Question

In my textbook (I'm using Stewarts calculus if you're interested pg 1066), it says the following equation:
r(t)= (1-t)r0+tr1. The "r" is a vector and 0 and 1 should be subscripts. It used this equation to create a parametric representation of the line segment (-5, -3) to (0,2), x=5t-5 and y=5t-3. How in the world did that happen? It's here, http://books.google.com/books?id=AavjDHGwGpIC&printsec=frontcover&dq=stewart+calculus&hl=en&sa=X&ei=2M3GT5uBHaq36gGQ9fnjCw&ved=0CE8Q6AEwAA#v=onepage&q=stewart%20calculus&f=false on page 1091...

Answer 1

ooh but the paremeter must be from 0<t<1 :)

Answer 2

Yes, I forgot to put that in. It's not really relevant to the question though (I don't believe).

Answer 3

well just plug in your numbers and the equation will work out

Answer 4

and it is relavant cuzz it works only for those points. 0<t<1

Answer 5

I don't get what your'e saying... What am I multiplying with what? Is it supposed to be scalar mutiplication?

Answer 6

you said yourself they are vectors. Use position vectors
(1-t)r0+tr1
(-5, -3) to (0,2)

r0 to r1
ro r1
here it is for x : (1-t)*(-5)+t*(0) -> -5+5*t -> 5t-5 look familiar?

Answer 7

Thanks man, that really helped me out. I was combining the vectors the wrong way . I really have to brush up on vectors...

Answer 8

for y
(1-t)r0+tr1

(1-t)*(-3)+t*2
| | distribute the -3
-3+3t + t*2
-3+3t + 2t rearrange
-3+5t combine like terms
5t-3 rearrange again look familiar?

Answer 9

yup just remember do everything in terms of one component y,x and z

dont combine them dont go like (1-t)*y+t*x
^ ^
instead always do this (1-t)*x0+t*x1