2016 ONE(c)

Merit

Question

A curve is defined by the parametric equations

x = 2\cos 2t

and

y = \tan^2 t

.

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

t = \dfrac{\pi}{4}

.

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

Official Answer

\dfrac{dx}{dt}=-4\sin 2t

\dfrac{dy}{dt}=2\tan t\sec^2 t

\dfrac{dy}{dx}=\dfrac{2\tan t\sec^2 t}{-4\sin 2t}

\quad\;=\dfrac{2\tan t}{-4\sin 2t\cos^2 t}

At

t=\dfrac{\pi}{4}

,

\dfrac{dy}{dx}=\dfrac{2}{-4\times\left(\dfrac{1}{\sqrt{2}}\right)^2}

\quad\;=\dfrac{2}{-2}=-1

Grading Criteria

Achievement (u)

Correct $\dfrac{dx}{dt}$ or $\dfrac{dy}{dt}$

Merit (r)

Correct solution with correct derivatives.

Excellence T1

-

Excellence T2

-

Exam Paper (2016)Assessment Schedule (2016)

Related Questions

2024 THREE(e)Excellence

The diagram below shows part of the graph of the function $f(x) = e^{-x^2}$, where $x \geq 0$. [DI...

2024 THREE(c)Merit

Find the $x$-value(s) of any stationary point(s) on the graph of the function $f(x)=\dfrac{x^2-5x+4...

2024 THREE(d)Merit

Jamie is doing some baking and pouring the flour to form a conical pile. The height o...

2024 TWO(e)Excellence

The graph of the function $y = \dfrac{xe^{3x}}{2x + k}$, where $k$ is a non-zero constant, has a sin...

2024 TWO(c)Merit

Show that $y = \sin(x^2) - \cos(x)$ is a solution to the equation $\dfrac{d^2y}{dx^2} + 4x^2\,y = 2...