Level 3 Differentiation Walkthrough

2021 NCEA Level 3 Differentiation Question 1(d)

2021 Paper

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Question

A curve is defined parametrically by the equations

\[ x=t^2+3t,\qquad y=t^2\ln(2t-3),\qquad t>\frac{3}{2}. \]

Find the gradient of the tangent to the curve at the point \((10,0)\).

First walkthrough idea

Focus to try first

For parametric curves, use \(\frac{dy}{dx}=\frac{dy/dt}{dx/dt}\), then find the matching value of \(t\).

Step 1

Differentiate x and y with respect to t

Find \(dx/dt\) and \(dy/dt\) separately.

Show the first step’s working
\[ \frac{dx}{dt}=2t+3 \] \[ \frac{dy}{dt}=2t\ln(2t-3)+t^2\cdot\frac{2}{2t-3} \] \[ \frac{dy}{dt}=2t\ln(2t-3)+\frac{2t^2}{2t-3} \]

Walkthrough overview

What this question practises

This 2021 walkthrough is part of AS91578 — Apply differentiation methods in solving problems.

Method: Parametric differentiation and evaluating a tangent gradient.

This is Question 1(d) from the 2021 NCEA Level 3 Differentiation paper for AS91578 — Apply differentiation methods in solving problems. Use the guided hints to practise the method before revealing the full working.

Calc.nz is an independent learning resource. Compare questions, diagrams, and assessment information with the official NZQA resources for AS91578.

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Learning summary

Review the method, not only the answer

This walkthrough helps you practise parametric differentiation and evaluating a tangent gradient. Use the hints to plan the method, then repeat the question without hints and check each step.

Common mistake to avoid

Use the derivative for the gradient and the original curve for the point before forming the tangent equation.

Continue practising

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