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
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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