Figure 2 shows a sketch of the curve C with parametric equations $x = ext{(}rac{ oot{3} ext{}}{2} ext{)} ext{sin }2t$, $y = 4 ext{cos}^2 t, 0 ext{ } ext{≤ }t ext{ ≤ } ext{π}$ (a) Show that $rac{dy}{dx} = k(rac{ oot{3}}{3}) an 2t$, where $k$ is a constant to be determined. (b) Find an equation of the tangent to C at the point where $t = rac{π}{3}$. Give your answer in the form $y = ax + b$, where $a$ and $b$ are constants. (c) Find a Cartesian equation of C.

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Figure 2 shows a sketch of the curve C with parametric equations $x = ext{(}rac{ oot{3} ext{}}{2} ext{)} ext{sin }2t$, $y = 4 ext{cos}^2 t, 0 ext{ } ext{≤ }t ext{ ≤ } ext{π}$ (a) Show that $rac{dy}{dx} = k(rac{ oot{3}}{3}) an 2t$, where $k$ is a constant to be determined - Edexcel - A-Level Maths Pure - Question 6 - 2012 - Paper 7

Question 6

$Figure-2-shows-a-sketch-of-the-curve-C-with-parametric-equations--$x-=---ext{(}-rac{-oot{3}-ext{}}{2}-ext{)}-ext{sin--}2t$,---$y-=-4--ext{cos}^2-t,---0--ext{--}-ext{≤--}t---ext{-≤--}-ext{π}$--(a)-Show-that-$-rac{dy}{dx}-=-k(-rac{-oot{3}}{3})-an-2t$,-where-$k$-is-a-constant-to-be-determined-Edexcel-A-Level Maths Pure-Question 6-2012-Paper 7.png$

Figure 2 shows a sketch of the curve C with parametric equations $x = ext{(}rac{ oot{3} ext{}}{2} ext{)} ext{sin }2t$, $y = 4 ext{cos}^2 t, 0 ext{ } ext{... show full transcript

Worked Solution & Example Answer:Figure 2 shows a sketch of the curve C with parametric equations $x = ext{(}rac{ oot{3} ext{}}{2} ext{)} ext{sin }2t$, $y = 4 ext{cos}^2 t, 0 ext{ } ext{≤ }t ext{ ≤ } ext{π}$ (a) Show that $rac{dy}{dx} = k(rac{ oot{3}}{3}) an 2t$, where $k$ is a constant to be determined - Edexcel - A-Level Maths Pure - Question 6 - 2012 - Paper 7

Step 1

Show that $\frac{dy}{dx} = k(\frac{\sqrt{3}}{3}) \tan 2t$

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Answer

To find $\frac{dy}{dx}$ , we start by calculating $\frac{dx}{dt}$ and $\frac{dy}{dt}$ :

Find $\frac{dx}{dt}$ :
[\frac{dx}{dt} = \frac{d}{dt}(\sqrt{3}\sin 2t) = \sqrt{3}\cdot 2\cos 2t = 2\sqrt{3}\cos 2t]\
Find $\frac{dy}{dt}$ :
[\frac{dy}{dt} = \frac{d}{dt}(4\cos^2 t) = 4\cdot 2\cos t(-\sin t) = -8\cos t \sin t]\
Find $\frac{dy}{dx}$ :
[\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{-8\cos t \sin t}{2\sqrt{3}\cos 2t}]\

Using the identity $\cos 2t = \cos^2 t - \sin^2 t$ , we simplify:
[\frac{dy}{dx} = \frac{-8\cos t \sin t}{2\sqrt{3}(\cos^2 t - \sin^2 t)}]\

Factoring out $-4$ gives:
[\frac{dy}{dx} = -\frac{4\sin(2t)}{\sqrt{3} \cos(2t)}]\

Now using $\tan(2t) = \frac{\sin(2t)}{\cos(2t)}$ , we express this as:
[\frac{dy}{dx} = -\frac{4}{\sqrt{3}}\tan 2t]\

Thus, we have:
[\frac{dy}{dx} = k(\frac{\sqrt{3}}{3})\tan 2t, \text{ where } k = -4]

Step 2

Find an equation of the tangent to C at the point where $t = \frac{π}{3}$

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Answer

First, we need to find the coordinates of the point when $t = \frac{π}{3}$ :

Calculate $x$ and $y$ :
[x = \sqrt{3}\sin(2 \cdot \frac{\pi}{3}) = \sqrt{3}\sin(\frac{2\pi}{3}) = \sqrt{3}\cdot \frac{\sqrt{3}}{2} = \frac{3}{2}]\
[y = 4\cos^2(\frac{\pi}{3}) = 4\cdot(\frac{1}{2})^2 = 4\cdot\frac{1}{4} = 1]\

Now, we find the slope of the tangent line:

Calculate the slope (m):
[m = \frac{dy}{dx} \text{ at } t = \frac{\pi}{3} = -\frac{4}{\sqrt{3}}\tan(2 \cdot \frac{\pi}{3})]

Since $\tan(\frac{2\pi}{3}) = -\sqrt{3}$ , we have:
[m = -\frac{4}{\sqrt{3}}(-\sqrt{3}) = 4]\

Use point-slope form to write the equation of the tangent:
The point is $\left(\frac{3}{2}, 1\right)$ and the slope is $4$ : [y - 1 = 4(x - \frac{3}{2})]
Simplifying gives:
[y = 4x - 6 + 1 = 4x - 5]\
Hence, in the form $y = ax + b$ : $a = 4$ , $b = -5$ .

Step 3

Find a Cartesian equation of C

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Answer

We start from the parametric equations:
[x = \sqrt{3}\sin 2t, \quad y = 4\cos^2 t]\

Express $\sin 2t$ in terms of $y$ :
From $y = 4\cos^2 t$ , we have $\cos^2 t = \frac{y}{4}$ . Using $\sin^2 t = 1 - \cos^2 t$ , we can write: [\sin^2 t = 1 - \frac{y}{4} = \frac{4-y}{4}]\

Since $\sin 2t = 2 \sin t \cos t$ , we need $\sin t$ and $\cos t$ : [\sin^2 t = \frac{4-y}{4}, \quad \cos^2 t = \frac{y}{4}]\

Use these in the equation for $\sin 2t$ :
[\sin 2t = 2\sqrt{\frac{4-y}{4}} \sqrt{\frac{y}{4}} = \frac{1}{2}\sqrt{(4-y)y}]\
Substituting in the equation:
Now, since $x = \sqrt{3}\sin 2t$ , we have
[x = \sqrt{3}\cdot\frac{1}{2}\sqrt{(4-y)y}]

Squaring gives:
[x^2 = \frac{3}{4}(4-y)y]\

Which can be rearranged to find the Cartesian equation.

Show that $\frac{dy}{dx} = k(\frac{\sqrt{3}}{3}) \tan 2t$

Find an equation of the tangent to C at the point where $t = \frac{π}{3}$

Find a Cartesian equation of C

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