# Finding the equation of a parametric curve

1. Dec 8, 2012

### rylz

1. if y(t)= (a/t, b/t, c/t)

2. Prove that this curve is a straight line. Find the equation of the line

3. i found the first part without a problem, i just am not sure how to find the equation f the line.

2. Dec 8, 2012

### LCKurtz

Apparently $y(t)$ is a vector instead of the second component of the right side? How did you show it is a straight line without finding its equation? And what does "the equation f " mean? What forms do you know for straight line equations in 3D?

3. Dec 8, 2012

### rylz

y(t) is the parametric curve, and i proved its a straight line by proving the curvature of the line. the "f" is supposed to be of

4. Dec 8, 2012

### Dick

Proving the curvature is 0 is probably the hard way. Like LCKurtz said, "What forms do you know for straight line equations in 3D?"

5. Dec 8, 2012

### rylz

x-xo/a=y-yo/b=z-zo/c

6. Dec 8, 2012

### Dick

That's a good one. It would be even better with pararentheses. Your parametric form is x=a/t, y=b/t, z=c/t. So?

7. Dec 8, 2012

### rylz

so do i sub in a point for t in the domain? and that will give <xo,yo, zo> and then how do i find the direction <a,b,c>.

8. Dec 8, 2012

### Dick

Try subbing <xo,yo,zo>=<0,0,0>.

9. Dec 8, 2012

### rylz

and then the direction (a,b,c) would be point on the line or would it just be (a,b,c)?

10. Dec 8, 2012

### Dick

Well, x/a=1/t, y/b=1/t, z/c=1/t. So x/a=y/b=z/c. Direction vector? Point on the line? Or BOTH?