Derivation of a curve that is not directly expressed

Question

I have read the question below and I couldn't figure out what it was asking:

Curve $C(x,y)$ is defined as below. Find $\frac{dy}{dx}$ expressed in $a$.
$$x=a-\sin(a)~, \quad y=1-\cos(a)$$

The problem of mine with that question is that I cannot rewrite the curve equation linking $y$ directly to $x$.

Every answer is appreciated in advance.

Answer 1

Hint: $\dfrac{\operatorname dy}{\operatorname dx}=\dfrac {\frac{\operatorname dy}{\operatorname da}}{\frac{\operatorname dx}{\operatorname da}}$. (This is Leibniz's suggestive notation.)

Answer 2

Quantity which is to be find $\cfrac{dy}{dx}$, $$ \cfrac{dy}{dx} = \cfrac{dy/da}{dx/da} = \cfrac{\sin(a)}{1 - \cos(a)} $$

Answer 3

Parametric Equation of Cycloid is $$x=a-\sin(a)~, \quad y=1-\cos(a)$$ Now from the above equation $$\cos(a)=1-y$$$$\implies\sin(a)=\sqrt{1-(1-y)^2}$$$$\implies\sin(a)=\sqrt{2y-y^2}$$ $$\implies a=\sin^{-1}\left(\sqrt{2y-y^2}\right)$$

So $$x=a-\sin(a)\implies x=\sin^{-1}\left(\sqrt{2y-y^2}\right)~-~\sqrt{2y-y^2}$$ $$\implies x~+~\sqrt{2y-y^2}=\sin^{-1}\left(\sqrt{2y-y^2}\right)$$

This is the relation between $~x~$ and $~y~$.

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Here $$x=a-\sin(a)~, \quad y=1-\cos(a)$$ Now $$\frac{dx}{da}=1-\cos(a)\qquad \text{and}\quad \frac{dy}{da}=\sin(a)$$ $$\implies \frac{dy}{dx}=\frac{\frac{dy}{da}}{\frac{dx}{da}}=\frac{\sin(a)}{1-\cos(a)}$$

Answer 4

You need to handle parametrization all the way ( both differentiation and integration) $$ y^{'}=\cfrac{dy}{dx} = \cfrac{dy/da}{dx/da} = \cfrac{\sin(a)}{1 - \cos(a)} =\tan(a/2)$$

$$ A= 2 \pi \int _{a1} ^{a2} y\sqrt{1+y^{'2}} dx =2\pi \int _{a1} ^{a2} y\sqrt{1+y^{'2}} \frac{dx}{da} da $$ $$=2\pi \int _{a1} ^{a2} (1-\cos \alpha) .\sec(a/2). (1-\cos \alpha). da $$

Can you express $a/2$ argument trig function in the middle in terms of $a$ and take the integration from there?