Question 19
A circular-based bowl has been positioned symmetrically on a Cartesian plane as shown in the diagram.
The bowl has a shape that can be generated by rotating the curve $y=\dfrac{4}{8-x}-1$ about the $y$-axis for $4\le x\le7.6$ cm.
The bowl is being filled with a liquid at the rate of $7\pi\ \text{cm}^{3}\,\text{s}^{-1}$.
Determine the rate at which the depth of liquid is increasing when the depth of liquid reaches one-third of the height of the bowl.
The wider the surface, the slower the level rises
Watch the bowl fill to one-third of its height, at the real rate. The orange curve is the one you rotate, and the orange layer is the surface at $h=3$.
The bowl has a flat base of radius 4, so the liquid starts as a wide disc, not a point.
Change the depth and watch the rate change
Move the slider to fill the bowl. The rate is $\frac{dh}{dt}=\frac{7\pi}{\pi x^{2}}=\frac{7}{x^{2}}$, where $x$ is the radius at the surface.
Work through the solution one mark at a time
Try each step yourself before you reveal it. There is no calculator, so it helps that you never have to integrate.
You rotate about the $y$-axis, so you need $x$ in terms of $y$.
$$y+1=\frac{4}{8-x}\quad\Rightarrow\quad8-x=\frac{4}{y+1}\quad\Rightarrow\quad x=8-\frac{4}{y+1}$$
This mark can be implied by your later working.
Marker: correctly expresses $x$ as the subject of the given relationship
Let $h$ be the depth of the liquid. The volume of liquid is the volume of revolution from $y=0$ up to $y=h$.
$$V=\pi\int_{0}^{h}\left(8-\frac{4}{y+1}\right)^{2}\,dy$$
You keep follow-through marks from here if you slipped earlier.
Marker: establishes the volume of the bowl as a definite integral
You do not need to evaluate the integral. Differentiating an integral with respect to its upper limit just gives you the integrand at $h$.
$$\frac{dV}{dh}=\pi\left(8-\frac{4}{h+1}\right)^{2}$$
This is the orange layer in the animation: a thin disc of radius $x$ adds $\pi x^{2}\,dh$ of volume. You can also integrate first and then differentiate, but it takes much longer.
Marker: determines an expression for $\frac{dV}{dh}$
Use the chain rule to link the rate you know, $\frac{dV}{dt}=7\pi$, to the rate you want.
$$\frac{dh}{dt}=\frac{dh}{dV}\cdot\frac{dV}{dt}=\frac{1}{\pi\left(8-\frac{4}{h+1}\right)^{2}}\times7\pi=\frac{7}{\left(8-\frac{4}{h+1}\right)^{2}}$$
Marker: uses a related rate of change equation to determine a relationship between $\frac{dh}{dt}$ and $h$
Find the height of the bowl from the top of the curve, at $x=7.6$.
$$y=\frac{4}{8-7.6}-1=\frac{4}{0.4}-1=9\ \text{cm}\qquad h=\frac{1}{3}\times9=3\ \text{cm}$$
Marker: correctly determines the instantaneous depth of liquid for the required calculation
Substitute $h=3$. The radius at the surface is $8-\frac{4}{4}=7$.
$$\left.\frac{dh}{dt}\right|_{h=3}=\frac{7}{(8-1)^{2}}=\frac{7}{49}=\frac{1}{7}\ \text{cm s}^{-1}$$
Marker: determines the required rate
The last mark is for how you set out your working. Define $h$, use proper integral notation for the volume, write the chain rule out in full, and link each line with an equals sign.
The depth is increasing at $\frac{1}{7}\ \text{cm s}^{-1}$.
Marker: shows logical organisation communicating key steps
Each phrase of the question gave you something
You have to combine a volume of revolution with related rates, and you are not told the height of the bowl. You find it from the end of the curve before you can work out one-third of it.
The step I would check first is where you go from $V$ to $\frac{dV}{dh}$. Without a calculator, integrating $\left(8-\frac{4}{y+1}\right)^{2}$ is long and easy to get wrong. The QCAA reward you for seeing that differentiating the integral with respect to $h$ skips all of that.
Only mark this done when you could do it without help. Reading the solution does not count.
Question wording and marking-guide steps are from the 2020 QCAA Specialist Mathematics external assessment, © State of Queensland (QCAA) 2020, licensed under CC BY 4.0, and have been adapted. Tangent Tuition is not affiliated with the QCAA.