# Parameterisation of a straight line to solve a line integral.

My question is concerned with the parameterisation of a straight line such that I can solve a line integral of the form:

$$\int_c \mathbf F\cdot d\mathbf r$$

where C is the straight line segment from $(0, 0)$ to $(1, 4)$ and $F = [ y^2, -x^2]$.

Thus,

$$\int_c \mathbf F\cdot d\mathbf r = \int_a^b \mathbf F\cdot \mathbf r'(t)dt$$ where $a = t_1, b = t_2$

would be the way to solve the problem. But how do I parameterise C, $y = 4x$, in the form $\mathbf r(t) = [?, ?]$ so that I can actually do the calculus?

Thank you.

• Have you tried starting with something like $x=t$? – amd Feb 7 '17 at 3:05

To generalize Anna's answer, if you want to parametrize the line between two point $a,b \in \mathbb{R}^n$, we can take the map $\gamma:[0,1] \to \mathbb{R}^n$ defined by $t \mapsto a(1-t)+ bt$.
The initial point is $(0,0)$ and the final point of the segment is $(1,4)$, so you can do $\mathbf{r}(t)=\left(x(t),y(t)\right)=(0,0)+t(1,4)$ and it will be $(0,0)$ at $t=0$ and $(1,4)$ at $t=1$. If you simplify it, $\mathbf{r}(t)=(t,4t)$, $t\in[0,1]$.