Second derivative of effective potential

In summary, the value of r in terms of l, k, and m for which the given function has a minimum is r = (l^2/km)^1/3. To prove that this is a minimum, we need to show that the second derivative of V(r) is positive at this point. To do this, we can use the second derivative test and plug in the value of r we found to show that it is positive. Additionally, we can practice derivatives to gain a better understanding of the concept, which will help us in solving problems like deriving Kepler's third law from Kepler's second law. Specifically, in the function f(x) = x^a, where a is a non-zero real number, the derivative
  • #1
dswatson
37
0
Determine the value of r in terms of l, k, and m for which the following function has a minimum.

V(r) = -(k/r) + (l^2/(2mr^2))

where l, k, and m are positive constants.

Prove that this is a minimum by showing that the second derivative of V(r) at the minimum is positive.

I have no idea how to even begin this...I am horrible at derivatives and am struggling in my physics class with them. Any help would be greatly appreciated.

I am then asked to derive Kepler's third law from Kepler's second law. So I feel I have a lot of work ahead of me.
 
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  • #2
Then practice derivates!

Try this one:

what is the derivative with respect to x in this function: [tex]f(x) = x^a[/tex], where a is a real number, (non zero).

That is all you need for this particlar problem :-)
 

What is the Second Derivative of Effective Potential?

The second derivative of effective potential is a mathematical concept used in physics to describe the curvature of a potential energy function. It is the rate of change of the first derivative of the effective potential function with respect to a given variable.

Why is the Second Derivative of Effective Potential Important?

The second derivative of effective potential is important because it can provide valuable information about the stability and behavior of a physical system. It can help determine the presence of local minima or maxima in the potential energy function, which can in turn affect the motion of particles in the system.

How is the Second Derivative of Effective Potential Calculated?

The second derivative of effective potential is calculated by taking the derivative of the first derivative of the effective potential function with respect to the given variable. This can be done using mathematical techniques such as the chain rule, product rule, or quotient rule.

What are Some Real-world Applications of the Second Derivative of Effective Potential?

The second derivative of effective potential has various applications in physics, particularly in the study of celestial mechanics and quantum mechanics. It is used to analyze the behavior of planets and satellites in orbit, as well as the behavior of subatomic particles in quantum systems.

Are there any Limitations to Using the Second Derivative of Effective Potential?

While the second derivative of effective potential can provide valuable insights into physical systems, it is important to note that it is only one of many factors that can affect the behavior of a system. It should be used in conjunction with other mathematical and physical principles to gain a more complete understanding of a system's behavior.

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