Q 63. - Calculus

Question

Prove that $\Vert \nabla f(x,y,z)\Vert =1$for every point in the domain of the function $f(x,y,z)=\sqrt{x^2+y^2+z^2}$ except the origin

Step-by-Step Solution

Answer

Solving $\nabla f=i\frac{\partial f}{\partial x}+j\frac{\partial f}{\partial y}+k\frac{\partial f}{\partial z}$ will prove the result.

Step 1: Given Information

It is given that

$f(x,y,z)=\sqrt{x^2+y^2+z^2}$

Step 2: Taking the divergence

Solving LHS

$\Vert \nabla f(x,y,z)\Vert$

$\nabla f=i\frac{\partial f}{\partial x}+j\frac{\partial f}{\partial y}+k\frac{\partial f}{\partial z}$

$\frac{\partial f}{\partial x}=\frac{2x}{2\sqrt{x^2+y^2+z^2}}$

$\frac{\partial f}{\partial x}=\frac{x}{\sqrt{x^2+y^2+z^2}}$

and

$\frac{\partial f}{\partial y}=\frac{2y}{2\sqrt{x^2+y^2+z^2}}$

$\frac{\partial f}{\partial y}=\frac{y}{\sqrt{x^2+y^2+z^2}}$

also

$\frac{\partial f}{\partial z}=\frac{2z}{2\sqrt{x^2+y^2+z^2}}$

$\frac{\partial f}{\partial z}=\frac{z}{\sqrt{x^2+y^2+z^2}}$

Step 3: Simplification

Solving, we get

$\Vert \nabla f(x,y,z)\Vert =||i\frac{\partial f}{\partial x}+j\frac{\partial f}{\partial y}+k\frac{\partial f}{\partial z}||$

$\Vert \nabla f(x,y,z)\Vert =||i\left(\frac{x}{\sqrt{x^2+y^2+z^2}}\right)+j\left(\frac{y}{\sqrt{x^2+y^2+z^2}}\right)+k\left(\frac{z}{\sqrt{x^2+y^2+z^2}}\right)||$

$\Vert \nabla f(x,y,z)\Vert =\sqrt{{\left(\frac{x}{\sqrt{x^2+y^2+z^2}}\right)}^2+{\left(\frac{y}{\sqrt{x^2+y^2+z^2}}\right)}^2+{\left(\frac{y}{\sqrt{x^2+y^2+z^2}}\right)}^2}$

$\Vert \nabla f(x,y,z)\Vert =\sqrt{\frac{x^2}{x^2+y^2+z^2}+\frac{y^2}{x^2+y^2+z^2}+\frac{z^2}{x^2+y^2+z^2}}$

$\Vert \nabla f(x,y,z)\Vert =\sqrt{\frac{x^2+y^2+z^2}{x^2+y^2+z^2}}$

$\Vert \nabla f(x,y,z)\Vert =1$