NCEA Calculus 2017 ONE(d - Solution & Explanation

Question

A curve is defined parametrically by the equations

x = \sqrt{t + 1}

and

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.*

Official Answer

\dfrac{dx}{dt}=\dfrac{1}{2}(t+1)^{-\frac{1}{2}}=\dfrac{1}{2\sqrt{t+1}}

\dfrac{dy}{dt}=2\cos 2t

\dfrac{dy}{dx}=\dfrac{dy/dt}{dx/dt}=\dfrac{2\cos 2t}{\dfrac{1}{2\sqrt{t+1}}}

=4\cos 2t\,\sqrt{t+1}

At

t=0

\dfrac{dy}{dx}=4\cos 0\times \sqrt{1}

=4

Grading Criteria

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Video Explanation

NCEA Level 3 Calculus Differentiation 2017 NZQA Exam - Worked Answers by infinityplusone(starts at 13:54)

2017 ONE(d)