Question

How would I evaluate the following vector-valued function:
r(t)=cos(t)i+2sin(t)j,
where r(t) is r(theta-pi)?

Answer 1

So far I've plugged in the value so my function looks like this:
cos(theta-pi)i+2sin(theta-pi)j

Answer 2

Yes, now the magnitude r is sqrt(cos(n)^2+sin(n)^2), or 1.

Answer 3

ok thanks.