# 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

4 years, 10 months ago

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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.

- 4 years, 10 months ago