Here is a question about the water trough, it goes like this: A farmer has a water trough of length 8m which has a semi-circular cross-section diameter 1m. Water is pumped to trough at a constant rate of $0.1m^3$ per minute.

enter image description here

Find the rate at which the water level is rising at the instant when the water is 25cm deep.

My attempt:

Information given:

${dV\over dt}=0.1$

$V=\theta-sin\theta$ then ${dV\over d\theta}=1-cos\theta$

water depth = 25cm =0.25m

Solving attempt:

Using ${dV\over dt}={dV\over d\theta}{d\theta\over dt}$

$$0.1=(1-cos\theta){d\theta\over dt}$$

$${d\theta\over dt}={0.1\over 1-cos\theta}$$

Solving for $\theta$

Since radius is 0.5m and remaining depth is 0.25m

$$cos\left({\theta\over 2}\right) ={0.25\over 0.5}=0.5$$

using the double angle formulae $cos 2\alpha = 2cos^2\alpha-1$


$$\theta = arccos(-0.5) = \frac{2\pi}{3}$$

Back to the differential equation ${d\theta\over dt}={0.1\over 1-cos\theta}$:

In the case where depth is 0.25m and $cos\theta =-0.5$

$${d\theta\over dt}={0.1\over 1-(-0.5)}=\frac{0.1}{1.5}=\frac{1}{15}$$

What I am confused about and am struggling with is how do I get to $\frac{dh}{dt}$ which is the rate of increase in depth at the point where the current depth is 0.25m.


Here is a simpler approach.

Let $V(h)$ be the volume in the tank at height $h$. Let $F$ be the flow rate into the tank. Let the length be $L$ and the radius $r$.

Then with $w(x) = \sqrt{r^2-x^2}$ we have $V(h) = L \int_{r-h}^r w(x)dx$. Then $V'(h) = L w(r-h)$.

Let $h \mapsto h(t)$ be the height at time $t$.

We have $(V \circ h)'(t) = F$, the chain rule gives $V'(h(t)) h'(t) = F$ and so $h'(t) = {F \over L w(r-h(t)) }$.

Simplifying gives $h'(t) = {F \over L \sqrt{2 r h(t)-h(t)^2} }$.

You want to compute $h'(t)$ for $h(t) = 0.25$.


Start with $$\frac{dV}{dt}=\frac{dV}{d\theta}\frac{d\theta}{dh}\frac{dh}{dt}\ .$$ Now

  • $dV/dt$ is given;
  • you have $dV/d\theta$ in terms of $\theta$, as well as the value of $\theta$ (I assume it is correct, I haven't checked your working);
  • you want $dh/dt$;
  • so you need to find $d\theta/dh$, or its reciprocal $dh/d\theta$.

From your diagram, $$h=\frac12-\frac12\cos\frac\theta2\ ,$$ and you know the value of $\theta$ so you can calculate $dh/d\theta$.

  • $\begingroup$ Makes sense that $h=\frac{1}{2}-\frac{1}{2}cos\left(\frac{\theta}{2}\right)$ given h is the remaining height from top of the trough to the water level. Now to get $\frac{d\theta}{dh}$ we have to make $\theta$ the subject. $\theta=2arccos(1-2h)$ then $\frac{d\theta}{dh}=\frac{4}{\sqrt{4h-4h^2}}$. Since h=0.25, $\frac{d\theta}{dh}=\frac{4}{\sqrt{1-0.25}}=\frac{4}{\sqrt{0.75}}$ $\endgroup$ – Kenny Guy Jan 18 '16 at 5:33
  • $\begingroup$ Haven't checked but should be OK. But that is doing it the hard way, finding $dh/d\theta$ is much easier. $\endgroup$ – David Jan 18 '16 at 5:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.