v=(x^2+y^2+z^2)6-1/2, prove that d^2v/dx^2+d^2v/dy^2+d^2v/dz^2=0
Answers
Answer:
If u=log
(x
2
+y
2
+z
2
)
then prove that (x
2
+y
2
+z
2
)(
dx
2
d
2
u
+
dy
2
d
2
u
+
dz
2
d
2
u
)=1.
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Answer
u=log
x
2
+y
2
+z
2
u=log(x
2
+y
2
+z
2
)
2
1
u=
2
1
log(x
2
+y
2
+z
2
)
⇒2u=log(x
2
+y
2
+z
2
)
Differentiating w.r.t. 'x' on both sides, we get
2
dx
du
=
x
2
+y
2
+z
2
1
⋅(2x)
⇒
dx
du
=
x
2
+y
2
+z
2
x
dx
2
d
2
u
=
(x
2
+y
2
+z
2
)
2
(x
2
+y
2
+z
2
)(
dx
d
(x))−x(
dx
d
(x
2
+y
2
+z
2
))
⇒
dx
2
d
2
u
=
(x
2
+y
2
+z
2
)
x
2
+y
2
+z
2
−x(2x)
=
(x
2
+y
2
+z
2
)
2
−x
2
+y
2
+z
2
Similarly,
dy
2
d
2
u
=
(x
2
+y
2
+z
2
)
2
−y
2
+x
2
+z
2
dz
2
d
2
u
=
(x
2
+y
2
+z
2
)
2
−z
2
+x
2
+y
2
(x
2
+y
2
+z
2
)(
dx
2
d
2
u
+
dy
2
d
2
u
+
dz
2
d
2
u
)=(x
2
+y
2
+z
2
)[
(x
2
+y
2
+z
2
)
2
x
2
+y
2
+z
2
]
=
(x
2
+y
2
+z
2
)
2
(x
2
+y
2
+z
2
)
=1.