Conservative Forces and Potential Energy

The Horizontal component of a conservative force and its relation to Potential Energy, $U$, of the system is $F_x = -\frac{\newcommand*\diff{\mathop{}\!\mathrm{d}} U}{\newcommand*\diff{\mathop{}\!\mathrm{d}} x}$. My physics book went further on to say that in 3 dimensions the equation becomes

$\vec{F} = -\frac{\partial U}{\partial x}\hat{i} - \frac{\partial U}{\partial y}\hat{j} - \frac{\partial U}{\partial z}\hat{k}$

Firstly why isn't it this instead

$\vec{F} = -\frac{\newcommand*\diff{\mathop{}\!\mathrm{d}} U}{\newcommand*\diff{\mathop{}\!\mathrm{d}} x}\hat{i} - \frac{\newcommand*\diff{\mathop{}\!\mathrm{d}} U}{\newcommand*\diff{\mathop{}\!\mathrm{d}} y}\hat{j} - \frac{\newcommand*\diff{\mathop{}\!\mathrm{d}} U}{\newcommand*\diff{\mathop{}\!\mathrm{d}} z}\hat{k}$

I'm quite new with partial derivates so a brief explanation of them would be helpful

Easy Math Editor

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Comments

Not sure about this, understand that the U here as 3 components, and so therefore it is impossible to use the d in the second equation because there are 3 dependent variables. As far as i know of, normal differentiation doesn't work when you have more than 1 dependent variable.

Tan Gian Yion - 7 years, 9 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$