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how can I determine what x and y is from this:
\[\hat r t=<t+sin\frac12 \pi t+cos \frac12 \pi t>\]
 one year ago
 one year ago
how can I determine what x and y is from this: \[\hat r t=<t+sin\frac12 \pi t+cos \frac12 \pi t>\]
 one year ago
 one year ago

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MathSofiyaBest ResponseYou've already chosen the best response.1
oh I wrote it wrong
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
\[\hat r t=<t+sin\frac12 \pi t,cos \frac12 \pi t>\]
 one year ago

mukushlaBest ResponseYou've already chosen the best response.0
still something wrong !!
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
yeah still wrong \[\hat r t=<t+sin\frac12 \pi t,t+cos \frac12 \pi t>\]
 one year ago

TuringTestBest ResponseYou've already chosen the best response.3
\[t\vec r=<t+\sin\frac12 \pi t,t+\cos \frac12 \pi t>\]makes more sense
 one year ago

TuringTestBest ResponseYou've already chosen the best response.3
\[t\vec r=\langle t+\sin(\frac\pi2 t),t+\cos(\frac\pi2t)\rangle\]is this right is 1/2pit the argument of the trig function
 one year ago

TuringTestBest ResponseYou've already chosen the best response.3
and is t being multiplied by r on the left? this notation is confuzzling me
 one year ago

TuringTestBest ResponseYou've already chosen the best response.3
\[\vec r(t)=\langle t+\sin(\frac\pi2 t),t+\cos(\frac\pi2t)\rangle\]correct?
 one year ago

TuringTestBest ResponseYou've already chosen the best response.3
\[\vec r(t)=\langle x(t),y(t)\rangle\]so that implies that\[\vec r(t)=\langle t+\sin(\frac\pi2 t),t+\cos(\frac\pi2t)\rangle\]leads to\[x(t)=t+\sin(\frac\pi2t)\]\[y(t)=t+\cos(\frac\pi2t)\]
 one year ago

TuringTestBest ResponseYou've already chosen the best response.3
so it is just a matter of identifying the components
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
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.
 one year ago

TuringTestBest ResponseYou've already chosen the best response.3
no prob, this is not one of the harder parts of vector calculus fortunately :)
 one year ago
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