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

For any curve, prove that: ((d^2 x/ds^2 ))^2 +((d^2 y/dx^2 ))^2 = (1/ρ^2 )

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

Answered by niroj last updated on 22/May/20

Sol^n : We know that , (dx/ds)= cos ψ ⇒ (d^2 x/ds^2 )= −sinψ(dψ/ds) and (dy/ds)= sinψ ⇒ (d^2 y/ds^2 )= cosψ (dψ/ds) Now, ((d^2 x/ds^2 ))^2 +((d^2 y/ds^2 )) = (−sinψ(dψ/ds))^2 +(codψ(dψ/ds))^2 = sin^2 ψ((dψ/ds))^2 +cos^2 ψ((dψ/ds))^2 = ((dψ/ds))^2 (sin^2 ψ+cos^2 ψ) = 1.(1/ρ^2 )=(1/ρ^2 )//.

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

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

outstanding��✍

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