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рдкреНрд░рд╢реНрди
If ЁЭТЩ = r cos θ and y= r sin θ prove that JJ-1=1.
рдЙрддреНрддрд░
Given ЁЭСе =r cos θ and y=r sin θ
i.e. x,y → f (r,θ)
`(delx)/(delr)=costheta` `(delx)/(deltheta)=-rsintheta`
`(dely)/(delr)=sintheta` `(dely)/(deltheta)=rcostheta`
`therefore "J"=(del(x,y))/(del(r,theta))=|((delx)/(delr),(delx)/(deltheta)),((dely)/(delr),(dely)/(deltheta))|=|(costheta,-rsintheta),(sintheta,rcostheta)|=r(cos^2theta+sin^2theta)=r.`
∴ J = r………………….. (1)
Now, to find values of r and θ
`therefore x^2+y^2=r^2(cos^2theta+sin^2theta)=r^2`
`thereforer=sqrt(x^2+y^2)` and `y/x=(rsintheta)/(rcostheta)=tantheta`
`thereforetheta=tan^(-1) y/x`
`therefore "J"'=(del(r,theta))/(del(x,y))=|((delx)/(delr),(delx)/(deltheta)),((dely)/(delr),(dely)/(deltheta))|=|(x/(sqrt(x^2+y^2)),y/(sqrt(x^2+y^2))),((-y)/(x^2+y^2),x/(x^2+y^2))|`
`=x^2/(x^2+y^2)^(3/2)+y^2/(x^2+y^2)^(3/2)`
`=(x^2+y^2)/(x^2+y^2)^(3/2)`
`=1/sqrt(x^2+y^2)= 1/r`……………….. (2)
From 1 and 2, we get
Hence, JJ'= r. `1/r`= 1
Hence proved
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