Show that $\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{2\sin\theta}{2\sin\theta-1}$

Question

Parametric equation of curve is $ x=2\sin\theta +\cos(2\theta) $ , $y=1+\cos(2\theta)$ , for $0\leq\theta\leq\pi/2$

Show that $$\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{2\sin\theta}{2\sin\theta-1}$$ I drived $x$ and $y$ seperatly for $x$,

$$ x=2\cos\theta+\sin\theta\pm\sin\theta$$ and for the

derivative of $y$ i got $y=-2\sin(2\theta)$

and then I did

$$dy/dx= y=-2\sin(2\theta), \qquad x=2\cos\theta+\sin\theta\pm \sin\theta$$

but I dont know if I did it wronge or I don't know how to continue. Any help would be much appreciated. Thank you!

Answer 1

$$\frac{dy}{dx}=\frac{dy}{d\theta}/\frac{dx}{d\theta}=\frac{-2\sin 2\theta}{2\cos \theta-2\sin 2 \theta}=\frac{-\sin \theta \cos \theta}{\cos \theta(1-2\sin \theta)}=\frac{\sin \theta}{2\sin \theta-1}.$$

Answer 2

You need to make use of the Chain Rule: https://en.wikipedia.org/wiki/Chain_rule.

$\frac{dy}{dx} = \frac{dy}{d\theta} \frac{d\theta}{dx}$.

The first term on the RHS is simple to calculate since you have been given $y$ as a function of $\theta$ but the second term requires a bit more attention. You have been given $x(\theta)$ and you want to get $\frac{d\theta}{dx}$.

$x = x(\theta)$

$\implies \frac{d(x)}{dx} = \frac{d(x(\theta))}{dx}$

$\implies \frac{d(x)}{dx} = \frac{d(x(\theta))}{d\theta} \frac{d\theta}{dx}$

$ \therefore \frac{d\theta}{dx} = \frac{1}{\frac{d(x(\theta))}{d\theta}}$

Apply these steps and also be mindful of you trigonometric identities. Good luck!

Answer 3

$$\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{dy/d \theta}{dx/d \theta}=\frac{-2\sin 2\theta}{2 \cos \theta-2\sin 2\theta}$$

Let $ \sin \theta =s, \cos \theta= c\;$ We have

$$ \dfrac{2\cdot-2 sc }{2c-2\cdot 2\cdot sc }=\dfrac{s}{2s-1}$$