derivatives of Kinetic Energy
I read that derivative of Kinetic Energy function = $F.v$ while I got $mv$ when I differentiated it with respect to velocity. The way I did it is:
$\frac{dK}{dv} = \frac{1}{2} m . \frac{d}{dv} v^2$
So I assumed that the mass is fixed and I differentiated the squared velocity by taking the (2) down and subtracting (1) from the exponent, which gave me $2v$ for $\frac{d}{dv} v^2$
Would you clarify the correct derivative of Kinetic Energy as well as the $F$ variable? And what does the derivative in this case represent?
$\endgroup$3 Answers
$\begingroup$The derivate of kinetic energy respect to the time $t$ is $Fv$:
$$K'=mvv'=mva=Fv$$
In general $v$ depends by time so the total derivative of $K$ is $Fv$, i.d. the instantaneous power.
$\endgroup$ 2 $\begingroup$The derivative of kinetic energy is momentum, which has the equation $p=mv$.
$\endgroup$ 3 $\begingroup$$$\frac{d v^2}{dt}=2v\frac{dv}{dt}$$
$$\implies \frac{dK}{dt}=\frac{1}{2}m\frac{d v^2}{dt}=mv\frac{dv}{dt}$$
but $$m\frac{dv}{dt}=F$$
thus
$$\frac{dK}{dt}=Fv$$
which represents the power of force $F$ ,( in Watt).
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