Question

how can I determine what x and y is from this:
\[\hat r t=\]

Answer 1

oh I wrote it wrong

Answer 2

\[\hat r t=<t+sin\frac12 \pi t,cos \frac12 \pi t>\]

Answer 3

much better!

Answer 4

got it

Answer 5

still something wrong !!

Answer 6

yeah still wrong
\[\hat r t=<t+sin\frac12 \pi t,t+cos \frac12 \pi t>\]

Answer 7

how about now?

Answer 8

\[t\vec r=<t+\sin\frac12 \pi t,t+\cos \frac12 \pi t>\]makes more sense

Answer 9

but wait, brb

Answer 10

\[t\vec r=\langle t+\sin(\frac\pi2 t),t+\cos(\frac\pi2t)\rangle\]is this right
is 1/2pit the argument of the trig function

Answer 11

?

Answer 12

and is t being multiplied by r on the left?
this notation is confuzzling me

Answer 13

no it r(t), my bad

Answer 14

\[\vec r(t)=\langle t+\sin(\frac\pi2 t),t+\cos(\frac\pi2t)\rangle\]correct?

Answer 15

yes

Answer 16

\[\vec r(t)=\langle x(t),y(t)\rangle\]so that implies that\[\vec r(t)=\langle t+\sin(\frac\pi2 t),t+\cos(\frac\pi2t)\rangle\]leads to\[x(t)=t+\sin(\frac\pi2t)\]\[y(t)=t+\cos(\frac\pi2t)\]

Answer 17

so it is just a matter of identifying the components

Answer 18

yep. It makes sense now. I initially wrote it without putting a comma in, and i forgot a "t" somewhere which made it more confusing. But yeah that makes perfect sense.

Answer 19

no prob, this is not one of the harder parts of vector calculus fortunately :)

Answer 20

:P