2013 ONE(d)

Merit

Question

A curve is defined by the parametric equations:

x = 5\sin t

and

y = 3\tan t

Find the gradient of the normal to the curve at the point where

t = \dfrac{\pi}{3}

.

*Show any derivatives that you need to find when solving this problem.*

Official Answer

\dfrac{dx}{dt}=5\cos t

\dfrac{dy}{dt}=3\sec^2 t=\dfrac{3}{\cos^2 t}

\dfrac{dy}{dx}=\dfrac{3}{5\cos^3 t}

At

t=\dfrac{\pi}{3}

,

\dfrac{dy}{dx}=\dfrac{24}{5}\ ( =4.8)

\therefore

gradient of normal

=\dfrac{-5}{24}\ (= -0.2083)

Grading Criteria

Achievement (u)

Correct $\dfrac{dx}{dt}$ and $\dfrac{dy}{dt}$ .

Merit (r)

Correct solution including all correct derivatives.

Excellence T1

-

Excellence T2

-

Exam Paper (2013)Assessment Schedule (2013)

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