Question and Answers Forum

All Questions      Topic List

Differential Equation Questions

Previous in All Question      Next in All Question      

Previous in Differential Equation      Next in Differential Equation      

Question Number 94954 by Mr.D.N. last updated on 22/May/20

$$\mathbf{For}\mathbf{any}\mathbf{curve},\mathbf{prove}\mathbf{that}:$$$${\left(\frac{{\mathrm{d}}^2\mathrm{x}}{{\mathrm{ds}}^2}\right)}^2+{\left(\frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2}\right)}^2=\frac{1}{{\rho }^2}$$$$$$

Answered by niroj last updated on 22/May/20

$${\mathrm{Sol}}^{\mathrm{n}}:$$$$\mathrm{We}\mathrm{know}\mathrm{that},$$$$\frac{\mathrm{dx}}{\mathrm{ds}}=\cos \psi \Rightarrow \frac{{\mathrm{d}}^2\mathrm{x}}{{\mathrm{ds}}^2}=-\sin \psi \frac{\mathrm{d}\psi }{\mathrm{ds}}$$$$\mathrm{and}$$$$\frac{\mathrm{dy}}{\mathrm{ds}}=\sin \psi \Rightarrow \frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{ds}}^2}=\cos \psi \frac{\mathrm{d}\psi }{\mathrm{ds}}$$$$\mathrm{Now},{\left(\frac{{\mathrm{d}}^2\mathrm{x}}{{\mathrm{ds}}^2}\right)}^2+\left(\frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{ds}}^2}\right)$$$$={(-\sin \psi \frac{\mathrm{d}\psi }{\mathrm{ds}})}^2+{\left(\mathrm{cod}\psi \frac{\mathrm{d}\psi }{\mathrm{ds}}\right)}^2$$$$={\sin }^2\psi {\left(\frac{\mathrm{d}\psi }{\mathrm{ds}}\right)}^2+{\cos }^2\psi {\left(\frac{\mathrm{d}\psi }{\mathrm{ds}}\right)}^2$$$$={\left(\frac{\mathrm{d}\psi }{\mathrm{ds}}\right)}^2({\sin }^2\psi +{\cos }^2\psi )$$$$=1.\frac{1}{{\rho }^2}=\frac{1}{{\rho }^2}//.$$$$$$

Commented by Mr.D.N. last updated on 22/May/20

outstanding��✍

Terms of Service

Privacy Policy

Contact: info@tinkutara.com