A water trough is 9 m long and has a cross-section in the shape of an isosceles trapezoid that is 30 cm wide at the bottom, 70 cm wide at the top, and has height 40 cm. If the trough is being filled with water at the rate of 0.2 m3/min how fast is the water level rising when the water is 20 cm deep?
[Math] How to solve this related rates problem with trapezoids
calculus
Related Solutions
Consider that every layer of water in the trough is rectangular, so it has an infinitesimal volume of $dV \ = \ A(y) \ dy $ , with $y$ being the "level" in the trough (you may choose to start from the bottom up or the top down) and $A(y)$ is the surface area. We'll measure from the bottom as $y = 0$ in what follows. The layer at $y = 0$ thus a surface area of $ \ 80 ยท 200 \ $ sq.cm.
We must consider how the surface area changes as $y$ increases. Since the trapezoid has straight (though inclined) sides, the width increases linearly from $w(y) = 80 \ $cm at $y = 0 \ $cm to $w(y) = 140 \ $cm at $\ y = Y \ $, this being the height of the trough. They were not quite so nice to you here, since they didn't tell you the full depth of the trough, so you will need to use a little trig to find $Y$. Then construct a "width function" $w(y)$ which varies linearly over the range from $y = 0$ to $y = Y$; you can then produce a "surface area function" $A(y) = w(y) \cdot 200 \ $sq.cm.
You will now have a volume function with respect to $y$ that you can use in your related rates problem.
EDIT: In response to your other question, water is entering the trough at the rate $\frac{dV}{dt} = 33,000 \frac{\text{cu.cm.}}{\text{sec}}$ . So find $\frac{dy}{dt}$ at $y = 20$ .
The change in the volume of the water is the difference in the fill and leak rates (taking the leak rate to be a positive number).
$$\frac{dV}{dt} = \frac{dF}{dt} - \frac{dL}{dt}$$
Thus, the leak rate is the difference in the fill rate and the actual rate of increase in the volume of the water.
$$\frac{dL}{dt} = \frac{dF}{dt} - \frac{dV}{dt}$$
The volume of the water when its height is $h$ and its radius is $r$ is
$$V = \frac{1}{3}\pi r^2h$$
Since the inverted conical tank has a diameter of $10~\text{ft}$ at the top, its radius at the top is $5~\text{ft}$.
By similar triangles, the ratio of the radius of the water to the height of the water in the cone is $$\frac{r}{h} = \frac{5~\text{ft}}{20~\text{ft}} = \frac{1}{4} \implies r = \frac{h}{4}$$ Thus, we can express the volume of the water as a function of $h$ by substituting $h/4$ for $r$, which yields $$V(h) = \frac{1}{3}\pi\left(\frac{h}{4}\right)^2h = \frac{1}{48}\pi h^3$$ Differentiating with respect to time yields $$\frac{dV}{dt} = \frac{1}{16}\pi h^2 \frac{dh}{dt}$$ We are given that $$\frac{dh}{dt} = 1~\frac{\text{in}}{\text{min}} = \frac{1}{12}~\frac{\text{ft}}{\text{min}}$$ when $h = 16~\text{ft}$. Thus, the rate at which the volume is increasing is $$\frac{dV}{dt} = \frac{1}{16}\pi (16~\text{ft})^2\left(\frac{1}{12}~\frac{\text{ft}}{\text{min}}\right) = \frac{4}{3}\pi~\frac{\text{ft}^3}{\text{min}}$$
Since the fill rate is $8~\text{ft}^3/\text{min}$, the leak rate is $$\frac{dL}{dt} = \frac{dF}{dt} - \frac{dV}{dt} = 8~\frac{\text{ft}^3}{\text{min}} - \frac{4}{3}\pi~\frac{\text{ft}^3}{\text{min}} = \left(8 - \frac{4}{3}\pi\right)~\frac{\text{ft}^3}{\text{min}}$$
Best Answer
You solve it in the usual way. You need to compute $V(h)$, the volume of water in the trough when the height of the water is $h$. Then invert it to get $h(V)$. You are asked for $\frac {dh}{dt}$ at a particular value of $h$ and are given $\frac {dV}{dt}$. Differentiate with respect to $t$, use the chain rule, etc.