line integrals and work

Question

I was doing practice problems for vector calculus when I jumped into some confusion. The question asked to evaluate a line integral which is general taking the integral of the function then multiplied with the magnitude of the velocity function. However, when I looked at the solution, instead of taking the $\int f(t)|v(t)| dt$, it used the work equation $\int f(t).dr$. What's the difference? How do I know to use $f.dr$ and $f(t)|v(t)|$?

Answer 1

When you compute a line integral of the form $$ \int_{C} f(x,y,z)\; dr $$ on a line $C$, typically you need to find a parametrization of $C$ of the form $$ \vec{r}(t)=x(t)\vec{i}+y(t)\vec{j}+z(t)\vec{k}, \quad t\in [a,b] $$ So that you can re-express the integral in terms of the parameter $t$: $$ \int_{C} f(x,y,z)\; dr = \int_a^b f(x(t),y(t),z(t))\;||\vec{r}'(t) ||\; dt $$ So as mentioned by @mathlover, both integrals are equivalent, they are just not expressed in the same way.