Consider a container of the shape obtained by revolving a segment of
the parabola $x=1+y^2$ about the $y-axis$ as shown below, the
container is initially empty. Water is poured in at a constant rate of
$1cm^3/s$. Let h(t) be the height of water in the container at time t.
find the time t when the rate of change of $h(t)$ is maximum
Source :- ISI UGB 2024
The maximum radius of the container is 2 and minimum is 1
I started by finding an expression for value
$dV= \pi r^2dy$, form the middle of the container
but $r^2=(1+y^2)^2$
so $V= \int_0^2 \pi (1+y^2)^2dy$
Or to a general height h
$V =\int_0^h \pi (1+y^2)^2dy$
so Actual volume is
$2V= 2\int_0^1 \pi (1+y)^2dy$
However, I'm not sure now to integrate wrt to the bottom which is what I think I need to do to solve this using differential calculus
Best Answer
You can use Cavalieri's principle to handle the integral as a one-dimensional integral, that is, if $A(y)$ is the area of the section at height $y$ of the solid, then the volume up to height $h$ is given by $$ V(h):=\int_{h_0}^{h}A(y)\,d y $$ where $h_0$ is the lowest height of the solid. Let write $h$ by $h(t)$, then you have that $(V\circ h)'(t)=1\mathrm{cm^3/s}$, what is equivalent to say that $(A\circ h)(t)h'(t)=1$. Differentiating implicitly we find that $$ h''(t)=0\iff (A'\circ h)(t)=0 \iff h(t)=0 $$ using that $A(h)=\pi (1+h^2)^2$. Therefore $h'$ is maximal at the time where $h=0$, what is intuitively obvious. Now from the relation $(V\circ h)'(t)=1$ it follows that $(V\circ h)(t)=t$, therefore the time we want to find is given by $V(0)$, that is, half the volume of the figure, so $$ t_{\mathrm{max}}=\pi\int_{h_0}^{0}(1+y^2)^2\,d y,\quad \text{ with }h_0<0\text{ such that }1+h_0^2=2 $$ This gives $t_{\mathrm{max}}=28\pi/15$.