Subject math GRE exam 0568 Q.47

The question and its answer is given in the following picture:

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

• $d\gamma = \frac{d\gamma}{dt} dt$ Jul 1 '17 at 22:49
• 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. Jul 1 '17 at 22:56
• The fork done by $\mathbf F$? A solution that doesn't look exactly right? Is this from a Kaplan® book?
– user307169
Jul 1 '17 at 23:17
• I do not think so.@tilper it is written by Charles Rambo.
– 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.

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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 $F=|\mathbf F| =1.$

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