Question

*Find dy/dx, If x = cos t - cos 2t and y = sin t - sin 2t* 1️⃣ (cos t - 2cos 2t)/(2sin 2t - sin t) 2️⃣ (cos 2t - 2 cost)/(2 sin 2t - sint) 3️⃣ (cos t - cos 2t)/(2 sin 2t - sint) 4️⃣ None of these​

Answer 1

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

Given parametric functions are

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

and

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

Now, Consider

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

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

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

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

We know,

$\red{\rm :\longmapsto\:\boxed{ \tt{ \: \dfrac{d}{dx}cosx = - \: sinx \: }}}$

So, using this identity, we get

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

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

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

$\rm \implies\:\boxed{ \tt{ \: \dfrac{dx}{dt} = 2sin2t - sint \: }}$

Now, Consider

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

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

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

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

We know

$\red{\rm :\longmapsto\:\boxed{ \tt{ \: \dfrac{d}{dx}sinx \: =\: cosx \: }}}$

So, using this identity, we get

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

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

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

Thus,

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

Now,

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

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

$\rm \:  =  \:\dfrac{cost - 2cos2t}{2sin2t - sint}$

$\red{\rm \implies\:\boxed{ \tt{ \: \dfrac{dy}{dx} \: = \:\dfrac{cost - 2cos2t}{2sin2t - sint} \: }}}$

• Hence, Option (1) is correct

More to know :-

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