00:02
So for this example, we're given a vector equation, and we need to prove the following statement.
00:10
So in order to prove the statement, we need to massage our vector equation to get it in the right form necessary to prove that statement.
00:20
So we know that r would be equal to the magnitude of the vector equation.
00:26
Therefore r would be the square root of x squared plus y squared plus z squared okay and we need to prove this statement so now we have an expression for r so therefore del of 1 over r is equal to so it be d d d x of 1 over r i plus d d y of 1 over r plus d, d, z, 1 over r.
01:07
Okay, now we just need to find our partial derivative, so d, d, d, d, d, y, and d, z.
01:16
Okay, so therefore, d, d, d, x is equal to d, d, d, x, sorry, of 1 over r, is equal to minus 1 over r squared, dr, dx, and partial with respect, to y will of course be the same thing except we have this other derivative that we have to find and d d z of 1 over r is equal to minus 1 over r squared d r d z okay so now we have the necessary components to put into our expression so therefore we can say that del of 1 over r is equal to and since all of the terms have a 1 over minus r squared then we can just factor that out and that just becomes one minus 1 over r squared of d r d x i plus d r d x i plus d r d y plus d z k okay so now we just need to find these partial derivatives.
02:46
Okay, so we know that r is equal to square root of x squared plus y squared plus z squared.
02:59
So therefore, dr d x is equal to the derivative of this, so that's x squared plus y squared plus z squared to the one half power, which is equal to 2x over 2 times square root of x squared plus y squared plus z squared.
03:29
Okay, twos are going to cancel.
03:32
And of course, this is just equal to r.
03:36
So then we can say that dr, dx is equal to x over r...