# Potential energy and Conservative Force

Gold Member

## Homework Statement

I have a force ##\vec{F} = a_x\vec{i}+2a_y\vec{j}+3a_z\vec{k}##. Find the potential

## The Attempt at a Solution

Lets suppose
And we know that ##\vec {F} = ∇U##

In this case I said that

##U_x=-\int F_xdx##
##U_y=-\int F_ydy##
##U_z=-\int F_zdz##
and then I said

##U= U_x+U_y+U_z##

Is this mathematically true ?

Normally I should have ##U(x,y,z) = -\int F_xdx## I guess..For example in this case I should have something like
##U(x,y,z) = -a_x^2+U(y,z)##
then I can say
##U(x,y,z) = \int (-a_x^2+ 2a_y)dy## etc. Which one is correct ?

Mileto Graziano

## Answers and Replies

haruspex
Science Advisor
Homework Helper
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2020 Award
Are ax etc. constants? If so, what are their integrals wrt x, y, z?
You can always check your answer using the differential equation you quoted.

Gold Member
Are ax etc. constants? If so, what are their integrals wrt x, y, z?
You can always check your answer using the differential equation you quoted.
##a## are constant. I find the answer its easy to find. I was asking about the mathematical part of it.Are both approaches are true ?

PeroK
Science Advisor
Homework Helper
Gold Member
2020 Award
Normally I should have ##U(x,y,z) = -\int F_xdx## I guess..For example in this case I should have something like
##U(x,y,z) = -a_x^2+U(y,z)##
then I can say
##U(x,y,z) = \int (-a_x^2+ 2a_y)dy## etc. Which one is correct ?

You should never integrate a constant ##a_x## and get ##\frac12 a_x^2##. That's a common mistake.

Instead, you have:

##\vec{F} = (a_x, 2a_y, 3a_z) = - \vec{\nabla}U(x, y, z) = -(\frac{\partial U}{\partial x}, \frac{\partial U}{\partial y}, \frac{\partial U}{\partial z}) ##

If we start with ##-\frac{\partial U}{\partial x} = a_x##, then this gives:

##U(x, y, z) = -a_xx + f(y, z)##

Now we have:

##\frac{\partial U}{\partial y} = \frac{\partial f}{\partial y} = -2a_y##

Which leads to ##f(y, z) = -2a_yy + g(z)##

And, finally we have:

##\frac{\partial U}{\partial z} = \frac{dg}{dz} = -3a_z##

Which gives ##g(z) = -3a_zz + C##

Your first method is just a shortcut for this.

Gold Member
You should never integrate a constant ##a_x## and get ##\frac12 a_x^2##. That's a common mistake.

Instead, you have:

##\vec{F} = (a_x, 2a_y, 3a_z) = - \vec{\nabla}U(x, y, z) = -(\frac{\partial U}{\partial x}, \frac{\partial U}{\partial y}, \frac{\partial U}{\partial z}) ##

If we start with ##-\frac{\partial U}{\partial x} = a_x##, then this gives:

##U(x, y, z) = -a_xx + f(y, z)##

Now we have:

##\frac{\partial U}{\partial y} = \frac{\partial f}{\partial y} = -2a_y##

Which leads to ##f(y, z) = -2a_yy + g(z)##

And, finally we have:

##\frac{\partial U}{\partial z} = \frac{dg}{dz} = -3a_z##

Which gives ##g(z) = -3a_zz + C##

Your first method is just a shortcut for this.
Yes sorry. Its a stupid mistake that I made...I see your point, thanks for clarifying it.