# Why is F=dU/dx=0 either side of inflexion point?

• I
• walking
In summary, the gradient of an inflexion point is zero, so when a particle is displaced either side of an inflexion point, even by a very small amount, there will immediately be a nonzero force on it and hence it will immediately lose equilibrium.
walking
In Tipler & Mosca 5th edition p173 it defines neutral equilibrium as a point in a U-x curve where ##\frac{dU}{dx}=0## and also ##\frac{dU}{dx}=0## for a small displacement either side of the point. However I do not understand why ##\frac{dU}{dx}## remains ##0## either side of the inflexion point. Surely the gradient of an inflexion point is just ##0## at the inflexion point itself, and either side of it the gradient is nonzero. If this is indeed the case, then when a particle is displaced either side of an inflexion point, even by a very small amount, there will immediately be a nonzero force on it and hence it will immediately lose equilibrium, meaning that ##\frac{dU}{dx}\ne 0##.

walking said:
In Tipler & Mosca 5th edition p173 it defines neutral equilibrium as a point in a U-x curve where ##\frac{dU}{dx}=0## and also ##\frac{dU}{dx}=0## for a small displacement either side of the point. However I do not understand why ##\frac{dU}{dx}## remains ##0## either side of the inflexion point. Surely the gradient of an inflexion point is just ##0## at the inflexion point itself, and either side of it the gradient is nonzero. If this is indeed the case, then when a particle is displaced either side of an inflexion point, even by a very small amount, there will immediately be a nonzero force on it and hence it will immediately lose equilibrium, meaning that ##\frac{dU}{dx}\ne 0##.

I suspect they are looking at a Taylor expansion of ##U## about an inflexion point. Remember that for an inflexion point the second derivative is zero, hence ##U## is approx constant (to second order in ##x##). Hence ##\frac{dU}{dx}## is apprx zero to first order in ##x##.

walking said:
In Tipler & Mosca 5th edition p173 it defines neutral equilibrium as a point in a U-x curve where ##\frac{dU}{dx}=0## and also ##\frac{dU}{dx}=0## for a small displacement either side of the point. However I do not understand why ##\frac{dU}{dx}## remains ##0## either side of the inflexion point.

The equilibrium point is a point where the first derivative is zero. An inflection point is a point where the second derivative is zero.

walking said:
In Tipler & Mosca 5th edition p173 it defines neutral equilibrium as a point in a U-x curve where ##\frac{dU}{dx}=0## and also ##\frac{dU}{dx}=0## for a small displacement either side of the point. However I do not understand why ##\frac{dU}{dx}## remains ##0## either side of the inflexion point. Surely the gradient of an inflexion point is just ##0## at the inflexion point itself, and either side of it the gradient is nonzero. If this is indeed the case, then when a particle is displaced either side of an inflexion point, even by a very small amount, there will immediately be a nonzero force on it and hence it will immediately lose equilibrium, meaning that ##\frac{dU}{dx}\ne 0##.

That's a definition of neutral equilibrium.

If you displace the system slightly but there is a restoring force which brings it back to the equilibrium point, that's stable equilibrium. Most equilibrium situations of interest have this character, for instance simple harmonic motion.

If you displace the system slightly and the resulting force takes it farther away from equilibrium, that's unstable equilibrium. For instance, something you've managed to balance on the top of a sphere.

If you displace the system slightly and there is no force at all, so the new point is also an equilibrium point, that's neutral equilibrium. The above is the definition of this condition. It doesn't say all equilibria are of this nature, it's defining a particular situation. An example is an object experiencing neutral buoyancy. You can move it around and it still stays in the same location. An even simpler example is an object sitting on the ground.

walking said:
Surely the gradient of an inflexion point is just ##0## at the inflexion point itself, and either side of it the gradient is nonzero.

You are restricting your thinking to a very narrow class of functions. Here's a counter example. Take a parabola. Slice off the bottom of it, replacing the curved portion with a horizontal line.

That function has the property that there is a finite region over which all derivatives are 0. If it represents a potential, there are no forces anywhere in that flat region.

Mister T said:
The equilibrium point is a point where the first derivative is zero. An inflection point is a point where the second derivative is zero.

Mister T

## 1. Why is F=dU/dx=0 at an inflexion point?

At an inflexion point, the curve of a function changes from concave up to concave down, or vice versa. This means that the slope of the curve, which is represented by the derivative dU/dx, changes from positive to negative or vice versa. Therefore, at the inflexion point, the derivative must equal zero.

## 2. How does the value of F=dU/dx change on either side of an inflexion point?

On one side of the inflexion point, the value of F=dU/dx will be positive, while on the other side it will be negative. This is because the slope of the curve changes from positive to negative (or vice versa) at the inflexion point.

## 3. What does F=dU/dx=0 at an inflexion point tell us about the function?

This condition indicates that the function is changing direction at the inflexion point. It also tells us that the function is neither increasing nor decreasing at that point, but rather has a point of inflection.

## 4. Is it possible for F=dU/dx to be zero at a point that is not an inflexion point?

Yes, it is possible. If the function has a local maximum or minimum at a point, the derivative will be equal to zero at that point. However, this does not necessarily mean that the point is an inflexion point.

## 5. How can we use the value of F=dU/dx at an inflexion point to analyze the behavior of a function?

The value of F=dU/dx at an inflexion point can help us determine the nature of the inflexion point. If the derivative is positive on one side and negative on the other, it is a point of inflection. If the derivative is zero on both sides, it is a horizontal point of inflection. This information can also be used to sketch the graph of the function.

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