Question

X=asin2t(1+cos2t) and y=b cos2t(1-cos2t) find dy/dx

Answer 1

I hope solve this answer

Answer 2

Answer:

$\dfrac{dy}{dx}=\dfrac{b\left[\sin{4t}-\sin{2t}\right]}{a\left[\cos{2t}+\cos{4t}\right]}$

Step-by-step explanation:

Given,

$x=a\sin{2t}(1+\cos{2t})\;\;and\;\;y=b\cos{2t}(1-\cos{2t})$

So, we have

$\dfrac{dx}{dt}=\dfrac{d}{dt}\left[a\sin{2t}(1+\cos{2t})\right]\\ \dfrac{dx}{dt}=a\left[\sin{2t}(-2sin{2t})+(1+\cos{2t})2\cos{2t}\right]\\ \dfrac{dx}{dt}=2a\left[\cos{2t}+\cos^2{2t}-\sin^2{2t}\right]\\ \dfrac{dx}{dt}=2a\left[\cos{2t}+\cos{4t}\right]$

and

$\dfrac{dy}{dt}=\dfrac{d}{dt}\left[b\cos{2t}(1-\cos{2t})\right]\\ \dfrac{dy}{dt}=b\left[\cos{2t}(2\sin{2t})-(1-\cos{2t})2\sin{2t}\right]\\ \dfrac{dy}{dt}=2b\left[\sin{2t}\cos{2t}-sin{2t}+sin{2t}\cos{2t}\right]\\ \dfrac{dy}{dt}=2b\left[2\sin{2t}\cos{2t}-sin{2t}\right]\\ \dfrac{dy}{dt}=2b\left[\sin{4t}-sin{2t}\right]$

so we have the solution:

$\dfrac{dy}{dx}=\dfrac{\frac{dy}{dt}}{\frac{dx}{dt}}\\ \dfrac{dy}{dx}=\dfrac{2b\left[\sin{4t}-sin{2t}\right]}{2a\left[\cos{2t}+\cos{4t}\right]}\\ \dfrac{dy}{dx}=\dfrac{b\left[\sin{4t}-\sin{2t}\right]}{a\left[\cos{2t}+\cos{4t}\right]}$