A-Level Edexcel Maths Pure Laws of Indices & Surds: A curve C has parametric equatio

A curve C has parametric equations $$ x = 2 ext{sin}t, y = 1 - ext{cos}2t \quad \left( -\frac{\pi}{2} \leq t \leq \frac{\pi}{2} \right)$$ (a) Find $ \frac{dy}{dx} $ at the point where $ t = \frac{\pi}{6} $ (b) Find a cartesian equation for C in the form $ y = f(x), \ -k \leq x \leq k $, stating the value of the constant k - Edexcel - A-Level Maths Pure - Question 5 - 2013 - Paper 9

Question 5

$A-curve-C-has-parametric-equations--$$-x-=-2-ext{sin}t,---y-=-1----ext{cos}2t-\quad-\left(--\frac{\pi}{2}-\leq-t-\leq-\frac{\pi}{2}-\right)$$--(a)-Find-$-\frac{dy}{dx}-$-at-the-point-where-$-t-=-\frac{\pi}{6}-$--(b)-Find-a-cartesian-equation-for-C-in-the-form-$-y-=-f(x),-\--k-\leq-x-\leq-k-$,-stating-the-value-of-the-constant-k-Edexcel-A-Level Maths Pure-Question 5-2013-Paper 9.png$

A curve C has parametric equations $$ x = 2 ext{sin}t, y = 1 - ext{cos}2t \quad \left( -\frac{\pi}{2} \leq t \leq \frac{\pi}{2} \right)$$ (a) Find \( \frac{dy}{... show full transcript

Worked Solution & Example Answer:A curve C has parametric equations $$ x = 2 ext{sin}t, y = 1 - ext{cos}2t \quad \left( -\frac{\pi}{2} \leq t \leq \frac{\pi}{2} \right)$$ (a) Find $ \frac{dy}{dx} $ at the point where $ t = \frac{\pi}{6} $ (b) Find a cartesian equation for C in the form $ y = f(x), \ -k \leq x \leq k $, stating the value of the constant k - Edexcel - A-Level Maths Pure - Question 5 - 2013 - Paper 9

Step 1

Find $ \frac{dy}{dx} $ at the point where $ t = \frac{\pi}{6} $

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Answer

To find ( \frac{dy}{dx} ), we can use the chain rule, with:

Differentiate ( x ) and ( y ) with respect to ( t ): $\frac{dx}{dt} = 2\cos t \text{ and } \frac{dy}{dt} = 2\sin t$
So, ( \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{2\sin t}{2\cos t} = \tan t$$
Substitute ( t = \frac{\pi}{6} ): $\frac{dy}{dx} = \tan \left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}}$

Thus, ( \frac{dy}{dx} ) at the point where ( t = \frac{\pi}{6} ) is ( \frac{1}{\sqrt{3}} ).

Step 2

Find a cartesian equation for C in the form $ y = f(x), \ -k \leq x \leq k $

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Answer

From the parametric equations:

We have ( x = 2\sin t ). Thus, ( \sin t = \frac{x}{2} ).
Substitute ( \sin t ) into the equation for ( y ): [ y = 1 - \cos(2t) = 1 - (1 - 2\sin^2 t) = 2\sin^2 t = 2\left(\frac{x}{2}\right)^2 = \frac{x^2}{2} ]
Therefore, the Cartesian equation is: [ y = \frac{x^2}{2} ]
The range of x as ( -2 \leq x \leq 2 ). Thus, the constants for k will be ( k = 2 ).

Step 3

Write down the range of $ f(x) $

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Answer

To determine the range of ( f(x) = \frac{x^2}{2} ):

Given that ( x ) ranges from ( -2 ) to ( 2 ):
- When ( x = -2 ): ( f(-2) = \frac{(-2)^2}{2} = 2 )
- When ( x = 2 ): ( f(2) = \frac{(2)^2}{2} = 2 )
Since the function is a parabola that opens upwards, the minimum value occurs at ( x = 0 ): ( f(0) = 0 ).

Thus, the range of ( f(x) ) is ( [0, 2] ).

Find \( \frac{dy}{dx} \) at the point where \( t = \frac{\pi}{6} \)

Find a cartesian equation for C in the form \( y = f(x), \ -k \leq x \leq k \)

Write down the range of \( f(x) \)

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