# Thread: Parameterize by a continuous map a polygonal curve

1. ## Parameterize by a continuous map a polygonal curve

Hi, some help with this problem in intro analysis please:

Parameterize by a continuous map f:[0,1]-->S in R^3 the polygonal curve L from (0,0,0) to (2,1,1) to (2,3,2) to (3,3,3).

Thanks a lot.

2. Originally Posted by AKTilted
Hi, some help with this problem in intro analysis please:

Parameterize by a continuous map f:[0,1]-->S in R^3 the polygonal curve L from (0,0,0) to (2,1,1) to (2,3,2) to (3,3,3).

Thanks a lot.

The line from (a,b,c) to (x,y,z) in R^3 is given by (a,b,c) + t*(x-a, y-b, z-c), with t in [0,1]. Since you need the parameter in [0,1] for the three poligonal lines you need to "scale" it:

== (a,b,c)) + 3t*((x-a), (y-b), (z-c)), t in [0,1/3] is from (a,b,c) to (x,y,z) ==> in your case we get:

(0,0,0) + 3t(2, 1, 1) = {(6t, 3t, 3t) , t in [0,1/3]}

== (x,y,z) + (3t-1)(p-x, q-y,r-z), with t in [1/3,2/3] is from (x,y,z) to (p,q,r), so in your case is...etc

Tonio

==

3. Is the last set of parameters:

== (p,q,r)+(3t-2)(d-p, e-q, f-r) with t in [2/3,1] is from (p,q,r) to (d,e,f)?
Thanks

4. Originally Posted by AKTilted
Is the last set of parameters:

== (p,q,r)+(3t-2)(d-p, e-q, f-r) with t in [2/3,1] is from (p,q,r) to (d,e,f)?
Thanks

Indeed. Very nice. Finally, instead (a,b,c), (x,y,z) etc. use your own vectors.

Tonio