Show That Vector Squared is Equal to 12r

  • Thread starter Thread starter Nat3
  • Start date Start date
  • Tags Tags
    Vector
Nat3
Messages
69
Reaction score
0

Homework Statement



\vec r = <x, y, z>, r = \left |\vec r \right |

Show that \nabla ^2 r^3 = 12r


Homework Equations



I don't understand how a vector can be squared. What does \nabla ^2 mean?
 
Physics news on Phys.org
del^2 is d^2/dx^2+d^2/dy^2+d^2/dz^2. It's the sum of the second derivatives. It's the laplacian.
 
Like Dick said nabla squared is called the laplacian. Don't forget that r=sqrt{x^2+y^2+z^2}, This problem is going to be a lot of work.
 
Do you know index notation? This is a 3-liner if you know it. Although come to think of it if you haven't seen the Laplacian, then you probably don't.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top