A trough is filled with water at a rate of 1 cubic meter per second. The trough has a trapezoidal cross section with the lower base of length half a meter and one meter sides opening outwards at an angle of $45^\circ$ from the base. The length of the trough is 4 meters. What is the rate at which the water level h is rising when h is one half meter?
The book's solution says that that the volume of the trough is given by $V = 4(h^2 + h/2)$, I want to know how and why?
Thanks,
Best Answer
No, the volume of the water when the water level is $h$ is $$V = 4(h^2 + h/2)$$ You can derive this by first finding the area of the trapezoidal cross-section of the water, when the height of the water is $h$, and it will be $h^2+\frac{h}{2}$, by this picture:
The area of the rectangular part is $\frac{1}{2}\times h$, and the area of the two triangles adds up to be the area of a square with side length $h$ (because the sides slope at $45^\circ$ angles), i.e. the area of the two triangles together is $h^2$. Thus, the total cross-sectional area is $h^2+\frac{h}{2}.$ Now, multiply by the length of the trough to get the volume, because we have that volume = base * height for these kinds of shapes.
Note that this implies that $$\frac{dV}{dh}=8h+2.$$ You will also want to use the fact that the trough is filled with water at a rate of 1 cubic meter per second, or in other words, $$\frac{dV}{dt}=1$$ Finally, using the chain rule, $$\frac{dV}{dt}=\frac{dV}{dh}\cdot\frac{dh}{dt}$$ You want to find $\frac{dh}{dt}$...