Find the work done by the force field $\vec{F}(x, y, z) = (x, y)$ when a particle is moved along the straight line-segment from $(0,0,1)$ to $(3,1,1)$
Attempt:
$\vec{C}(t) = (3t, t, 1), 0 \leq t \leq 1$
Work done is
$\int_{C} \vec{F}(\vec{C}(t)) = \int_{C} \vec{F}(\vec{C}(t)) \cdot \vec{C}'(t) dt = \int_{0}^{1} (3t, t, 0)(3,1,0) dt = \int_{0}^{1} (9t + t) = 5$
yay or nay?
Best Answer
Your answer is correct, nicely done. Just as some feedback:
Comments:
To be correct you should write $F(x,y,z)=(x,y,0)$. You can't omit the third coordinate in $\mathbb{R}^3$.
As far as the computation is concerned, a more correct notation might be as follows: let $\vec{r}(t)=(3t,t,1)$ for $0\le t \le 1$ denote the parametrization you proposed:
$$\int_C \vec{F}\cdot d\vec{r}=\int_C\vec{F}(\vec{r}(t))\cdot d\vec{r}=\int_C\vec{F}(\vec{r}(t))\cdot \vec{r}'(t)dt=\int_0^1(3t,t,0)\cdot (3,1,0)dt=\int_0^19t+tdt=5.$$ Sometimes it is useful to distinguish between the curve $C$ and its parametrization $\vec{r}(t)$.