The question and its answer is given in the following picture:enter image description here

I do not exactly from where the second equality in the following line came, could anyone explain it for me?

enter image description here

  • 3
    $\begingroup$ $d\gamma = \frac{d\gamma}{dt} dt$ $\endgroup$
    – Ben G.
    Jul 1 '17 at 22:49
  • $\begingroup$ Check any vector calculus text (or a multivariate calculus text) for more info on line integrals. This stuff also happens when you deal with differential forms. $\endgroup$ Jul 1 '17 at 22:56
  • 2
    $\begingroup$ The fork done by $\mathbf F$? A solution that doesn't look exactly right? Is this from a Kaplan® book? $\endgroup$
    – user307169
    Jul 1 '17 at 23:17
  • $\begingroup$ I do not think so.@tilper it is written by Charles Rambo. $\endgroup$
    – user426277
    Jul 1 '17 at 23:25

This solution doesn't look exactly right. First, let us define work, the work done by a force $\bf{F}$ along a path $C$ (a curve in the space), is equal to line integral

$$W = \int_C \mathbf{F} \cdot d\mathbf l.$$ Where, $d\mathbf{l}$ is a vector line element.

By definition, we have, as u can see here: wiki

$$\int_C \mathbf{F} \cdot d\mathbf l = \int_a^b \mathbf{F}(r(t)) \cdot \mathbf{r}'(t) dt,$$ where $\mathbf r$ is a parametrization of the curve $C$, such that $\mathbf r(a)$ and $\mathbf r(b)$ match the endpoints of C.

It's the right form of the second equality.


By the way, the force is not equal to the vector $(-1,0,1)$, actually it is $\frac{1}{\sqrt{2}}(-1,0,1)$, since $F=|\mathbf F| =1.$

  • $\begingroup$ why$ F = |\mathbb{F}| =1$? $\endgroup$
    – user426277
    Jul 1 '17 at 23:32
  • 2
    $\begingroup$ The question assert that $\mathbf F$ is a constant unit force. $\endgroup$
    – user301959
    Jul 1 '17 at 23:36

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