Question

How do you parameterize a curve? For example a line segment from (0,2) to (1,4).

Answer 1

do you understand how to parameterize?

Answer 2

i mean what it is?

Answer 3

Yeah i think so

Answer 4

I was looking over the calc III stuff, and I needed to parameterize a curve. and they used the formula (1-t)<x,y>+t<x,y> something like this. And I didn't understand this

Answer 5

the formula 1-t<x,y> i'm in calc 3 myself but we never parameterized anything like this what was the curve?

Answer 6

?

Answer 7

if it's above find the line that these points make

Answer 8

It's not really a curve in this case. It's a line but how do you parameterize it? is it that formula? 1-t<x,y>?

Answer 9

when parameterizing you can use any paremeter

Answer 10

literally in my book, this section for the book didn't even have answers

Answer 11

if they give you a parameter it's a little bit different

Answer 12

Yeah, it was way back in chapter 13. I got it now.

Answer 13

At the point (0,2) the vector looks like 0i+2j. When parameterizing a curve I set it up as (0i+2j+t)(i+2j). The i+2j comes from going from the point (0,2) to (1,4). The difference is 1i and 2j. Therefore, the parameterized curves are ti+(2+2t)j where x=t and y=2+2t

Answer 14

lol i did the slope wrong ... it's late yes... jt is correct it should be 2t+2... i did x/y ...fail

Answer 15

m = 2
y-2=2x

y=2x+2

x=t
y=2t+2

Answer 16

for trig usually you'll need to use an identity