Question Number 44120 by peter frank last updated on 21/Sep/18
Answered by ajfour last updated on 21/Sep/18
![(dy/dx)=((dy/dθ)/(dx/dθ))=((asin θ)/(a+acos θ)) =((sin θ)/(1+cos θ)) (d^2 y/dx^2 )=[(d/dθ)((dy/dx))]/((dx/dθ)) =((cos θ(1+cos θ)−sin θ(−sin θ))/((1+cos θ)^2 ×a(1+cos θ))) = ((cos θ+1)/((1+cos θ)^2 ×a(1+cos θ))) ⇒ (d^( 2) y/dx^2 ) = (1/(a(1+cos θ)^2 )) (∀ θ≠2nπ+π).](https://www.tinkutara.com/question/Q44121.png)
$$\frac{{dy}}{{dx}}=\frac{{dy}/{d}\theta}{{dx}/{d}\theta}=\frac{{a}\mathrm{sin}\:\theta}{{a}+{a}\mathrm{cos}\:\theta}\:=\frac{\mathrm{sin}\:\theta}{\mathrm{1}+\mathrm{cos}\:\theta} \\ $$$$\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }=\left[\frac{{d}}{{d}\theta}\left(\frac{{dy}}{{dx}}\right)\right]/\left(\frac{{dx}}{{d}\theta}\right) \\ $$$$\:\:\:=\frac{\mathrm{cos}\:\theta\left(\mathrm{1}+\mathrm{cos}\:\theta\right)−\mathrm{sin}\:\theta\left(−\mathrm{sin}\:\theta\right)}{\left(\mathrm{1}+\mathrm{cos}\:\theta\right)^{\mathrm{2}} ×{a}\left(\mathrm{1}+\mathrm{cos}\:\theta\right)} \\ $$$$\:\:\:=\:\frac{\mathrm{cos}\:\theta+\mathrm{1}}{\left(\mathrm{1}+\mathrm{cos}\:\theta\right)^{\mathrm{2}} ×{a}\left(\mathrm{1}+\mathrm{cos}\:\theta\right)} \\ $$$$\Rightarrow\:\:\frac{{d}^{\:\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }\:=\:\frac{\mathrm{1}}{{a}\left(\mathrm{1}+\mathrm{cos}\:\theta\right)^{\mathrm{2}} }\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\forall\:\theta\neq\mathrm{2}{n}\pi+\pi\right). \\ $$
Commented by peter frank last updated on 21/Sep/18
$${the}\:{second}\:{line}\:{am}\:{not}\:{understood}\:{why}\:{you}\:{divide}\:\frac{{dx}}{{d}\theta}? \\ $$$$ \\ $$
Commented by ajfour last updated on 21/Sep/18
$$\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }\:=\:\frac{{d}}{{dx}}\:\left(\frac{{dy}}{{dx}}\right)\:=\:\frac{{d}}{{d}\theta}\left(\frac{{dy}}{{dx}}\right)×\frac{{d}\theta}{{dx}} \\ $$$$\:\:\:\:\:\:\:=\:\frac{{d}}{{d}\theta}\left(\frac{{dy}}{{dx}}\right)\:/\left(\frac{{dx}}{{d}\theta}\right)\:. \\ $$
Commented by peter frank last updated on 21/Sep/18
$$\mathrm{okay}\:\mathrm{sir}\:\mathrm{thank}\:\mathrm{you} \\ $$