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$F$ is a function of $U$, which is a 3-dimensional function of $r$, so we were trying to prove that 3-dimensional functions can be considered 1-dimensional in calculations. So they found the gradient of the function $U$, I don't understand how did they take the gradient. \begin{align} F & = -\left( \frac{\partial U}{\partial x}, \frac{\partial U}{\partial y}, \frac{\partial U}{\partial z} \right) \\ F & = -\frac{\mathrm d U}{\mathrm dr}\left( \frac{\partial r}{\partial x}, \frac{\partial r}{\partial y}, \frac{\partial r}{\partial z} \right) \end{align} I thought when we were taking partial derivatives of functions, we would add differentials in front of them and the $\mathrm dU/\mathrm dr$ would have also been a partial derivative.

Emilio Pisanty
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Megan mars
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    Who is "they"? Is there more context here, like what the function $U$ was, for example? – probably_someone Dec 03 '19 at 15:34
  • Note that we use MathJax to typeset mathematics; you can find a good tutorial here. I have edited it in for you, but you should do it yourself in future. – Emilio Pisanty Dec 03 '19 at 15:38
  • Thank you for editing my question Emilio! I wasn't aware of typing mathematics in a different font. I was referring to the mark scheme as 'they', sorry I wasn't explicit enough! Also U was a potential function. – Megan mars Dec 03 '19 at 23:24

1 Answers1

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If $U$ is a function (Potential) in $3$ dimensions, let

$$U = U(x,y,z) = U(r, \phi, \theta) = U(r).$$

Then the force is definded as

$$ \vec{F} = - \vec{\nabla}U(x,y,z).$$

Now using the chain rule

$$\dfrac{\partial U}{\partial a} = \dfrac{\partial U}{\partial r} \dfrac{\partial r}{\partial a},$$

where $a \in \{x,y,z\}.$ If you plug that into the equation for $F$ you obtain

$$\vec{F} = \dfrac{\partial U}{\partial r}(\partial_x r, \partial_y r, \partial_z r)^T,$$

which should be what you are looking for.

Tera
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  • This makes sense, thank you! – Megan mars Dec 03 '19 at 23:27
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    @Meganmars Note that the use of the chain rule here only works if $U(r,\phi,\theta)=U(r)$ is independent of the angular variables $\phi$ and $\theta$. If it does depend on them then further terms need to be included. – Emilio Pisanty Dec 04 '19 at 15:15
  • @EmilioPisanty Didn't think of that when writing, but yes you are of course right. Thanks for the correction. – Tera Dec 04 '19 at 16:01
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    @Tera When these types of corrections are pointed out, and you agree with them, the site etiquette is to edit the corrections into the post itself, so that the post is readable in its full form without needing to parse the comment thread underneath. Note also that there is a full version history for all posts (accessible by clicking the 'edited XXX ago' link to the left of your signature in the post), so there is no need to explicitly mark any edits, unless it really is a major overhaul. – Emilio Pisanty Dec 04 '19 at 16:04
  • Thank you. I hope you agree with my now edited answere. – Tera Dec 04 '19 at 17:52