IB Maths AI HL 5.14 Differential equation modelling

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This question explores models for the height of water in a cylindrical container as water drains out.
The diagram shows a cylindrical water container of height 3.2 metres and base radius 1 metre. At the base of the container is a small circular valve, which enables water to drain out.

Figure for Question IB Maths AI HL Ahl 5 14 Hl Differential Equation Modelling Questions question 1 — IB Maths AI HL

Eva closes the valve and fills the container with water.
At time t=0, Eva opens the valve. She records the height, h metres, of water remaining in the container every 5 minutes.

Table for Question IB Maths AI HL Ahl 5 14 Hl Differential Equation Modelling Questions question 1 — IB Maths AI HL

Eva first tries to model the height using a linear function, h(t)=a t+b, where $a, b \in \mathbb{R}$ .

(a)

Find the value of k.

[ 2 ]

$\quad 0.002 t^{2}-0.174 t+3.2=0$
26.4 (26.4046...)

(b)

Show that $\frac{\mathrm{d} h}{\mathrm{~d} t}=-R^{2} \sqrt{70560 h}$ .

[ 3 ]

$\quad V=\pi(1)^{2} h$

EITHER
$\frac{\mathrm{d} V}{\mathrm{dt}}=\pi \frac{\mathrm{d} h}{\mathrm{dt}}$

OR
attempt to use chain rule
$\frac{\mathrm{d} h}{\mathrm{~d} t}=\frac{\mathrm{d} h}{\mathrm{~d} V} \times \frac{\mathrm{d} V}{\mathrm{~d} t}$
THEN

\begin{aligned} & \frac{\mathrm{d} h}{\mathrm{~d} t}=\frac{1}{\pi} \times-\pi R^{2} \sqrt{70560 h} \\ & \frac{\mathrm{~d} h}{\mathrm{~d} t}=-R^{2} \sqrt{70560 h} \end{aligned}

(c)

By solving the differential equation $\frac{\mathrm{d} h}{\mathrm{~d} t}=-R^{2} \sqrt{70560 h}$ , show that the general solution is given by $h=17640\left(c-R^{2} t\right)^{2}$ , where $c \in \mathbb{R}$ .

Eva measures the radius of the valve to be 0.023 metres. Let T be the time, in minutes, it takes for all the water to drain out of the container.

[ 5 ]

attempt to separate variables M1

\begin{aligned} & \int \frac{1}{\sqrt{70560 h}} \mathrm{~d} h=\int-R^{2} \mathrm{~d} t \\ & \frac{2 \sqrt{h}}{\sqrt{70560}}=-R^{2} t+c \end{aligned}

Note: Award A1 for each correct side of the equation.

\sqrt{h}=\frac{\sqrt{70560}}{2}\left(c-R^{2} t\right)

Note: Award the final A1 for any correct intermediate step that clearly leads to the given equation.

h=17640\left(c-R^{2} t\right)^{2}

(d)

Use the general solution from part (d) and the initial condition h(0)=3.2 to predict the value of T.

Eva wants to use the container as a timer. She adjusts the initial height of water in the container so that all the water will drain out of the container in 15 minutes.

[ 4 ]

$t=0 \Rightarrow 3.2=17640 c^{2}$

c=0.0134687 \ldots

substituting h=0 and their non-zero value of c

\begin{aligned} & T=\frac{c}{R^{2}}=\frac{0.0134687 \ldots}{0.023^{2}} \\ & =25.5(\text { minutes })(25.4606 \ldots) \end{aligned}

(e)

Find this new height.

Eva has another water container that is identical to the first one. She places one water container above the other one, so that all the water from the highest container will drain into the lowest container. Eva completely fills the highest container, but only fills the lowest container to a height of 1 metre, as shown in the diagram.

diagram not to scale

At time t=0 Eva opens both valves. Let H be the height of water, in metres, in the lowest container at time t.

[ 3 ]

$\quad h=0 \Rightarrow c=R^{2} t$ $c=0.023^{2} \times 15(=0.007935)$ $t=0 \Rightarrow h=17640\left(0.023^{2} \times 15\right)^{2}$
h=1.11 (metres) (1.11068...)

(f)

Show that $\frac{\mathrm{d} H}{\mathrm{~d} t} \approx 0.2514-0.009873 t-0.1405 \sqrt{H}$ , where $0 \leq t \leq T$ .

[ 4 ]

let h be the height of water in the highest container from parts (d) and (e) we get

\begin{aligned} & \frac{\mathrm{d} h}{\mathrm{~d} t}=-35280 R^{2}\left(0.0134687 \ldots-R^{2} t\right) \\ & \text { so } \frac{\mathrm{d} H}{\mathrm{~d} t}=35280 R^{2}\left(0.0135-R^{2} t\right)-R^{2} \sqrt{70560 H} \\ & \left(\frac{\mathrm{~d} H}{\mathrm{~d} t}=18.6631 \ldots(0.0134687 \ldots-0.000529 t)-0.000529 \sqrt{70560 H}\right) \\ & \left(\frac{\mathrm{d} H}{\mathrm{~d} t}=0.251367 \ldots-0.0987279 \ldots-0.140518 \ldots \sqrt{H}\right) \\ & \frac{\mathrm{d} H}{\mathrm{~d} t} \approx 0.2514-0.009873 t-0.1405 \sqrt{H} \end{aligned}

IB Maths AI HL Ahl 5 14 Hl Differential Equation Modelling Questions question 1

Question (a)

Question (b)

Question (c)

Question (d)

Question (e)

Question (f)