Question

If v=(x^2+y^2+z^2 )^(-1/2) then (∂^2 v)/(∂x^2 )+(∂^2 v)/(∂y^2 )+(∂^2 v)/(∂z^2 )= ?

Answer 1

Answer:

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Answer 2

The value of (∂^2v)/(∂x^2) + (∂^2v)/(∂y^2) + (∂^2v)/(∂z^2) is 0.

Concept :

To find the value of (∂^2v)/(∂x^2) + (∂^2v)/(∂y^2) + (∂^2v)/(∂z^2), we need to first find the second partial derivatives of v with respect to x, y, and z.

Let's Solve It !

We have:

↝ v = (x^2 + y^2 + z^2)^(-1/2)

Taking the partial derivative of v with respect to x, we get:

↝ ∂v/∂x = (-1/2)(x^2 + y^2 + z^2)^(-3/2) * 2x

↝ -x(x^2 + y^2 + z^2)^(-3/2)

Taking the partial derivative of ∂v/∂x with respect to x again, we get:

↝ ∂^2v/∂x^2 = (-1)(x^2 + y^2 + z^2)^(-3/2) + (-3/2)(-x)(-1/2)(x^2 + y^2 + z^2)^(-5/2) * 2x^2

↝ (x^2 - y^2 - z^2)(x^2 + y^2 + z^2)^(-5/2)

Now, taking the second partial derivative of v with respect to y and z, we get:

↝ ∂^2v/∂y^2 = (y^2 - x^2 - z^2)(x^2 + y^2 + z^2)^(-5/2)

↝ ∂^2v/∂z^2 = (z^2 - x^2 - y^2)(x^2 + y^2 + z^2)^(-5/2)

Therefore, we have:

↝ (∂^2v)/(∂x^2) + (∂^2v)/(∂y^2) + (∂^2v)/(∂z^2)

↝ (x^2 - y^2 - z^2)(x^2 + y^2 + z^2)^(-5/2) + (y^2 - x^2 - z^2)(x^2 + y^2 + z^2)^(-5/2) + (z^2 - x^2 - y^2)(x^2 + y^2 + z^2)^(-5/2)

Simplifying, we get:

(∂^2v)/(∂x^2) + (∂^2v)/(∂y^2) + (∂^2v)/(∂z^2) = 0

Therefore, the value of (∂^2v)/(∂x^2) + (∂^2v)/(∂y^2) + (∂^2v)/(∂z^2) is zero/0.