Math, asked by slnchary2031, 5 months ago

u=f(r,s,t) where r=x/y, s=y/z, t=z/x
prove that xdu/dx+ydu/dy+zdu/dz=0

Answers

Answered by amitnrw
8

Given : U = f(r , s , t)  

r = x/y

s = y/z

t = z/x

To find : Show that x∂u/∂x +  y∂u/∂y  + z∂u/∂z  = 0

Solution:

∂u/∂x  = (∂u/∂r)(∂r/∂x) +  (∂u/∂s)(∂s/∂x) +  (∂u/∂t)(∂t/∂x)

=> ∂u/∂x  = (∂u/∂r)(∂(x/y)/∂x) +  (∂u/∂s)(∂(y/z)/∂x) +  (∂u/∂t)(∂(z/x)/∂x)

=> ∂u/∂x  = (∂u/∂r)(1/y)  +  (∂u/∂s)(0) +  (∂u/∂t)(-z/x²)

=> ∂u/∂x  = (∂u/∂r)(1/y) - (∂u/∂t)(z/x²)

∂u/∂y  = (∂u/∂r)(∂r/∂y) +  (∂u/∂s)(∂s/∂y) +  (∂u/∂t)(∂t/∂y)

=> ∂u/∂y = (∂u/∂r)(∂(x/y)/∂y) +  (∂u/∂s)(∂(y/z)/∂y) +  (∂u/∂t)(∂(z/x)/∂y)

=> ∂u/∂y  = (∂u/∂r)(-x/y²)  +  (∂u/∂s)(1/z) +  (∂u/∂t)(0)

=> ∂u/∂y  = -(∂u/∂r)(x/y²) +  (∂u/∂s)(1/z)

∂u/∂z  = (∂u/∂r)(∂r/∂z) +  (∂u/∂s)(∂s/∂z) +  (∂u/∂t)(∂t/∂z)

=> ∂u/∂z = (∂u/∂r)(∂(x/y)/∂z) +  (∂u/∂s)(∂(y/z)/∂z) +  (∂u/∂t)(∂(z/x)/∂z)

=> ∂u/∂z  = (∂u/∂r)(0)  +  (∂u/∂s)(-y/z²) +  (∂u/∂t)(1/x)

=> ∂u/∂z  = -(∂u/∂r)(y/z²) +  (∂u/∂t)(1/x)

LHS = x∂u/∂x +  y∂u/∂y  + z∂u/∂z

= x( (∂u/∂r)(1/y) - (∂u/∂t)(z/x²)) + y(-(∂u/∂r)(x/y²) + (∂u/∂s)(1/z) + z( -(∂u/∂r)(y/z²) +  (∂u/∂t)(1/x))

= (x/y)∂u/∂r  -(z/x)∂u/∂t -(x/y)(∂u/∂r) + (y/z) (∂u/∂s) - (y/z)(∂u/∂r + (z/x)(∂u/∂t)

=  (x/y)∂u/∂r  - (x/y)(∂u/∂r) - (z/x)∂u/∂t + (z/x)∂u/∂t + (y/z) (∂u/∂s) - (y/z)(∂u/∂r

= 0

= RHS

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