Question

parameterized Equations

Answer 1

I'm curious. How do you do it?

Answer 2

whoops i need help with question #1

Answer 3

@Mr.Math maybe he knows how to do it

Answer 4

I think the first graph is \(y=x^2\), but I don't know the parameterization.
Maybe \(y=t^2\),\(x=t\). I'm assuming parameterization has something to do with parametric equation.

Answer 5

Thats incorrect i think

Answer 6

Still you gave me a medal?

Answer 7

Joemath is here, he probably knows this stuff.

Answer 8

Like i thought y=t^2 and x= is smth with cost

Answer 9

But if it were like that then, \(y = x^2 + C\).
For x = -1 or 1, \(y = 1 + C\). But in the graph y is just 1, so C has to be Zero.

Answer 10

i cant think of a good way to explain it >.< the first one is:\[x=\cos(t+\pi),y=\sin(t+\pi)+1, 0\le t\le \pi\]

Answer 11

joe is kinda on the button

Answer 12

Like this is the books answer

Answer 13

its the bottom half of a circle shifted up one unit. A circle is normally x = cos t, y = sin t. if you move it up one unit, that adds one to the y coordinate. and you want the bottom half, so you add pi to the angle to start on the other side.

Answer 14

Joe is always right! He is the Professor.

Answer 15

x=cost and y=1+sint pi till 2pi

Answer 16

my answer is the same as that. instead of putting (t + pi) and going from 0 to pi, they just put (t), and went from pi to 2 pi. its the same thing.