A water trough is 8 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 0.1 m3/min how fast is the water level rising when the water is 30 cm deep?
Well, if the water level is rising, it's safe to say things are getting pretty wet in there. But let's not dive right into the water just yet, and approach this problem step by step.
First, let's convert all the measurements to meters, just to avoid any confusion later on. The bottom width of the trapezoid is 0.4 m, the top width is 1 m, and the height is 0.6 m. So far so good?
Now, we need to find the volume of the water in the trough. Since the trough is being filled at a rate of 0.1 m³/min, we can express the volume as a function of time: V(t) = 0.1t.
Next, we need to find an expression for the area of the cross section of the water in terms of the water level. Can't have an isosceles trapezoid without a little geometry, right?
Let's denote the water level as h (in meters). The area of a trapezoid is given by the formula A = (a + b)h/2, where a and b are the lengths of the parallel sides. In this case, a = 0.4 m, b = 1 m, and h = 0.3 m.
Plugging in these values, we get A(h) = (0.4 + 1) * 0.3 / 2 = 0.45 m².
Now it's time to put it all together. We know that the volume of water is increasing with time, so we can differentiate V(t) with respect to time to find the rate at which the water level is rising.
dV/dt = d(V(t))/dt = d(0.1t)/dt = 0.1 m³/min.
We also know that the volume is related to the area and height by the equation V = A(h) * h.
Differentiating V with respect to t, we get dV/dt = d(A(h) * h)/dt.
Using the product rule, we have dV/dt = A(h) * dh/dt + h * d(A(h))/dt.
Since we're looking for dh/dt (the rate at which the water level is rising), we can solve for it:
dh/dt = (dV/dt - h * d(A(h))/dt) / A(h).
Plugging in the previously determined values, we get:
dh/dt = (0.1 - 0.3 * 0) / 0.45 = 0.1/0.45 ≈ 0.2222 m/min.
So, the water level is rising at a rate of approximately 0.2222 meters per minute when the water is 30 cm deep. And that's some deep water, my friend!
To find how fast the water level is rising, we need to find the rate of change of the volume of the water with respect to time.
Let's start by calculating the volume of water in the trough. Since the cross-section of the trough is in the shape of an isosceles trapezoid, we can use the formula for the area of a trapezoid to find the cross-sectional area at a certain water level. The formula for the area of a trapezoid is:
Area = (a + b) * h / 2
where 'a' and 'b' are the lengths of the parallel sides of the trapezoid, and 'h' is the height of the trapezoid.
In this case, a = 0.4 m (since the trough width is given in centimeters, we need to convert it to meters), b = 1 m (since the top width is also given in centimeters and needs to be converted to meters), and h is the depth of the water.
Using the formula:
Area = (0.4 + 1) * h / 2 = 0.7h (m^2)
To find the volume of water in the trough, we multiply the cross-sectional area by the length of the trough:
Volume = 0.7h * 8 = 5.6h (m^3)
Now, we know that the rate of change of the volume of water with respect to time is 0.1 m^3/min. We can differentiate the volume equation with respect to time to find the rate at which the volume is changing:
dV/dt = 5.6 * dh/dt
where dV/dt is the rate of change in volume, and dh/dt is the rate of change in water level.
We are given that the rate of change of the volume is 0.1 m^3/min, so we can substitute that into the equation:
0.1 = 5.6 * dh/dt
Finally, let's solve for dh/dt:
dh/dt = 0.1 / 5.6 = 0.0179 m/min
Therefore, the water level is rising at a rate of approximately 0.0179 m/min when the water is 30 cm deep.
To find the rate at which the water level is rising, we need to first determine the volume V of the water in the trough as a function of the water depth h. Once we have this volume function, we can differentiate it and evaluate the derivative dV/dt with respect to time t to get the rate of change of volume, which will tell us how fast the water level is rising.
To find the volume of the water, we need to determine the area of the cross-section at each height h and integrate it with respect to h.
Let's start by finding the area of the cross-section at a general height h in the trapezoid.
The formula to find the area of a trapezoid is:
Area = (a + b) * h / 2
where a and b are the lengths of the parallel sides and h is the height.
In this case, the bottom width of the trapezoid is 40 cm and the top width is 100 cm. We need to convert these to meters:
40 cm = 0.4 m
100 cm = 1 m
Now, let's denote the current height of the water as h. For the given problem, we are interested in finding the rate at which the water level is rising when the water has a depth of 30 cm, which means h = 0.3 m.
To find the volume V of the water, we integrate the area with respect to h from 0 to the given height h:
V = ∫ [(a + [(b - a) / H] * h) * h / 2] dh
where H is the total height of the trapezoid (60 cm or 0.6 m).
Simplifying this expression gives:
V = ∫ [(0.4 + [(1 - 0.4) / 0.6] * h) * h / 2] dh
V = ∫ [(0.4 + (0.6/0.6) * h) * h / 2] dh
V = ∫ [(0.4 + h) * h / 2] dh
V = ∫ [(0.2h + h² / 2)] dh
Now, we can integrate this expression:
V = 0.1h² + (h³ / 6) + C
where C is the constant of integration.
Next, we differentiate this volume function with respect to time t using the chain rule:
dV/dt = (dV/dh) * (dh/dt)
We are given that the rate of change of volume dV/dt is 0.1 m³/min.
Substituting the values we know:
0.1 = (0.2h + h² / 2) * (dh / dt)
We can solve this equation for dh/dt to find the rate at which the water level is rising:
dh/dt = 0.1 / [(0.2h + h² / 2)]
Now, we can plug in the known value for h (0.3 m) to find the rate at which the water level is rising when the water depth is 30 cm:
dh/dt = 0.1 / [(0.2*0.3 + 0.3² / 2)]
dh/dt = 0.1 / (0.06 + 0.045)
dh/dt = 0.1 / 0.105
dh/dt ≈ 0.952 m/min
Therefore, when the water is 30 cm deep, the water level is rising at a rate of approximately 0.952 m/min.
The real question is how wide is the tank when it is half full?
I draw a rectangle in the cross section of the tank, 40 wide and 60 high.
them at the top and bottom the sides stick out (100-40)/2 = 30cm on each side
halfway up they stick out 15 on each side.
so the width halfway up is 40 + 30 = 70 cm
Now
the area of the surface is 70cm * 800 cm = .7 m * 8 m = 5.6 m^2
now finally calculus
dV/dt = A * dh /dt
0.1 m^3/min = 5.6 m^2 * dh/dt
dh/dt = 0.1 / 5.6 meters/ minute = 0.0179 meters/minute = 1.79 cm/minute