Level 3 Differentiation Walkthrough

2018 NCEA Level 3 Differentiation Question 3(b)

2018 Paper

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Question

A curve is defined parametrically by

\[x=5e^{2t},\qquad y=2e^{5t}.\]

Find the gradient of the tangent to this curve at the point where \(t=0\).

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

First walkthrough idea

Focus to try first

For a parametric curve, divide \(dy/dt\) by \(dx/dt\), then substitute \(t=0\).

Step 1

Differentiate both coordinates

Differentiate \(x\) and \(y\) separately with respect to \(t\).

Show the first step’s working
\[\frac{dx}{dt}=10e^{2t},\qquad \frac{dy}{dt}=10e^{5t}\]

Walkthrough overview

What this question practises

This 2018 walkthrough is part of AS91578 — Apply differentiation methods in solving problems.

Method: Finding a parametric gradient.

This is Question 3(b) from the 2018 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 finding a parametric gradient. Use the hints to plan the method, then repeat the question without hints and check each step.

Common mistake to avoid

Check each step against the original condition, preserve signs and restrictions, and confirm that the final result answers the question asked.

Continue practising

← Question 3(a) Question 3(c) →