = A * x = (0.7h)(x)
4. Now, we need to differentiate the volume V with respect to time (t) to find the rate of change of volume with time:
dV/dt = d/dt((0.7h)(x))
5. We are given that the trough is being filled at a rate of 0.3 m^3/min, so the rate of change of volume with respect to time is 0.3 m^3/min.
Therefore, we have:
dV/dt = 0.3
6. Finally, we need to find the rate at which the water level is rising (dh/dt) when the water is 40 cm deep, which is equivalent to h = 0.4 m.
Plugging in the values:
dV/dt = (0.7)(0.4)(dx/dt)
Rearranging the equation:
dx/dt = (dV/dt) / ((0.7)(0.4))
Plugging in the given value of dV/dt:
dx/dt = 0.3 / ((0.7)(0.4))
Evaluating the expression:
dx/dt = approximately 1.07 m/min
Therefore, the water level is rising at a rate of approximately 1.07 m/min when the water is 40 cm deep.
To find the rate at which the water level is rising, we need to use related rates. Let's start by identifying the known values and what we're looking for:
Given:
- Length of the trough (l) = 7 m
- Bottom width of the trapezoid (b1) = 40 cm
- Top width of the trapezoid (b2) = 100 cm
- Height of the trapezoid (h) = 60 cm
- Rate of water filling the trough (dV/dt) = 0.3 m^3/min
- Depth of the water (y) = 40 cm
Unknown:
- Rate at which the water level is rising (dh/dt)
Now let's use the formula for the volume of a trapezoidal trough to relate the variables:
V = (1/2) * (b1 + b2) * h * l
Next, differentiate both sides of the equation with respect to time (t) using implicit differentiation:
dV/dt = (1/2) * [(db1/dt + db2/dt) * h * l + (b1 + b2) * dh/dt + h * (dl/dt)]
Since the trough's dimensions (b1, b2, and l) are constant, their rates of change with respect to time (t) are all zero. Hence, the equation simplifies to:
dV/dt = (1/2) * (b1 + b2) * dh/dt
Now we can substitute the given values:
0.3 m^3/min = (1/2) * (40 cm + 100 cm) * dh/dt
Simplify the equation:
0.3 m^3/min = (1/2) * 140 cm * dh/dt
Now, let's convert the units to be consistent:
0.3 m^3/min = (1/2) * 1.4 m * (dh/dt) * 100 cm/min
dh/dt = (0.3 m^3/min) / [(1/2) * 1.4 m * 100 cm/min]
Now, divide to calculate the value for dh/dt:
dh/dt = 0.3 m^3/min / (0.7 m * 100 cm/min)
Simplify the units:
dh/dt = 0.3 m^2/ min / (0.7 m * 100 * 10^-2 min)
= 0.3 m^2/ min / (0.7 m * 10 cm/min)
= 0.3 m^2/ min / 7 m * 10 cm/min
= 0.3 / 7 m/cm
= 0.0429 m/min
Thus, the water level is rising at a rate of approximately 0.0429 meters per minute when the water is 40 cm deep.