How does the kinematic energy appear in the work integral?

[AntoineCompagnie]

Homework Statement

We call Kinetic energy Ec=1/2 mv^2

its deferral into the work integral gives:

\begin{align*}
w_{12}&=\int\limits_1^2\delta W\\
& = \int\limits_1^2 \vec F \wedge d\vec r\\
&\underbrace{=\int\limits_1^2dE_c}_{why?}
\end{align*}

But Why does the kinematic energy appears in the last equation?

Homework Equations

According to Newton's second law,

$$\vec{F}=\frac{d\vec{P}}{dt}$$

The Attempt at a Solution

Multiplicating it by $$d\vec{r}$$

\begin{align*}
\vec F \wedge d\vec r &= \frac{d\vec P}{dt} \wedge d\vec r\\
&= d\vec P \wedge \frac{d\vec r}{dt}\\
&= d\vec P \wedge \vec v\\
&= d(m\vec v) \wedge \vec v\\
\end{align*}I'm so sorry but I don't know from there..!

 

[cnh1995]

But Why does the kinematic energy appears in the last equation?

Work done on the body results in change in it's kinetic energy.

 

[PeroK]

Homework Statement

We call Kinetic energy Ec=1/2 mv^2

its deferral into the work integral gives:

\begin{align*}
w_{12}&=\int\limits_1^2\delta W\\
& = \int\limits_1^2 \vec F \wedge d\vec r\\
&\underbrace{=\int\limits_1^2dE_c}_{why?}
\end{align*}

But Why does the kinematic energy appears in the last equation?

Homework Equations

According to Newton's second law,

$$\vec{F}=\frac{d\vec{P}}{dt}$$

The Attempt at a Solution

Multiplicating it by $$d\vec{r}$$

\begin{align*}
\vec F \wedge d\vec r &= \frac{d\vec P}{dt} \wedge d\vec r\\
&= d\vec P \wedge \frac{d\vec r}{dt}\\
&= d\vec P \wedge \vec v\\
&= d(m\vec v) \wedge \vec v\\
\end{align*}I'm so sorry but I don't know from there..!

If it's any consolation, none of those integrals makes mathematical sense to me. Instead:

$\vec{F}.d\vec{r} = m\frac{d\vec{v}}{dt}.\frac{d\vec{r}}{dt}dt = m\frac{d\vec{v}}{dt}.\vec{v}dt = \frac{1}{2}m\frac{d}{dt}(\vec{v}.\vec{v})dt$

And, then if you do your line integral with $t$ as your parameter it comes out.

 

[PeroK]

$\int_C \vec{F}.d\vec{r} = \int_{t_1}^{t_2}\frac{1}{2}m \frac{d}{dt}(v^2)dt$

That's the mathematical format that I'm used to and makes sense to me. A line integral depends on the curve C not just its end- points.