Setting up a path for a line integral

[hivesaeed4]

If $${C}$$ is the straight line that connects points in the plane $${(x_i, y_i)}$$ and $${(x_f, y_f)}$$, find a path $${\tmmathbf{r} (t)}$$ that traces out $${C}$$ starting at the initial point $${(x_i, y_i)}$$ and ending at $${(x_f, y_f)}$$ as $${t}$$ goes from zero to 1.

Now the path that I've been able to come up with is

r=(xf-xi)cos(t) i +(yf-yi)sin(t) j

Note r is a vector giving the path and i and j are the unit vectors.

Is my path correct or is their some error in it?

 

[DonAntonio]

If $${C}$$ is the straight line that connects points in the plane $${(x_i, y_i)}$$ and $${(x_f, y_f)}$$, find a path $${\tmmathbf{r} (t)}$$ that traces out $${C}$$ starting at the initial point $${(x_i, y_i)}$$ and ending at $${(x_f, y_f)}$$ as $${t}$$ goes from zero to 1.

Now the path that I've been able to come up with is

r=(xf-xi)cos(t) i +(yf-yi)sin(t) j

Note r is a vector giving the path and i and j are the unit vectors.

Is my path correct or is their some error in it?

I must be misunderstanding something as the path you described above is NOT a straight one, as you required at the beginning...

Question: why not simply take the straight line connecting the two points??

DonAntonio

 

[hivesaeed4]

So the path should be something like:

r=(xf-xi)i + (yf-yi) j ?

 

[muscaria]

The problem is you have simply defined a vector and not how one would move along the line the vector is describing (which is quite straight forward, how would you do this?). If you're having trouble i came across a lecture on parametric equations for lines by denis auroux at mit (in youtube lecture 5 mit course on multivariable calculus, the first 10 mins will do for you)

Hope this helps.

 

[HallsofIvy]

So the path should be something like:

r=(xf-xi)i + (yf-yi) j ?

That is a vector from (xi, yi) to (xf, yf)- which doesn't have a variable "t". The path itself would be x= (xf- xi)t+ xi, y= (yf- yi)t+ yi. You can see that is right by taking t= 0 and t= 1.