Symbolic manipulation inside integral

147 Views Asked by At

I'm an undergrad who has just completed the standard calculus sequence (1, 2, and multivariable). I've done well in the courses, however, things like the following, which is a derivation of kinetic energy, still confuse me:

$$ \mathbf{F} \cdot \mathrm{d}\mathbf{x} = \mathbf{F} \cdot \mathbf{v} \mathrm{d} t = \frac{\mathrm{d}\mathbf{p}}{\mathrm{d}t} \cdot \mathbf{v} \mathrm{d} t = \mathbb{v} \cdot \mathrm{d} \mathbf{p} = \mathbf{v} \cdot \mathrm{d}(m \mathbf{v}).$$

Taken from here.

I want to understand the symbolic manipulation that often occurs when making meaningful integrations. I was taught that the ending 'dx' term simply signifies the variable to be integrated over. However, it is commonly used, for example, as a term to cancel things out. In general, I see a lot of symbolic manipulation with differential elements that I want to understand. Could you recommend something I could read to better understand this stuff?

Thank you.

1

There are 1 best solutions below

0
On

It is actually possible to define these kind of concepts only when every statement is meaningful: the example you reported lacks of sense (or many passages are in the best case implied) and rigour although it is a very common way to present introductory physics. Just for the case you mentioned: given the position of a material point with respect to time as $\mathbf{r}=\mathbf{r}\left(t\right)$, the velocity vector is $\mathbf{v}=\mathbf{v}\left(t\right)=\dot{\mathbf{r}}\left(t\right)$. One can define the power of a force $\mathbf{F}=\mathbf{F}\left(\mathbf{r},\mathbf{v},t\right)$ acting on the material point as

$\mathbf{F}\cdot\mathbf{v}=\mathbf{F}\left(\mathbf{r},\mathbf{v},t\right)\cdot\mathbf{v}\left(t\right)$

The second law of dynamics gives $\mathbf{F}$ (intended as the resultant of all the forces acting on the mp) as $ \mathbf{F}\left(\mathbf{r},\mathbf{v},t\right)=\dot{\mathbf{p}}\left(t\right) $ If we assume that mp has constant mass $m\left(t\right)=m$ then $ \mathbf{F}\cdot\mathbf{v}\left(t\right)=m\dot{\mathbf{v}}\cdot\mathbf{v}=\frac{m}{2}\frac{\mathrm{d}}{\mathrm{d}t}\left(\mathbf{v}\cdot\mathbf{v}\right)=\frac{\mathrm{d}}{\mathrm{d}t}\left(\frac{1}{2}m\left|\mathbf{v}\left(t\right)\right|^{2}\right) $ We can now integrate wrt time to obtain $ \int_{t_{1}}^{t_{2}}\mathrm{d}t\left(\mathbf{F}\left(\mathbf{r},\mathbf{v},t\right)\cdot\mathbf{v}\left(t\right)\right)=\frac{1}{2}m\left(\left|\mathbf{v}\left(t_{2}\right)\right|^{2}-\left|\mathbf{v}\left(t_{1}\right)\right|^{2}\right) $ which is the well known relationship between work of the force $\mathbf{F}$ and difference of kinetic energy $T\left(t\right)=\frac{1}{2}m\left|\mathbf{v}\left(t\right)\right|^{2}$. We can now proficiently restrict to positional forces (e.g. $\mathbf{F}=\mathbf{F}\left(\mathbf{r}\right)$) and to known trajectory $\gamma$ for mp (e.g. because of constraints); we can also choose a coordinate $s=s\left(t\right)$ along the curve to obtain $\int_{t_{1}}^{t_{2}}\mathrm{d}t\left(\mathbf{F}\left(\mathbf{r}\left(t\right)\right)\cdot\mathbf{v}\left(t\right)\right)=\int_{s_{1}=s\left(t_{1}\right)}^{s_{2}=s\left(t_{2}\right)}\mathrm{d}s\left(\mathbf{F}\left(s\right)\cdot\boldsymbol{\tau}\right)=\int_{s_{1}}^{s_{2}}\mathrm{d}s\left(F_{\parallel}\left(s\right)\right)$ where $\boldsymbol{\tau}\left(s\right)=\frac{\mathrm{d}\mathbf{r}}{\mathrm{d}s}$ is the tangent vector to the curve and the last result is the very common way to introduce "work of a force" but, as you can see, is only a particular case and formal manipulation as that of $\mathbf{F}\cdot\mathrm{d}\mathbf{x}$ are by the way only sensed inside integrals (with particular care, also!) as you were taught.