Question

Can someone please help me with part three of this problem attached in the image below. It's not accepting my answer even though i'm pretty sure it's correct. Any help would be appreciated.

Thank You.

Answer 1

are you starting at 9,3 ?

Answer 2

when t=0, x=9, y=3

that is not what your solutions give us

Answer 3

sin(u) = 1 when u = pi/2

if u = t/2 + k, when t=0, what is k?

Answer 4

I get what you're saying. I see the problem but I am unsure exactly what you're asking. How do I change the starting points?

Answer 5

by shifting the phase of the function

Answer 6

so inside the sin function it will be something along the lines of (t/2+c) where c is a constant that shifts the function to 9,3.

Answer 7

yes

or, replacing the constants

Answer 8

x = 9+0 = 9
y = 9-6 = 3

maybe,

Answer 9

i thinkphase shift might be the way to approch it

Answer 10

yeah i tried just changing constants didnt work

Answer 11

oksy yeah by adding (pi/2) inside each trig function shifted it thank you

Answer 12

:)