# Derivative of kinetic energy

1. Jan 12, 2010

### Taturana

Hi.

I differentiated the equation of kinetic energy

$$E(v) = \tfrac{1}{2} mv^2$$

and I got:

$$E'(v) = mv$$.

I'm new to the concept of derivatives so I only know that it means that m*v is the slope of a tangent line to the graph of E at a given point.

But I can't really understand the exact meaning of it (if there are any).

Thank you

2. Jan 12, 2010

### Chewy0087

If you're not familiar with calculus a "real" derivation is difficult, in this case what you've just done (while correct) is of little use at all;

I'll outline the more traditional approach, if you've not covered integration i'm afraid you might struggle, the use of the $$\int ^{x1}_{x2} f(x)dx$$ indicates that you're trying to find the area under the curve of a graph (y = f(x)) between two arbitrary points on the x-axis, x1 and x2 (x1 > x2)

The "Work - Energy Theorum" states that the net work done is the change in kinetic energy, so the work done is given by; (Work done is the force multiplied by the distance it has acted over)

$$W = \int ^{x1}_{x2} F =\int ^{x1}_{x2} m\frac{dv}{dt} dx$$

now we also know that $$\frac{dx}{dt}= v$$ so we can make a substitution showing that $$dx = v dt$$

so putting that in;

$$W = \int ^{x1}_{x2} m\frac{dv}{dt}v dt =\int ^{v1}_{v2} mv dv = 0.5 m(v1^2-v2^2)$$ = 0.5 m(v1^2-v2^2)

What i will say is that if you've not encountered this stuff before, DON'T be intimidated, this is not a particularly easy derivation and when you've had an introduction to calculus this will be bread-and-butter stuff.

About the tangeant line, see this link

http://www.intmath.com/Differentiation/3_Derivative-first-principles.php

it's very clear and concise, differentiation is a wonderful operation

Last edited: Jan 12, 2010
3. Jan 12, 2010

### rcgldr

This would only make sense if V is a function of time (t) or distance across some path (s). In those cases the derivative is the slope of the function V(t) or V(s). Chewy explained the meaning of this when V is a function of time.

4. Jan 12, 2010

### Chewy0087

You COULD reverse-engineer my proof above, starting with d(e)/dv = mv, then multiplying by dv and substituting (to find the work done - again referring to the work-energy theorum) but that's not very clear at all.

5. Jan 12, 2010

### Taturana

Thank you very much Chewy0087 for the explanation.

I didn't fully understand it because I'm not familiar with the concept of integration, but I'll study integration (I'm studying through MIT OpenCouseware) and try to understand your reply.

I'll take a look in the link you sent too...

Thank you

You guys can continue posting things if you want... Every help is welcome =D

6. Jan 12, 2010

### Chewy0087

You should definateley study differentiation before integration, the MIT courseware is great - have a look at the 18.01 set of lectures (single variable calculus), they're accelerated but good.

Good luck!

7. Jan 12, 2010

### GRDixon

Are you differentiating the kinetic energy wrt time? If so, then
d(mv^2/2)/dt = mv(dv/dt) = mva = mav = Fv. Hope this helps.

8. Jan 12, 2010

### Taturana

Hey I have another question but I won't open new topic.

In the website that Chewy0087 linked to this topic, at the lesson 4 it says:

Now a question appears: to know how fast the temperature is increasing, we need to know the temperature at least in two moments to compare. I know that derivative tell us how fast the temperature is increasing right now. But, how derivatives do it with just one input? There shouldn't be at least two inputs?

Can someone explain me this?

9. Jan 12, 2010

### GRDixon

The way I think of this problem is as follows: Record the temperature at times t-dt/2 and t+dt/2. Then the MEAN rate of change during the time interval from t-dt/2 to t+dt/2 is (T(t+dt/2) - T(t-dt/2))/dt. Now begin shrinking the size of dt. (T(t+dt/2) - T(t-dt/2)) will also get smaller. Take it to the limit where dt becomes infinitesimal. (T(t+dt/2) - T(t-dt/2)) will also be infinitesimal. And the ratio of two infinitesimal values can be a perfectly finite quotient. You call this ratio the rate of change of T AT time the single time t.