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Derivative of kinetic energy 
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#1
Jan1210, 02:21 PM

P: 109

Hi.
I differentiated the equation of kinetic energy [tex]E(v) = \tfrac{1}{2} mv^2[/tex] and I got: [tex]E'(v) = mv[/tex]. 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). Could someone explain me more about this? Thank you 


#2
Jan1210, 02:39 PM

P: 373

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 [tex]\int ^{x1}_{x2} f(x)dx[/tex] indicates that you're trying to find the area under the curve of a graph (y = f(x)) between two arbitrary points on the xaxis, 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) [tex] W = \int ^{x1}_{x2} F =\int ^{x1}_{x2} m\frac{dv}{dt} dx [/tex] now we also know that [tex] \frac{dx}{dt}= v [/tex] so we can make a substitution showing that [tex] dx = v dt [/tex] so putting that in; [tex] W = \int ^{x1}_{x2} m\frac{dv}{dt}v dt =\int ^{v1}_{v2} mv dv = 0.5 m(v1^2v2^2) [/tex] = 0.5 m(v1^2v2^2) 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 breadandbutter stuff. About the tangeant line, see this link http://www.intmath.com/Differentiati...principles.php it's very clear and concise, differentiation is a wonderful operation 


#3
Jan1210, 02:41 PM

HW Helper
P: 7,049

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
Jan1210, 02:44 PM

P: 373

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



#5
Jan1210, 02:59 PM

P: 109

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
Jan1210, 03:03 PM

P: 373

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
Jan1210, 03:07 PM

P: 250

d(mv^2/2)/dt = mv(dv/dt) = mva = mav = Fv. Hope this helps. 


#8
Jan1210, 03:33 PM

P: 109

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: Can someone explain me this? 


#9
Jan1210, 04:04 PM

P: 250




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