Question

@Loser66 fun question Parametrize the line segment \[y=2x \text{ where } x \in [1,3] \text{ on the interval } t \in [a,b]\]

Answer 1

I made up this question. Hopefully the wording is correct. I think it is.

Answer 2

Could we just use x=a+bt, y=d+te, where we start at x = a and y = d, then move in the direction of (b,e) from what I remember at least when parametrize lines xD

Answer 3

x=t then y=2t :)

Answer 4

Yeah, x = t, then y = g(t)

Answer 5

hehe, yep. the easy way out

Answer 6

This is what I did:
\[x(t)=ct+d \\ 1 \le ct+d \le 3 \\ \text{ assume } c>0 \\ \frac{1-d}{c} \le t \le \frac{3-d}{c} \\ \text{ choose } a=\frac{1-d}{c} \\ \text{ and } b=\frac{3-d}{c}\]
\[ac=1-d \\ bc=3-d \\ \text{ subtract these equations giving } \\ (a-b)c=-2 \\ c=\frac{-2}{a-b}=\frac{2}{b-a} , \text{ then go back and solve for } d \text{ in terms of } a \text{ and } b \]
\[ac=1-d \\ ac-1=-d \\ d=1-ac \\ d=1-a \cdot \frac{2}{b-a}=1-\frac{2a}{b-a} \\ \text{ so we have } \\ x(t)=\frac{ 2 t}{b-a}+(1-\frac{2a}{b-a}) \\ \text{ or } \\ x(t)=\frac{2}{b-a}(t-a)+1 \\ \text{ and so } \\ y(t)=\frac{4}{b-a}(t-a)+2 \\ (x,y)=(\frac{2}{b-a}(t-a)+1, \frac{4}{b-a}(t-a)+2) \text{ where } t \in [a,b]\]

Answer 7

I think x(t)=t and y(t)=2t
would have worked if I wanted t to be between 1 and 3 like x is

Answer 8

https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=3&cad=rja&uact=8&ved=0CCoQFjACahUKEwiNiOqO6fDIAhXKNj4KHVP5DSQ&url=http%3A%2F%2Fwww.math.uh.edu%2F~jiwenhe%2FMath1432%2Flectures%2Flecture14_handout.pdf&usg=AFQjCNHse8F_z_EXocM7UKXS43_fFIXwPw&sig2=XZn3Y1tXYC_9Xc4wphEm9g
@loser66 and I were studying this site
see page 2

Answer 9

oh, you want the limits as t goes to a to equal the corresponding things?

Answer 10

yes I wanted t to be between a and b

Answer 11

ah, I thought we got to pick a and b.

Answer 12

That is very cool!

Answer 13

It is just kind of fun. And I just wanted to see if @Loser66 could get there. :p

Answer 14

Ah yeah, sorry about intruding

Answer 15

It is fine. He isn't here. I'm glad you guys came so I could have someone to play with.

Answer 16

I'm looking at that pdf, the archimedes spiral looks very fun

Answer 17

Looks fun, yep. We didn't go that far.
I think I seen that spiral thing before somewhere... Not sure I thought I remember something about Fibonacci sequence.

Answer 18

yea, if illustrating the fibonacci sequence you can use a spiral

Answer 19

Reminds me of @FibonacciChick666

Answer 20

hehe yes

Answer 21

here check it out, https://en.wikipedia.org/wiki/Golden_spiral

Answer 22

:D

Answer 23

Look at you @FibonacciChick666 you are famous

Answer 24

I guess those spirals are a little different.

Answer 25

what can I say, I chose my name well. and yea, that's an archimedes spiral

Answer 26

when I google freckles I think I see a lot more results come up then when I google fibonacci though :p

Answer 27

freckles=19,700,000
fibonacci=16,500,000
freckles>fibonacci

Answer 28

you beat astrophysics though
astrophysics=14,400,000

Answer 29

Haha :c

Answer 30

dan815 comes in last place
dan815=20,900

Answer 31

lol

Answer 32

this is how i did it :)

we travel from A to B in (b-a) seconds,
so unit vector with w.r.t time = AB/(b-a) =< 2, 4> /(b-a)


Now we parametrize with t' and velocity (b-a)


parametirc eq C(t' )=\[C=~~<1,2> + \frac{ <2,~4 > }{ b-a }t'\]

but t'=t-a

because 'a' seconds have already passed when we reach pt A. (same logic as shifting origin in graphs

thus
\[C=~~<1,2> + \frac{ <2,~4 > }{ b-a }(t-a)\]

Answer 33

*<2,4> is vector AB ( i forgot to mention)

Answer 34

cute stuff

Answer 35

:p

Answer 36

it might seem easier than what I did :p

Answer 37

jk it is

Answer 38

* when i said "unit vector w.r.t time"
i meant "velocity vector"

:( i worded this very poorly, sorry

Answer 39

its k
I think I'm a bad worder too