Question

x=2cost -cos2t y=2sint -sin2t find dy/dx

Answer 1

$\large\underline{\sf{Solution-}}$

Given that,

$\rm \: x = 2cost - cos2t$

and

$\rm \: y = 2sint - sin2t$

Now, Consider

$\rm \: x = 2cost - cos2t$

On differentiating both sides w. r. t. t, we get

$\rm \: \dfrac{d}{dt}x = \dfrac{d}{dt}(2cost - cos2t)$

$\rm \: \dfrac{dx}{dt} = 2\dfrac{d}{dt}cost - \dfrac{d}{dt}cos2t$

$\rm \: \dfrac{dx}{dt} = - 2sint - ( - 2sin2t)$

$\rm \: \dfrac{dx}{dt} = - 2sint + 2sin2t$

$\rm \: \dfrac{dx}{dt} = 2( - sint + sin2t )$

We know,

$\boxed{\sf{  \:sinx - siny = 2cos\bigg[\dfrac{x + y}{2} \bigg]sin\bigg[\dfrac{x - y}{2} \bigg] \: }}$

So, using this result, we get

$\rm \: \dfrac{dx}{dt} = 4cos\bigg[\dfrac{2t + t}{2} \bigg]sin\bigg[\dfrac{2t - t}{2} \bigg]$

$\rm \: \dfrac{dx}{dt} = 4cos\bigg[\dfrac{3t}{2} \bigg]sin\bigg[\dfrac{t}{2} \bigg]$

Now, Consider

$\rm \: y = 2sint - sin2t$

On differentiating both sides w. r. t. t, we get

$\rm \: \dfrac{d}{dt}y = \dfrac{d}{dt}(2sint - sin2t)$

$\rm \: \dfrac{dy}{dt} = 2 \dfrac{d}{dt}sint - \dfrac{d}{dt}sin2t$

$\rm \: \dfrac{dy}{dt} = 2cost - 2cos2t$

$\rm \: \dfrac{dy}{dt} = 2(cost - cos2t)$

We know,

$\boxed{\sf{  \:cosx - cosy = 2sin\bigg[\dfrac{x + y}{2} \bigg]sin\bigg[\dfrac{y - x}{2} \bigg] \: }}$

So, using this result, we get

$\rm \: \dfrac{dy}{dt} = 4sin\bigg[\dfrac{2t + t}{2} \bigg]sin\bigg[\dfrac{2t - t}{2} \bigg]$

$\rm \: \dfrac{dy}{dt} = 4sin\bigg[\dfrac{3t}{2} \bigg]sin\bigg[\dfrac{ t}{2} \bigg]$

Thus,

$\rm \: \dfrac{dy}{dx}$

$\rm \: =  \: \: \dfrac{dy}{dt} \div \dfrac{dx}{dt}$

$\rm \: = 4sin\bigg[\dfrac{3t}{2} \bigg]sin\bigg[\dfrac{ t}{2} \bigg] \: \div \: 4cos\bigg[\dfrac{3t}{2} \bigg]sin\bigg[\dfrac{ t}{2} \bigg]$

$\rm \: =  \: \: tan\bigg(\dfrac{3t}{2} \bigg)$

Hence,

$\rm\implies \:\rm \:\dfrac{dy}{dx} =  \: \: tan\bigg(\dfrac{3t}{2} \bigg)$

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Formulae Used :-

$\boxed{\sf{  \:\rm \: \dfrac{d}{dx}sinx = cosx \: }}$

$\boxed{\sf{  \:\rm \: \dfrac{d}{dx}cosx = \: - \: sinx \: }}$

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Additional Information :-

$\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \dfrac{d}{dx}f(x) \\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf 0 \\ \\ \sf sinx & \sf cosx \\ \\ \sf cosx & \sf - \: sinx \\ \\ \sf tanx & \sf {sec}^{2}x \\ \\ \sf cotx & \sf - {cosec}^{2}x \\ \\ \sf secx & \sf secx \: tanx\\ \\ \sf cosecx & \sf - \: cosecx \: cotx\\ \\ \sf \sqrt{x} & \sf \dfrac{1}{2 \sqrt{x} } \\ \\ \sf logx & \sf \dfrac{1}{x}\\ \\ \sf {e}^{x} & \sf {e}^{x} \end{array}} \\ \end{gathered}$

Answer 2

Step-by-step explanation:

$\color{brown} \begin{aligned} \tt\frac{d x}{d t} & \tt=-2 \sin t+2 \sin 2 t \\ \\ & \tt=2(\sin 2 t-\sin t) \\ \\ & \tt=2\left(2 \cos \frac{3 t}{2} \sin \frac{t}{2}\right) \\ \\ & \tt =4 \cos \frac{3 t}{2} \sin \frac{t}{2} \\ \\ \tt \: y & \tt=2 \sin t-\sin 2 t \end{aligned}$

$\color{red} \begin{aligned} \tt\frac{d y}{d x} & \tt=2 \cos t-2 \cos 2 t \\ \\ & \tt=2(\cos t-\cos 2 t) \\ \\ & \tt=2\left(2 \sin \frac{3 t}{2} \sin \frac{t}{2}\right) \\ \\ & \tt=4 \sin \frac{3 t}{2} \sin \frac{t}{2} \end{aligned}$

$\color{darkcyan} \begin{aligned} \tt\frac{d y}{d x} & \tt=\frac{d y / d t}{d x / d t} \\ \\ & \tt=\dfrac{4 \sin \dfrac{3 t}{2} \sin \dfrac{t}{2}}{4 \cos \dfrac{3 t}{2} \sin \dfrac{t}{2}} \\ \\ & \blue{\tt=\tan \frac{3 t}{2} } \rule{1pt}{2pt}\end{aligned}$