# Find force from potential energy

If we have a force whose component along $$\vec{r}$$ is $$F_r$$. where $$\vec{r} = x \hat{i} + y \hat{j} + z \hat{k}$$. then the force is = derivative of $$U$$ (potential energy) wrt $$r$$.

So my question is : if $$F_r = \frac{dU}{dr}$$
then: $$F_r = \frac{\partial{U}}{\partial{x}} + \frac{\partial{U}}{\partial{y}}+ \frac{\partial{U}}{\partial{z}}$$ ...is this correct? where $$\partial$$ is partial differential

• $F=-\nabla U$. This formula encodes force being equal to the negative of the gradient of the potential and is enough to calculate what you are asking for. This is basically equal to what you have written but with proper unit vectors added which are missing in the equation given above.
– Lost
Oct 14 at 8:34
• @Lost ok , thanks a lot. Oct 14 at 8:59

the force is $$\begin{equation} \vec{F} = - \frac{\partial U} {\partial{\vec{r}}} =-\left(\frac{\partial U} {\partial{x}}; \frac{\partial U} {\partial{y}}; \frac{\partial U} {\partial{z}}\right) \end{equation}$$