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.
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.
Best Answer
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 $h(t) = 0.25$.