Question

Eliminate the parameter t from the parametric equations. x=-1+sint, y=3+cost

Answer 1

hint : \(\sin^2t + \cos^2t = 1\)

Answer 2

i still cant get t eliminated

Answer 3

solve for sin t and for cos t then square and add

Answer 4

getting a high number which isn't possible

Answer 5

\[(x+1)^2 + (y-3)^2 = 1\]

Answer 6

circle with radius 1 centered at (-1, 3)

Answer 7

i keep getting 6-dont know

Answer 8

how?

Answer 9

\[x=-1+\sin t \Rightarrow \sin t = x + 1\]\[y=3+\cos t \Rightarrow \cos t = y -3\]

Answer 10

thx

Answer 11

do i have to solve? this is something we havent learned yet, just ec

Answer 12

you have to do what you have to do...

Answer 13

k

Answer 14

remember
\[\sin^2 t + \cos^2 t = 1\]then substitue in and no more t!

Answer 15

(2x+2) + (2y-9)=1...y=5 x=-1

Answer 16

?

Answer 17

i guess i am confused

Answer 18

after eliminating t, i thought i needed to solve further?

Answer 19

\[\sin t = (x+1) \Rightarrow \sin^2 t = (x+1)^2\]

Answer 20

likewise for cos t and then you have the curve in terms of x and y only, not t

Answer 21

ok, so once t is removed there is no need to solve further?

Answer 22

yeah, you look to see what you have... if it's a curve you recognize (hopefully).

Answer 23

ah, ok

Answer 24

\[(x+1)^2 + (y-3)^2 = 1\] is a circle with radius 1 , centered at (-1, 3)