NCEA Calculus 2022 ONE(d - Solution & Explanation

Question

A curve is defined parametrically by the equations:

x = 2 + 3t

and

y = 3t - \ln(3t - 1)

where

t > \dfrac{1}{3}

.

Find the coordinates,

(x,y)

, of any point(s) on the curve where the tangent to the curve has a gradient
of

\dfrac{1}{2}

.

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

Official Answer

\dfrac{dx}{dt}=3

\dfrac{dy}{dt}=3-\dfrac{3}{3t-1}

=\dfrac{3(3t-1)-3}{3t-1}

=\dfrac{9t-6}{3t-1}

\dfrac{dy}{dx}=\dfrac{9t-6}{3t-1}\times\dfrac{1}{3}

=\dfrac{3t-2}{3t-1}

\dfrac{dy}{dx}=\dfrac{1}{2}

\dfrac{3t-2}{3t-1}=\dfrac{1}{2}

6t-4=3t-1

3t=3

t=1

x=5\quad y=3-\ln 2\text{ or }2.307

Grading Criteria

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Video Explanation

2022 NCEA L3 Calculus Exam Walkthrough by infinityplusone(starts at 11:19)

2022 ONE(d)