Edexcel Pure Paper 1 2024 Question 14

Edexcel Pure Paper 1 2024 Question 14 – Differential Equations and Modelling

❓ The Question

🧠 Before you start

This is one of those questions where everything is quite standard… but it can still go wrong if you rush it.

Nothing here is especially difficult:

translate words → equation
separate variables
integrate
substitute

The mistakes tend to come from mixing up what each value represents.

So just go step by step.

✏️ Working

Part (a)

“Inversely proportional to the square root of the radius”

That should immediately suggest:

$\frac{dr}{dt} \propto \frac{1}{\sqrt{r}}$

Bring in a constant:

$\frac{dr}{dt} = \frac{k}{\sqrt{r}}$

That’s all that’s needed here.

Part (b)

Now we actually use the numbers given.

At $t = 10$ , we have:

$r = 16$
$\frac{dr}{dt} = 0.9$

Substitute into:

$\frac{dr}{dt} = \frac{k}{\sqrt{r}}$

So:

$0.9 = \frac{k}{\sqrt{16}}$

$0.9 = \frac{k}{4}$

$k = 3.6$

Now go back to the differential equation:

$\frac{dr}{dt} = \frac{3.6}{\sqrt{r}}$

Separate variables:

$\sqrt{r},dr = 3.6,dt$

Integrate both sides:

$\int r^{1/2} , dr = \int 3.6 , dt$

$\frac{2}{3}r^{3/2} = 3.6t + c$

Now use the same condition again:
$t = 10, ; r = 16$

Substitute:

$\frac{2}{3}(16)^{3/2} = 3.6(10) + c$

Now carefully:

$16^{3/2} = 64$

So:

$\frac{2}{3} \times 64 = 36 + c$

$\frac{128}{3} = 36 + c$

Convert 36:

$36 = \frac{108}{3}$

So:

$\frac{128}{3} = \frac{108}{3} + c$

$c = \frac{20}{3}$

Now put that back:

$\frac{2}{3}r^{3/2} = 3.6t + \frac{20}{3}$

Multiply everything by 3:

$2r^{3/2} = 10.8t + 20$

Divide by 2:

$r^{3/2} = 5.4t + 10$

That matches what we were asked to show.

Part (c)

Now just use it.

Put $t = 20$ :

$r^{3/2} = 5.4(20) + 10$

$= 108 + 10 = 118$

So:

$r = 118^{2/3}$

If you work that out:

$r \approx 24.1 \text{ cm}$

Convert to mm:

$24.1 \text{ cm} = 241 \text{ mm}$

That’s what they want.

Part (d)

This is just interpretation.

The model assumes the balloon keeps expanding forever.

That’s not realistic — eventually it bursts.

🎯 Where the Marks Are

Part (a) is just 1 mark — but easy to lose.

You need the constant:

$\frac{dr}{dt} = \frac{k}{\sqrt{r}}$

Part (b) carries most of it.
- find $k$ correctly
- separate variables
- integrate properly
- use the condition
Miss one of those and marks drop quickly.

Part (c) is just substitution — provided part (b) is correct.

Part (d) is a simple modelling comment, but it needs to be specific.

⚠️ What Went Wrong

A lot of errors here were small but important.

In part (a), quite a few answers missed the constant $k$ .

That’s enough to lose the mark immediately.

In part (b), the biggest issue was not using the value $\frac{dr}{dt} = 0.9$ properly.

Some treated 0.9 as $k$ , which then breaks everything that follows.

Another common mistake was in the integration step.

People wrote:

$\int r^{1/2} dr = r^{3/2}$

and forgot the factor $\frac{2}{3}$ .

That small slip affects the constant and the final result.

There were also a few cases where students didn’t substitute the initial condition early enough.

That left both $k$ and $c$ unknown, which made things harder than necessary.

Part (c) was mostly fine, but accuracy mattered.

Some left the answer as $118^{2/3}$ or rounded poorly.

Part (d) was generally answered, but vague responses like “the model is unrealistic” didn’t always score.

You needed to say why.

💡 The takeaway

Nothing here is conceptually difficult.

But it’s very easy to lose marks through small slips.

Keep it tidy. Do one step at a time.

🚀 If This Felt Difficult

If this didn’t feel smooth, it’s usually because differential equations need repetition to feel natural.

Working through more questions like this in a full A level maths course helps build that consistency.

If you need more focused practice, getting A level maths help online can make these steps feel much more routine.

🔗 Next Steps

← Question 13
→ Question 15

👨‍🏫Author Bio

This solution written by S Mahandru is a tutor who focuses on making longer exam questions feel manageable.

The aim is to show how marks are picked up step by step — not just how to reach the answer, but how to avoid losing marks along the way.

❓ Frequently Asked Questions

📌 Why do we include k?

Because “proportional to” always introduces a constant.

📌 What’s the key step in part (b)?

Separating variables before integrating.

📌 Why is r to the power 3/2 used?

It comes from integrating $r^{1/2}$ .

📌 Where do most marks get lost?

Usually in small slips during integration or substitution.

Edexcel Pure Paper 1 2024 Question 14