Question #143100
Using the transformation T=z^2/2, reduce the equation f(x,y,z,p,q)= x(y^2+z^2×q^2)- zy^2×p=0 to a form f(P,x)= g(Q,y) where p= ∂T/∂x, Q= ∂T/∂y
1
Expert's answer
2020-11-09T20:06:03-0500

Given equation is,

f(x,y,z,p,q)=x(y2+z2×q2)zy2p=xy2+xz2q2zy2pf(x,y,z,p,q) = x(y^2+z^2×q^2)- zy^2p= xy^2+xz^2q^2- zy^2p


Since, Transformation is, T=z22T = \frac{z^2}{2}


Now,

z=2Tz = \sqrt{2T}

p=zx=212T1/2Tx=12TTx=12TPp =\frac{\partial z}{\partial x} = \sqrt{2}\frac{1}{2} T^{-1/2}\frac{\partial T}{\partial x} =\frac{1}{\sqrt{2T}} \frac{\partial T}{\partial x} =\frac{1}{\sqrt{2T}} P


q=zy=212T1/2Ty=12TTy=12TQq=\frac{\partial z}{\partial y} = \sqrt{2}\frac{1}{2} T^{-1/2}\frac{\partial T}{\partial y} =\frac{1}{\sqrt{2T}} \frac{\partial T}{\partial y} =\frac{1}{\sqrt{2T}} Q


Putting value of z, p and q in the given equation,

xy2+x(2T)(12TQ)22Ty2(12TP)=0xy^2+x(2T)(\frac{1}{\sqrt{2T}} Q)^2 -\sqrt{2T} y^2 (\frac{1}{\sqrt{2T}} P)=0


xy2+x(2T)(12TQ2)y2(P)=0xy^2+x(2T)(\frac{1}{{2T}} Q^2) - y^2 ( P)=0

xy2+xQ2y2P=0xy^2+xQ^2 - y^2 P=0

y2(xP)=xQ2y^2(x-P) = -xQ^2


y2Q2=x(Px)\frac{y^2}{Q^2} = \frac{x}{(P-x)}

It can written as,

g(Q,y)=f(P,x)g(Q,y) = f(P,x)

which is the required result.


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