NCEA Calculus 2018 ONE(e - Solution & Explanation

Question

A curve is defined by the parametric equations

x = t^3 + 1

y = t^2 + 1

Show that

\dfrac{\dfrac{d^2y}{dx^2}}{\left(\dfrac{dy}{dx}\right)^4}

is a constant.

Official Answer

\dfrac{dx}{dt}=3t^2\qquad \dfrac{dy}{dt}=2t

\dfrac{dy}{dx}=\dfrac{dy}{dt}\times\dfrac{dt}{dx}=2t\cdot\dfrac{1}{3t^2}=\dfrac{2}{3t}

\dfrac{d^2y}{dx^2}=\dfrac{d\left(\dfrac{dy}{dx}\right)}{dt}\times\dfrac{dt}{dx}

=\dfrac{-2}{3t^2}\times\dfrac{1}{3t^2}=\dfrac{-2}{9t^4}

\left(\dfrac{dy}{dx}\right)^4=\left(\dfrac{2}{3t}\right)^4

\dfrac{\dfrac{d^2y}{dx^2}}{\left(\dfrac{dy}{dx}\right)^4}=\dfrac{-2}{9t^4}\div\left(\dfrac{2}{3t}\right)^4

=\dfrac{-2}{9t^4}\times\dfrac{81t^4}{16}

=\dfrac{-9}{8}\text{ or }-1.125

Grading Criteria

-

Video Explanation

NCEA Level 3 Calculus Differentiation 2018 NZQA Exam - Worked Answers by infinityplusone(starts at 14:30)

2018 ONE(e)