Filling and Emptying Containers | CAIE 9709 Pure Mathematics 1 Notes

Pure Mathematics 1

Mathematics9709

The water radius changes with its depth

An inverted cone has fixed height 20 cm and top radius 10 cm. Water enters at $2\pi\text{ cm}^3/\text{s}$ . Let water depth be $h$ cm and water-surface radius be $r$ cm.

The water occupies a smaller similar cone. Equal angles give equal corresponding side ratios:

\frac rh=\frac{10}{20}\quad\Rightarrow\quad r=\frac h2

MF19 gives a cone’s volume as one third of base area × perpendicular height. Substitute the changing radius before differentiating:

V=\tfrac13\pi r^2h=\frac{\pi h^3}{12}

Common mistake

The container’s top radius is fixed, but the water’s radius is not 10 cm at every depth.

Water depth h5 cm

Constant inflow: 2π cm³/s. Water radius r = h/2.

Volume filled: 1.5625%
Depth rate dh/dt: 0.32 cm/s

As the water gets deeper, the surface gets wider and its upward motion slows.

Constant inflow does not mean constant depth rate

\begin{gathered}\frac{dV}{dh}=\frac{\pi h^2}{4}\\\frac{dh}{dt}=\frac{2\pi}{\pi h^2/4}=\frac8{h^2}\end{gathered}

At $h=5$ , the water rises at $8/25=0.32\text{ cm/s}$ . At twice that depth the horizontal water area is four times as large, so the same incoming volume raises the surface at one quarter of the rate.

This formula applies while filling, for $0<h<20$ . It is undefined at the empty tip, and must not be continued past the top as though the container kept filling.

Half full means half the volume

Since volume is proportional to $h^3$ , half the volume is not half the depth. Compare with the full depth before substituting into the rate:

\left(\frac h{20}\right)^3=\frac12\quad\Rightarrow\quad h=\frac{20}{\sqrt[3]2}

Keep that exact depth in $8/h^2$ . The rate when half full is $0.0317\text{ cm/s}$ to 3 significant figures.

Your turnPractice

A different inverted cone has top radius 10 cm and height 30 cm. Water leaves at 4 cm³/s. Find the signed depth rate when the water is 20 cm deep.

Show worked answer

Similarity gives $r=h/3$ . Thus $V=\pi h^3/27$ , $dV/dh=\pi h^2/9$ . Outflow gives $dV/dt=-4$ .

Answer

\begin{aligned}\frac{dh}{dt}&=-\frac9{100\pi}\\&\approx-0.0286\text{ cm/s}\end{aligned}

Choose from the information given

A prism has the same cross-section along its length, so volume is cross-sectional area × fixed length. If a volume formula is supplied, use it directly; a new geometric derivation is unnecessary.

Your turnPractice

A tank has $V=12h^2$ cm³. Water enters at 24 cm³/s. Find the depth rate at $h=2$ cm.
A curve has $f'(x)=(x-1)(x-3)$ . Where does it decrease, and which stationary input gives a maximum?
A model has $A=12x-x^2$ for $1\le x\le4$ . Find the greatest allowed area.

Show worked answer

$dV/dh=24h=48$ , so $dh/dt=24/48=0.5\text{ cm/s}$ .
Negative derivative for $1<x<3$ ; positive-to-negative at $x=1$ gives the maximum.
The stationary input 6 is excluded. The model increases throughout the allowed interval, so $A(4)=32$ is greatest.

The water radius changes with its depth

An inverted cone has fixed height 20 cm and top radius 10 cm. Water enters at $2\pi\text{ cm}^3/\text{s}$ . Let water depth be $h$ cm and water-surface radius be $r$ cm.

The water occupies a smaller similar cone. Equal angles give equal corresponding side ratios:

\frac rh=\frac{10}{20}\quad\Rightarrow\quad r=\frac h2

MF19 gives a cone’s volume as one third of base area × perpendicular height. Substitute the changing radius before differentiating:

V=\tfrac13\pi r^2h=\frac{\pi h^3}{12}

Common mistake

The container’s top radius is fixed, but the water’s radius is not 10 cm at every depth.

Water depth h5 cm

Constant inflow: 2π cm³/s. Water radius r = h/2.

Volume filled: 1.5625%
Depth rate dh/dt: 0.32 cm/s

As the water gets deeper, the surface gets wider and its upward motion slows.

Constant inflow does not mean constant depth rate

\begin{gathered}\frac{dV}{dh}=\frac{\pi h^2}{4}\\\frac{dh}{dt}=\frac{2\pi}{\pi h^2/4}=\frac8{h^2}\end{gathered}

This formula applies while filling, for $0<h<20$ . It is undefined at the empty tip, and must not be continued past the top as though the container kept filling.

Half full means half the volume

Since volume is proportional to $h^3$ , half the volume is not half the depth. Compare with the full depth before substituting into the rate:

\left(\frac h{20}\right)^3=\frac12\quad\Rightarrow\quad h=\frac{20}{\sqrt[3]2}

Keep that exact depth in $8/h^2$ . The rate when half full is $0.0317\text{ cm/s}$ to 3 significant figures.

Your turnPractice

A different inverted cone has top radius 10 cm and height 30 cm. Water leaves at 4 cm³/s. Find the signed depth rate when the water is 20 cm deep.

Show worked answer

Similarity gives $r=h/3$ . Thus $V=\pi h^3/27$ , $dV/dh=\pi h^2/9$ . Outflow gives $dV/dt=-4$ .

Answer

\begin{aligned}\frac{dh}{dt}&=-\frac9{100\pi}\\&\approx-0.0286\text{ cm/s}\end{aligned}

Choose from the information given

A prism has the same cross-section along its length, so volume is cross-sectional area × fixed length. If a volume formula is supplied, use it directly; a new geometric derivation is unnecessary.

Your turnPractice

A tank has $V=12h^2$ cm³. Water enters at 24 cm³/s. Find the depth rate at $h=2$ cm.
A curve has $f'(x)=(x-1)(x-3)$ . Where does it decrease, and which stationary input gives a maximum?
A model has $A=12x-x^2$ for $1\le x\le4$ . Find the greatest allowed area.

Show worked answer

$dV/dh=24h=48$ , so $dh/dt=24/48=0.5\text{ cm/s}$ .
Negative derivative for $1<x<3$ ; positive-to-negative at $x=1$ gives the maximum.
The stationary input 6 is excluded. The model increases throughout the allowed interval, so $A(4)=32$ is greatest.