parametrize the following curve

Question

sweetburger · Answer

$$y=\sin(x)+\cos(x)$$
im guessing i let t=x or something but that produces r(t)=(t,sin(t)+cos(t)) idk if that works

sweetburger · Answer

might be afk feel free to respond

sillybilly123 · Answer

yer already done :) https://www.mathsisfun.com/definitions/parameter.html

sweetburger · Answer

oh cool. I was looking at solutions for the problem set im working on and the question im doing isn't quite the same but the solution says let $t=\frac{ \pi }{ 2 }$ and that is the entire solution haha

sweetburger · Answer

not really sure what the implication of that is given the instructions are purely to parametrize the curve

sillybilly123 · Answer

$y=\sin(x)+\cos(x) =\sqrt{2} \left(  \dfrac {1}{\sqrt {2}} \sin(x)+ \dfrac{1}{\sqrt{2}} \cos(x) \right) =\sqrt{2}  \sin(x + \dfrac{\pi}{4} ) $

but that's just simplifying.  it's already single variable ... so you must wonder what **to parameterise** actually means.  hint

sillybilly123 · Answer

you get this stuff in vector representations too, in which case you are totally correct.

ie $r(x) = <x, \sin x + \cos x>$

you're still not doing anything substantive, though; just using new notation

sweetburger · Answer

Yeah i realize that its just new notation. The vector representation is what I assumed was desired. I'm just still considering what you have written above there and why the solution to this question was t=pi/2

sweetburger · Answer

actual question was y=cos(x)-sin(x) with a solution of t=3pi/2
was just trying to figure out how to do these with out directly doing the one from my problem set

sillybilly123 · Answer

grab the original qu and post it here makes most sense IMHO  :)