2013 THREE(d)

Merit
Question
A curve is defined by the parametric equations:

x=t2−tx = t^2 - t and y=t3−3ty = t^3 - 3t

Find the coordinates of the point(s) on the curve for which the normal to the curve is parallel
to the yy-axis.

*You must use calculus and clearly show your working, including any derivatives you need to
find when solving this problem.*
Official Answer
For the curve, dydx=3t2−32t−1\dfrac{dy}{dx}=\dfrac{3t^2-3}{2t-1}

Normal parallel to the yy-axis means tangent parallel to the xx-axis.

⇒dydx=0\Rightarrow\dfrac{dy}{dx}=0

3t2=33t^2=3

t=±1t=\pm 1

t=1⇒t=1\Rightarrow point (0,−2)(0,-2)

t=−1⇒t=-1\Rightarrow point (2,2)(2,2)
Grading Criteria

Achievement (u)

  • Correct expression for dydx\dfrac{dy}{dx}

Merit (r)

  • Correct solution with correct derivative.

Excellence T1

-

Excellence T2

-
Exam Paper (2013)Assessment Schedule (2013)
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