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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.

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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 $cos 2\alpha = 2cos^2\alpha-1$

$$cos\theta=2(0.5)^2-1=-0.5$$

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

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.

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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 $h(t) = 0.25$.

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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 $dh/d\theta$.

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  • $\begingroup$ 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

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