Question

don't understand why i'm getting this wrong...

(x-3)^2+(x-1)^2=4

if x=3+2cos(t) then y = ?

i said 1+2sin(t), but webwork is saying i'm getting it wrong...not sure why...can someone explain it to me?

Answer 1

you mean
(x-3)^2+(y-1)^2=4
right ?
and y = 1+2sin t seems to be correct
maybe they need the answer in another form ?

Answer 2

i don't know what other form to put it in...it's the same form as the x is. i've tried 2sin(t)+1....can't see why it's wrong.

Answer 3

(y-1)^2 = 4 sin^2 t
y-1 = \(\pm\) 2 sin t
so maybe try 1 - 2sin t ?
but 1+2 sin t should also work...

Answer 4

(x-3)^2+(x-1)^2=4
x=3+2cost what is y is your question but there is no y in first or second equation how are you meant to substitute how did you get the answer

Answer 5

i think she means (x-3)^2+(y-1)^2=4

Answer 6

just to try to work something out ...

(x-3)^2+(y-1)^2=4

(3 + 2cos(t) -3)^2+(y-1)^2=4

4cos^2(t)+(y-1)^2=4

(y-1)^2=4 - 4cos^2(t)

(y-1)^2=4(1 - cos^2(t))

(y-1)^2=4sin^2(t)

y-1= +- 2sin(t)

y= 1 +- 2sin(t)

yeah, just making sure in my own head :)

Answer 7

ok so in that case (3+2cost-3)^2+(y-1)^2=4
(y-1)^2=4-(2cost)^2 ...yeah well you beat me to it :) amistre64

Answer 8

It might accept this answer: \[\Large y = 1 \pm 2\sin(t)\]

Answer 9

there is no way to do a +/-, but i'll try minus...

Answer 10

ok, the minus was correct...wonder why it wasn't accepting +...i'll have to email my teacher and ask why both answers aren't correct...

Answer 11

ohh :O
atleast one of them worked :P

Answer 12

maybe it's just webwork being fickle. i don't know. cause yes, if it's a circle, then it is +/-.