Level 3 Differentiation Walkthrough

2017 NCEA Level 3 Differentiation Question 1(d)

2017 Paper

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Question

A curve is defined parametrically by the equations

\[x=\sqrt{t+1},\qquad y=\sin(2t).\]

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

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

First walkthrough idea

Focus to try first

Differentiate both coordinates with respect to the common parameter, then divide the two rates to obtain the curve's gradient.

Step 1

Differentiate x with respect to t

Rewrite the radical as a power and apply the chain rule.

Show the first step’s working
\[x=(t+1)^{1/2}\] \[\frac{dx}{dt}=\frac12(t+1)^{-1/2}=\frac{1}{2\sqrt{t+1}}\]

Walkthrough overview

What this question practises

This 2017 walkthrough is part of AS91578 — Apply differentiation methods in solving problems.

Method: Parametric differentiation and evaluating the gradient.

This is Question 1(d) from the 2017 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 the 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 1(c) Question 1(e) →