Why do physics books perform this "trick" in integration?

Question

I have the following question, because general physics books like Sears and Serway and others, if not the vast majority, simplify differentials when integrated. Formally I interpret that this is justified by the first fundamental theorem of calculus, but I'm not sure. I would appreciate your support, to clarify this doubt, I attach an example.

Let $v=\frac{dr}{dt}$ denote the velocity vector, as let $r$ denote the position vector. Let $F=m\frac{dv}{dt}$ the projection of the net force with the displacement and denote the force at a specified position. Let the mass $m$ be constant. Then the work done along a curve $C$ is defined as,

$$W=\int_{C} F \cdot dr=m \int_{C} \frac{dv}{dt} \cdot dr$$

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The next steps seem sloppy to me.

$$=m \int_{v_i}^{v_f} \frac{dr}{dt} \cdot \frac {dv} {dr} dr$$ HERE, justification for the "cancellation" of the differential $$=m \int_{v_i}^{v_f} \frac{dr}{dt} \cdot dv$$ $$=m \int_{v_i}^{v_f} v \cdot dv$$

$$=m \int_{|v_i|}^{|v_f|} |v| \,d|v|$$

$$=\frac{1}{2}m|v_f|^2-\frac{1}{2}m|v_i|^2$$

$$=\Delta \text{KE}$$

Answer 1

This makes no sense at all to me. $\vec r$ is a vector variable here, so I have no idea what $\dfrac{d\vec v}{d\vec r}d\vec r$ is supposed to mean. Nor does it make sense for the velocities to become the limits of a $d\vec r$ integration.

It's easy enough to make it correct. \begin{align*} \int_C \frac{d\vec v}{dt}\cdot d\vec r &= \int_{t=a}^{t=b} \frac{d\vec v}{dt}\cdot \frac{d\vec r}{dt}dt = \int_{t=a}^{t=b} \frac{d\vec v}{dt}\cdot \vec v\, dt \\ &= \int_{t=a}^{t=b} \frac12 \frac d{dt}\big(\vec v\cdot \vec v\big)\,dt = \frac12 \|\vec v\|^2 \Big|_{t=a}^{t=b}. \end{align*} The first equality is the definition of the line integral, and the final equality is the Fundamental Theorem of Calculus.

Answer 2

$ \newcommand\diff\underline \newcommand\adj\overline \newcommand\DD[2]{\frac{\mathrm d#1}{\mathrm d#2}} $

We can make this work, though I agree that the way it's presented is very sloppy and perhaps outright wrong.

If $v = v(r(t))$, then by the chain rule $$ \DD vt = \diff{v\circ r} = \diff v\circ\diff r = \diff v\left(\DD rt\right) $$ where $\diff F$ is the total differential of $F$, i.e. the matrix of the linear map $\diff F : \mathbb R^n \to \mathbb R^m$ in the standard basis is the Jacobian matrix. It follows that $$ \DD vt\cdot dr = \diff v\left(\DD rt\right)\cdot dr = \DD rt\cdot\adj v(dr) $$ where $\adj v$ rs the adjoint of $\diff v$ under the inner product. But $\adj v(dr) = dv$, i.e. $\adj v^{-1}$ gives the transformation of $dr$ under the change-of-variables $r \mapsto v(r)$. To be more precise, $$ \adj v(dr) := \adj v(w(r))ds = dv $$ where $w(r)$ is the unit tangent to the curve at $r$ and $ds$ is the scalar arc-length element. So altogether we've shown $$ \DD vt\cdot dr = \DD rt\cdot dv. $$