# Derivative of the kinetic energy

1. Feb 10, 2008

### watsup91749

Derivative of the kinetic energy....

1. The problem statement, all variables and given/known data
If you take the derivative of the kinetic energy of a particle with respect to its velocity you get...
A) force
B) momentum
C) acceleration
D) mass
E) potential energy

2. Relevant equations
K= .5mv^2

3. The attempt at a solution
Im in pre calc right now and i have no idea how to take the derivative of anything, quick google searching led me to believe that its a calculus thing, so it would be really helpfull if someone could explain how to do this..

2. Feb 10, 2008

### rock.freak667

If you have

$$y=kx^n$$

and you want to find the derivative of that with respect to x,denoted as $\frac{dy}{dx}$

it is simply

$$\frac{dy}{dx}=knx^{n-1}$$

Where k is a constant

3. Feb 10, 2008

### watsup91749

I think the answer is B momentum, is that right? momentum has p=mv and kinetic energy has the same variables, so it must be B right?
Also K=mv^(2-1) would be the same as K=mv which is the the same as momentum, is this correct?

4. Feb 10, 2008

### rock.freak667

$$\frac{dK}{dv}=mv$$ which is momentum so you are correct.

5. Feb 10, 2008

### awvvu

You could also exploit the notation and think about it in terms of units.

$$dK/dv = [\frac{kg * m^2/s^2}{m/s}] = [kg * m/s]$$