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.

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.

anonymous · Answer

attachment

amistre64 · Answer

are you starting at 9,3 ?

amistre64 · Answer

when t=0, x=9, y=3

that is not what your solutions give us

amistre64 · Answer

sin(u) = 1 when u = pi/2

if u = t/2 + k,  when t=0, what is k?

anonymous · Answer

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?

amistre64 · Answer

by shifting the phase of the function

anonymous · Answer

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.

amistre64 · Answer

yes

or, replacing the constants

amistre64 · Answer

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

amistre64 · Answer

i thinkphase shift might be the way to approch it

anonymous · Answer

yeah i tried just changing constants didnt work

anonymous · Answer

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

amistre64 · Answer

:)