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Question: 4] a) If x = u+v+w, y = uv+vw+uw, z = uvw and $\omega$ is a function of x,y and z. Prove that
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$x\frac{∂φ}{∂x} + 2y\frac{∂φ}{∂y} + 3z\frac{∂φ}{∂z} = u\frac{∂φ}{∂u} + v\frac{∂φ}{∂v} + w\frac{∂φ}{∂w}$

Subject:- Applied Mathematics

Marks:- 3

Mumbai Unversity>FE>Sem1>Applied Maths1

m1(81) • 8 views
 modified 4 days ago  • written 4 days ago by
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φ is a function of x,y and z and x,y,z are themselves functions of u,v,w.

$∴∂φ/∂u = ∂φ/∂x.∂x/∂u + ∂φ/∂y.∂y/∂u + ∂φ/∂z.∂z/∂u = ∂φ/∂x.1+∂φ/∂y (v+w) + ∂φ/∂z vw$

And $∂φ/∂v = ∂φ/∂x.∂x/∂v + ∂φ/∂y.∂y/∂v + ∂φ/∂z.∂z/∂v$

$=∂φ/∂x.1 + ∂φ/∂y (u+w) + ∂φ/∂z.uw$

And $∂φ/∂w = ∂φ/∂x.∂x/∂w + ∂φ/∂y.∂y/∂w + ∂φ/∂z.∂z/∂w$

$= ∂φ/∂x.1 + ∂φ/∂y (v+u) + ∂φ/∂z.uv$

Multiplying (1) by u , (2) by v, (3) by w and add

$∴u∂φ/∂u + v∂φ/∂v + w∂φ/∂w = (u+v+w)∂φ/∂x + [(uv+uw) + (vu+vw)+(wv + wu)]∂φ/∂y + 3uvw∂φ/∂z$

$= (u+v+w)∂φ/∂x + [2(uv+vw+uw)]∂φ/∂y + 3uvw∂φ/∂z$

$= x∂φ/∂x + 2y∂φ/∂y + 3z∂φ/∂z$

$∴ x∂φ/∂x + 2y∂φ/∂y + 3z∂φ/∂z = u∂φ/∂u + v∂φ/∂v + w∂φ/∂w$