Question

find (d²y/dx²)if x=a(theta+sin(theta))

,y=a(1-cos(theta))

Answer 1

$$\begin{gathered} x=a\left(\theta +\sin \theta \right)andy=a\left(1-\cos \theta \right) \\ Differentiatew.r.t.,\theta ,wehave, \\ \frac{dx}{d\theta }=a\frac{d}{d\theta }\left(\theta +\sin \theta \right)=a\left(1+\cos \theta \right) \\ \frac{dy}{d\theta }=a\frac{d}{d\theta }\left(1-\cos \theta \right)=a\sin \theta \\ So, \\ \frac{dy}{dx}=\frac{\frac{dy}{d\theta }}{\frac{dx}{d\theta }}=\frac{a\sin \theta }{a\left(1+\cos \theta \right)} \\ Differentiatingw.r.t.x, \end{gathered}$$
$$\begin{gathered} \frac{d^{2}y}{d{x}^2}=\frac{d}{dx}\left(\frac{a\sin \theta }{a\left(1+\cos \theta \right)}\right) \\ =\frac{a\left(1+\cos \theta \right)\frac{d}{dx}a\sin \theta -a\sin \theta \frac{d}{dx}a\left(1+\cos \theta \right)}{{\left(a\left(1+\cos \theta \right)\right)}^2} \\ =\frac{a^2\left(1+\cos \theta \right)\cos \theta \frac{d\theta }{dx}-a\sin \theta \left(-a\sin \theta \right)\frac{d\theta }{dx}}{{\left(a\left(1+\cos \theta \right)\right)}^2} \\ =\frac{a^2\left(1+\cos \theta \right)\cos \theta \times {\displaystyle \frac{1}{a\left(1+\cos \theta \right)}}-a\sin \theta \left(-a\sin \theta \right)\times \frac{1}{a\left(1+\cos \theta \right)}}{{\left(a\left(1+\cos \theta \right)\right)}^2} \\ =\frac{a^2\cos \theta \left(1+\cos \theta \right)+a^2{\sin }^2\theta }{{\left(a\left(1+\cos \theta \right)\right)}^3} \\ =\frac{a^2\cos \theta +\left(a^2{\cos }^2\theta +a^2{\sin }^2\theta \right)}{{\left(a\left(1+\cos \theta \right)\right)}^3} \\ =\frac{a^2\cos \theta +a^2}{{\left(a\left(1+\cos \theta \right)\right)}^3} \\ =\frac{1}{a{\left(1+\cos \theta \right)}^2} \\ \Rightarrow \frac{d^{2}y}{d{x}^2}=\frac{1}{a{\left(1+\cos \theta \right)}^2} \end{gathered}$$