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Question Number 54880 by Otchere Abdullai last updated on 14/Feb/19

$$If{x}^2-y^2=a^{2}find\frac{d^{2}y}{{dx}^2}ifais$$$$constant.$$

Answered by kaivan.ahmadi last updated on 14/Feb/19

$$\mathrm{f}(\mathrm{x},\mathrm{y})={\mathrm{x}}^2-{\mathrm{y}}^2-{\mathrm{a}}^2=0$$$${\mathrm{y}}_{\mathrm{x}}^{{}^{\prime }}=\frac{\mathrm{dy}}{\mathrm{dx}}=-\frac{{\mathrm{f}}_{\mathrm{x}}}{{\mathrm{f}}_{\mathrm{y}}}=-\frac{2\mathrm{x}}{-2\mathrm{y}}=\frac{\mathrm{x}}{\mathrm{y}}$$$$\frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2}=\frac{1\times \mathrm{y}-{\mathrm{y}}_{\mathrm{x}}^{{}^{\prime }}\times \mathrm{x}}{{\mathrm{y}}^2}=\frac{\mathrm{y}-\frac{{\mathrm{x}}^2}{\mathrm{y}}}{{\mathrm{y}}^2}=\frac{{\mathrm{y}}^2-{\mathrm{x}}^2}{{\mathrm{y}}^3}$$

Commented by Otchere Abdullai last updated on 14/Feb/19

$$Thankssir!$$

Answered by tanmay.chaudhury50@gmail.com last updated on 14/Feb/19

$$y^2=x^2-a^2$$$$2y\frac{dy}{dx}=2x\to \frac{dy}{dx}=\frac{x}{y}$$$$y\times \frac{d^{2}y}{{dx}^2}+{\left(\frac{dy}{dx}\right)}^2=1$$$$y\left(\frac{d^{2}y}{{dx}^2}\right)=1-{\left(\frac{x}{y}\right)}^2$$$$\frac{d^{2}y}{{dx}^2}=\frac{y^2-x^2}{y^3}\to \frac{d^{2}y}{{dx}^2}=\frac{-a^2}{y^3}$$$$\frac{d^{2}y}{{dx}^2}=\frac{-a^2}{{(x^2-a^2)}^{\frac{3}{2}}}$$$$$$

Commented by Otchere Abdullai last updated on 14/Feb/19

$$Thanksprof$$

Commented by tanmay.chaudhury50@gmail.com last updated on 14/Feb/19

$$iamnotaprofessorbut49yearsoldindianinterestd$$$$inmathematicsphysicsandmythology,metaphysics$$$$unsolvedmystry...$$

Commented by Otchere Abdullai last updated on 14/Feb/19

$$yesigaveyoutheprofessorbecause$$$$youhavedonegreatcontributionto$$$$thegroup$$

Commented by $@ty@m last updated on 14/Feb/19

$$Yeah.$$$$Hedeserevesthetitleof‘‘Honorary{Professor}^{″}.$$

Commented by Otchere Abdullai last updated on 14/Feb/19

$$Yesyouareright{don}^{\prime }tforgetMrW$$$$wasthefirsthonouredprofessor$$$$soialwaysuseprofWinsteredofMr.$$

Commented by rahul 19 last updated on 14/Feb/19

$$Notonlyhim,butMrWsir,Mjssir,$$$$Ajfoursirtoooo!$$

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