Level 3 Differentiation Walkthrough

2016 NCEA Level 3 Differentiation Question 1(c)

2016 Paper

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Question

A curve is defined by the parametric equations

\[x=2\cos(2t)\quad\text{and}\quad y=\tan^2t.\]

Find the gradient of the tangent to the curve at the point where \(t=\dfrac{\pi}{4}\).

You must use calculus and show any derivatives that you need to find when solving this problem.

First walkthrough idea

Focus to try first

Differentiate both coordinates with respect to \(t\), then divide \(dy/dt\) by \(dx/dt\).

Step 1

Differentiate x with respect to t

The input \(2t\) contributes a chain-rule factor of two.

Show the first step’s working
\[x=2\cos(2t)\] \[\frac{dx}{dt}=2\bigl(-\sin(2t)\bigr)(2)=-4\sin(2t).\]

Walkthrough overview

What this question practises

This 2016 walkthrough is part of AS91578 — Apply differentiation methods in solving problems.

Method: Parametric differentiation and a tangent gradient.

This is Question 1(c) from the 2016 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 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

← Question 1(b) Question 1(d) →