NCEA Calculus 2021 ONE(d - Solution & Explanation

Question

A curve is defined parametrically by the equations

x = t^2 + 3t

and

y = t^2 \ln(2t - 3)

, for

t > \dfrac{3}{2}

.

Find the gradient of the tangent to the curve at the point

(10,0)

.

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

Official Answer

x=t^2+3t

\dfrac{dx}{dt}=2t+3

y=t^2\ln(2t-3)

\dfrac{dy}{dt}=2t\ln(2t-3)+\dfrac{2t^2}{2t-3}

\dfrac{dy}{dx}=\dfrac{2t\ln(2t-3)+\dfrac{2t^2}{2t-3}}{2t+3}

At

(10,0):\ t^2+3t=10

t^2+3t-10=0

(t+5)(t-2)=0

t=-5\ \text{or}\ t=2

Since

t>\dfrac{3}{2},\ t=2

\dfrac{dy}{dx}=\dfrac{4\ln(1)+8}{7}

\dfrac{dy}{dx}=\dfrac{8}{7}

Grading Criteria

-

Video Explanation

NCEA Level 3 Calculus Differentiation 2021 NZQA Exam - Worked Answers by infinityplusone(starts at 13:19)

2021 ONE(d)