The derivative of Kinetic Energy

[Appleton]

From my very limited knowledge of calculus I would have thought that if mass is assumed constant d/dt (1/2mv^2) = 1/2m2v. This seems to be corroborated by one website http://www.Newton.dep.anl.gov/askasci/phy05/phy05008.htm and seems to follow the general rule d/dt (x^n) = nx^(n-1)
However in the Feynman lectures (Volume 1 13-1 (13.2)) http://student.fizika.org/~jsisko/Knjige/Opca%20Fizika/Feynman%20Lectures%20on%20Physics/Vol%201%20Ch%2013%20-%20Work%20and%20Potential%20Energy%201.pdf he states that if mass is assumed constant d/dt (1/2mv^2) = 1/2m2v(dv/dt).
Could someone help explain what I have most likely misunderstood?

 

[Hootenanny]

Welcome to Physics Forums.

From my very limited knowledge of calculus I would have thought that if mass is assumed constant d/dt (1/2mv^2) = 1/2m2v. This seems to be corroborated by one website http://www.Newton.dep.anl.gov/askasci/phy05/phy05008.htm and seems to follow the general rule d/dt (x^n) = nx^(n-1)
However in the Feynman lectures (Volume 1 13-1 (13.2)) http://student.fizika.org/~jsisko/Knjige/Opca%20Fizika/Feynman%20Lectures%20on%20Physics/Vol%201%20Ch%2013%20-%20Work%20and%20Potential%20Energy%201.pdf he states that if mass is assumed constant d/dt (1/2mv^2) = 1/2m2v(dv/dt).
Could someone help explain what I have most likely misunderstood?

Your expression is incorrect and does not agree with either reference.

You should note that in the first reference you quote, they are taking the derivative with respect to velocity, whilst in the second Feynman is taking the derivative with respect time time. In the former case you can indeed simply apply the "power rule". However, in the latter case because v depends on time, in the latter case, you need to use the chain rule in conjunction with the power rule.

Simply put, the first reference finds the rate of change of kinetic energy with respect to velocity. Whilst, the second the reference find the rate of change of kinetic energy with respect to time.

Does that make sense?

 

[Appleton]

Yes, I understand what I have overlooked now, thanks

 

[Ken G]

Incidentally, the interesting derivative of kinetic energy is with respect to distance, not time. If you take d/dx (mv²/2) = mv dv/dx, and you note that v dv/dx = a if you imagine the curve v(x) and want the "a" at some "x", then we have:
d/dx (mv²/2) = ma = F = -d/dx V
where V is potential energy for a force F that can be written that way (a "conservative" force). This is entirely the basis of the concept of potential energy to track the changes in kinetic energy.

 

[PJC01]

The derivative of kinetic energy is a fundamental concept in physics and is derived from the relationship between force and velocity. It is important to note that the derivative of kinetic energy is not a simple equation and can vary depending on the assumptions made.

In the first scenario, where mass is assumed to be constant, the derivative of kinetic energy can be simplified to 1/2m2v, as you correctly stated. This follows the general rule of differentiating a power function, as you mentioned.

However, in the second scenario presented in the Feynman lectures, the assumption of constant mass is not made. Instead, the derivative of kinetic energy takes into account the changing mass of the object as it accelerates. This is represented by the term dv/dt, which represents the rate of change of velocity with respect to time. This is a more accurate representation of the derivative of kinetic energy, as it considers the dynamic nature of the system.

It is important to note that both equations are correct in their respective assumptions. The first equation is a simplification that is often used in introductory physics courses, while the second equation is a more precise representation that is used in more advanced physics. It is also important to understand the assumptions being made when using these equations in order to apply them correctly in different scenarios.