The derivative of Kinetic Energy


by Appleton
Tags: derivative, energy, kinetic
Appleton
Appleton is offline
#1
Sep21-11, 04:31 AM
P: 30
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/askasc...5/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/Kn...Energy%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?
Phys.Org News Partner Physics news on Phys.org
Information storage for the next generation of plastic computers
Scientists capture ultrafast snapshots of light-driven superconductivity
Progress in the fight against quantum dissipation
Hootenanny
Hootenanny is offline
#2
Sep21-11, 04:49 AM
Emeritus
Sci Advisor
PF Gold
Hootenanny's Avatar
P: 9,789
Welcome to Physics Forums.
Quote Quote by Appleton View Post
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/askasc...5/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/Kn...Energy%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
Appleton is offline
#3
Sep21-11, 05:14 AM
P: 30
Yes, I understand what I have overlooked now, thanks

Ken G
Ken G is online now
#4
Sep21-11, 06:02 PM
PF Gold
P: 3,071

The derivative of Kinetic Energy


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.


Register to reply

Related Discussions
Rotational kinetic energy versus translation kinetic energy? Classical Physics 1
difference between relatavistic kinetic energy and normal kinetic energy Special & General Relativity 2
Derivative of kinetic energy Classical Physics 8
Ratio kinetic energy of alpha particle / kinetic energy of proton Introductory Physics Homework 2
Derivative of the kinetic energy.... Introductory Physics Homework 4