Computing the Magnitude of F Using Gradient

[JD_PM]

Homework Statement

Homework Equations

$$F = \nabla \phi$$

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

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.

 

[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]

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:

 

[DrClaude]

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]

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?

 

[DrClaude]

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