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Euclidean norm

263
21
1. The problem statement, all variables and given/known data
Screenshot (148).png


2. Relevant equations

$$F = \nabla \phi$$

$$| F | = \sqrt{F \cdot F}$$


3. The attempt at a solution

I want to compute ##| F | = \sqrt{F \cdot F}##

$$| F | = \sqrt{F \cdot F} = \sqrt{\frac{(km)^2 (r-r_0)^2}{|r-r_0|^6}} = \sqrt{\frac{(km)^2}{(r-r_0)}} = \frac{km}{(r-r_0)^{1/2}} = \frac{km}{|r-r_0|} $$

That seems to be OK. I have one doubt here; would the following equality hold?

$$(r-r_0)^2 = ((x-x_0) - (y-y_0) - (z-z_0))^2$$

I do not need it for getting | F | but I want to be sure of this.

Thanks.
 

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mjc123

Science Advisor
740
322
No, (r-r0)2 = (x-x0)2 + (y-y0)2 + (z-z0)2
There also seems to be a mistake in your algebra. (r-r0)2/(r-r0)6 = 1/(r-r0)4, so you should end up with (r-r0)2 in the denominator (any other power would be dimensionally wrong).
 
263
21
No, (r-r0)2 = (x-x0)2 + (y-y0)2 + (z-z0)2
There also seems to be a mistake in your algebra. (r-r0)2/(r-r0)6 = 1/(r-r0)4, so you should end up with (r-r0)2 in the denominator (any other power would be dimensionally wrong).
OK I see. Now I get:

$$| F | = \sqrt{F \cdot F} = \sqrt{\frac{(km)^2 (r-r_0)^2}{|r-r_0|^6}} = \sqrt{\frac{(km)^2}{(r-r_0)^4}} = \frac{km}{(r-r_0)^{2}} = \frac{km}{|r-r_0|^2}$$

But the solution is ##\frac{km}{|r-r_0|}## so I must be missing something...
 
263
21
OK I see. Now I get:

$$| F | = \sqrt{F \cdot F} = \sqrt{\frac{(km)^2 (r-r_0)^2}{|r-r_0|^6}} = \sqrt{\frac{(km)^2}{(r-r_0)^4}} = \frac{km}{(r-r_0)^{2}} = \frac{km}{|r-r_0|^2}$$

But the solution is ##\frac{km}{|r-r_0|}## so I must be missing something...
Basically, I want to understand how to compute the derivative of the length of a vector function with respect to one variable, using:

Screenshot (149).png


The stated solution is:

Screenshot (150).png


NOTE: The scalar field is:

Screenshot (151).png
 

Attachments

DrClaude

Mentor
6,867
3,007
OK I see. Now I get:

$$| F | = \sqrt{F \cdot F} = \sqrt{\frac{(km)^2 (r-r_0)^2}{|r-r_0|^6}} = \sqrt{\frac{(km)^2}{(r-r_0)^4}} = \frac{km}{(r-r_0)^{2}} = \frac{km}{|r-r_0|^2}$$

But the solution is ##\frac{km}{|r-r_0|}## so I must be missing something...
That solution is the scalar field ##\phi##, not ##F##.
 
263
21
That solution is the scalar field ##\phi##, not ##F##.
I was computing the gradient of the wrong field! Actually, I have to compute the gradient of ##\phi##.Thanks for pointing that out.

Now we have two ways of computing it:

1):

Screenshot (152).png


I get that one.

What I do not get is the second method:

Screenshot (150).png


I am trying to get the encircled step using formula on #4 but don't get it. May you show it me?
 

Attachments

DrClaude

Mentor
6,867
3,007
That comes from the rule
$$
\frac{d}{dx} f(x)^{n} = n f(x)^{n-1} \frac{d}{dx} f(x)
$$
with ##n=-1##.
 

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