A-Level Edexcel Maths Pure Modelling with Functions: Figure 2 shows a sketch of the c

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

Question 7

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

Figure 2 shows a sketch of the curve C with parametric equations $x = oldsymbol{rac{oldsymbol{ ext{ ext{√3}}}}{ ext{sin} oldsymbol{2t}}}, oldsymbol{y} = 4 old... show full transcript

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

Step 1

Show that $\frac{dy}{dx} = k\left(\frac{\sqrt{3}}{\tan 2t}\right)$, where k is a constant to be determined.

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Answer

To find $\frac{dy}{dx}$ , we first need to calculate $\frac{dx}{dt}$ and $\frac{dy}{dt}$ .

Differentiate $x$ with respect to $t$ :

$x = \sqrt{3} \sin 2t \implies \frac{dx}{dt} = \sqrt{3} \cdot 2 \cos 2t = 2\sqrt{3} \cos 2t$
Differentiate $y$ with respect to $t$ :

$y = 4 \cos t \implies \frac{dy}{dt} = -4 \sin t$
Now, use the chain rule to find $\frac{dy}{dx}$ :

$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{-4 \sin t}{2\sqrt{3} \cos 2t} = \frac{-2 \sin t}{\sqrt{3} \cos 2t}$
Use the identity $\sin 2t = 2 \sin t \cos t$ and $\tan 2t = \frac{\sin 2t}{\cos 2t} = 2 \frac{\sin t \cos t}{\cos 2t}$ :

So we will rewrite it as:

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

To find $k$ , we can compare and determine that $k = \frac{2}{3}$ .

Step 2

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

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Answer

Calculate the coordinates at $t = \frac{\pi}{3}$ :

$x = \sqrt{3} \sin(2 \cdot \frac{\pi}{3}) = \sqrt{3} \sin\left(\frac{2\pi}{3}\right) = \sqrt{3} \cdot \frac{\sqrt{3}}{2} = \frac{3}{2}$

$y = 4 \cos\left(\frac{\pi}{3}\right) = 4 \cdot \frac{1}{2} = 2$

So the point is $\left(\frac{3}{2}, 2\right)$ .
Find the slope of the tangent line:

$m = \frac{dy}{dx} \bigg|_{t = \frac{\pi}{3}} = \frac{\sqrt{3}}{3} \tan\left(\frac{2\pi}{3}\right) = -2$
Use the point-slope form to find the equation of the tangent line:

$y - 2 = -2\left(x - \frac{3}{2}\right)$

Simplifying gives:

$y = -2x + 3 + 2 = -2x + 5$

Step 3

Find a Cartesian equation of C.

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Answer

To eliminate the parameter $t$ , start with:

$x = \sqrt{3} \sin 2t \\ \Rightarrow\ sin 2t = \frac{x}{\sqrt{3}}$

We need to use the corresponding parametric relation:

$y = 4 \cos t$

Knowing that $\sin^2 t + \cos^2 t = 1$ , we substitute:

$\sin^2 t = 1 - \left(\frac{y}{4}\right)^2$

Using double angle identity:

$\sin^2 2t = 4\sin^2 t \cos^2 t$

The Cartesian equation simplifies to:

$x^2 = 12 \left(1 - \frac{y^2}{16}\right)\\ = 12 - \frac{3y^2}{4}$

Show that $\frac{dy}{dx} = k\left(\frac{\sqrt{3}}{\tan 2t}\right)$, where k is a constant to be determined.

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

Find a Cartesian equation of C.

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