A-Level Edexcel Maths Pure Differential Equations: 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 6 - 2013 - Paper 9

Question 6

$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 6-2013-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... 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 6 - 2013 - Paper 9

Step 1

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

96%

114 rated

Answer

To find ( \frac{dy}{dx} ), we will first calculate ( \frac{dx}{dt} ) and ( \frac{dy}{dt} ).

Differentiate ( x = 2 \text{sin} , t ): $\frac{dx}{dt} = 2 \cos t$
Differentiate ( y = 1 - \text{cos} , 2t ): $\frac{dy}{dt} = 2 \sin 2t$
Substitute these into ( \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} ): $\frac{dy}{dx} = \frac{2 \sin 2t}{2 \cos t} = \frac{\sin 2t}{\cos t}$
Now, substitute ( t = \frac{\pi}{6} ):
- Calculate ( \sin 2 \left( \frac{\pi}{6} \right) = \sin \left( \frac{\pi}{3} \right) = \frac{\sqrt{3}}{2} )
- Calculate ( \cos \left( \frac{\pi}{6} \right) = \frac{\sqrt{3}}{2} )
Thus: $\frac{dy}{dx} = \frac{\frac{\sqrt{3}}{2}}{\frac{\sqrt{3}}{2}} = 1$

Step 2

Find a cartesian equation for C in the form $ y = f(x), \ -k \leq x \leq k $ and state the value of the constant k.

99%

104 rated

Answer

We start from the parametric equations: $x = 2 \sin t \implies \sin t = \frac{x}{2}$

Substituting this into the equation for y: $y = 1 - \text{cos} \, 2t$ Using the double angle formula: $\text{cos} \, 2t = 1 - 2\sin^2 t$

Substituting ( \sin t ): $\text{cos} \, 2t = 1 - 2\left( \frac{x}{2} \right)^2 = 1 - \frac{x^2}{2}$

Thus: $y = 1 - \left( 1 - \frac{x^2}{2} \right) = \frac{x^2}{2}$

The resulting Cartesian equation is: $y = \frac{x^2}{2}$

To find the range of x, given ( -\frac{\pi}{2} \leq t \leq \frac{\pi}{2} ):

At ( t = -\frac{\pi}{2}, x = 2 \sin \left( -\frac{\pi}{2} \right) = -2 )
At ( t = \frac{\pi}{2}, x = 2 \sin \left( \frac{\pi}{2} \right) = 2 )

Thus, ( k = 2 ) and the range is ( -2 \leq x \leq 2 ).

Step 3

Write down the range of $ f(x) $.

96%

101 rated

Answer

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

Since the minimum value of ( x^2 ) is 0 (when ( x = 0 )), the minimum value of ( f(x) ) is: $f(0) = \frac{0^2}{2} = 0$
The maximum value at ( x = \pm 2 ): $f(2) = \frac{2^2}{2} = 2$

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 \( y = f(x), \ -k \leq x \leq k \) and state the value of the constant k.

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

Join the A-Level students using SimpleStudy...