# Euclidean norm

#### JD_PM

1. The problem statement, all variables and given/known data 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

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).

• JD_PM

#### JD_PM

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...

#### JD_PM

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: The stated solution is: NOTE: The scalar field is: #### Attachments

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#### DrClaude

Mentor
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 $F$.

• JD_PM

#### JD_PM

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): I get that one.

What I do not get is the second method: I am trying to get the encircled step using formula on #4 but don't get it. May you show it me?

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#### DrClaude

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

• JD_PM

"Euclidean norm"

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