Edexcel Pure Paper 2 2024 Question 10 Solution

Edexcel Pure Paper 2 2024 Question 10 – Parametric Differentiation

❓ The Question

🧠 Before you start

This is one of those questions where the method is clear — but only if you actually follow what it’s asking.

It explicitly says to use parametric differentiation. A few people try to convert everything into x and y first, but that usually makes things longer than necessary.

So it’s worth sticking with $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$ from the start.

✏️ Working

We are given:

$x = (t+3)^2$
$y = 1 - t^3$

Part (a)

We are told that point $P(4,2)$ lies on the curve.

Step 1: Find t

Use the x-coordinate:

$4 = (t+3)^2$

So:

$t+3 = \pm 2$

$t = -1 \quad \text{or} \quad t = -5$

Now check which value works using y:

If $t = -1$ :

$y = 1 - (-1)^3 = 1 + 1 = 2$ ✅

If $t = -5$ :

$y = 1 - (-125) = 126$ ❌

So:

$t = -1$

Step 2: Differentiate

Differentiate both equations with respect to t.

For x:

$\frac{dx}{dt} = 2(t+3)$

For y:

$\frac{dy}{dt} = -3t^2$

Step 3: Form $\frac{dy}{dx}$

$\frac{dy}{dx} = \frac{-3t^2}{2(t+3)}$

Step 4: Substitute t = -1

$\frac{dy}{dx} = \frac{-3(-1)^2}{2(-1+3)} = \frac{-3}{4}$

So the gradient is:

$-\frac{3}{4}$

Step 5: Equation of tangent

Use point-slope form:

$y – 2 = -\frac{3}{4}(x – 4)$

Expand:

$y – 2 = -\frac{3}{4}x + 3$

$y = -\frac{3}{4}x + 5$

Rearrange:

$3x + 4y = 20$

Final answer (part a):

$3x + 4y = 20$

Part (b)

We are asked for the greatest height of the slide.

The height is given by:

$y = 1 - t^3$

From the domain:

$-2 \le t \le 1$

Now check endpoints.

If $t = -2$ :

$y = 1 - (-8) = 9$

If $t = 1$ :

$y = 1 - 1 = 0$

Maximum height:

$9 \text{ metres}$

🎯 Where the Marks Are

Part (a) carries most of the marks.

finding the correct t value
using parametric differentiation correctly
forming the tangent equation

Part (b) is just recognising where the maximum occurs.

⚠️ What Went Wrong

From the examiner report, most students handled part (a) quite well — but small slips did show up.

A common mistake was using $t = -5$ instead of checking which value actually fits the point.

Some also confused $\frac{dy}{dx}$ with $\frac{dx}{dy}$ , or didn’t complete the chain rule properly.

In part (b), quite a few answers didn’t check the endpoints properly. Some tried setting $\frac{dy}{dx} = 0$ , which isn’t needed here.

💡 One Small Tip

If you get two values for t, always check which one matches the point fully.

It only takes a few seconds, and it avoids bigger mistakes later.

🚀 If This Felt Difficult

If this felt slightly awkward, it’s usually because parametric questions mix a few steps together.

You’re not just differentiating — you’re interpreting the parameter as well.

Working through more of these helps you master A level maths, especially when switching between different representations.

And if you want to feel more consistent across questions like this, having some structured A level maths study support can make these steps feel much more routine.

🔗 Next Steps

← Question 9 Solution
→ Question 11 Solution

👨‍🏫Author Bio

S Mahandru specialises in helping students handle multi-step exam questions with confidence. The focus is on keeping methods clear and making sure each step connects logically under exam conditions.

❓ Frequently Asked Questions

📌Why use dy/dt over dx/dt?

Because that’s how gradients are found for parametric equations.

📌Why reject t = -5?

Because it doesn’t give the correct y-value for point P.

📌Do I need to expand the tangent fully?

Yes — the question asks for a specific form.

📌Why check endpoints in part (b)?

Because the maximum occurs at the boundary of the interval.