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

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MathSofiya
 2 years ago
Best ResponseYou've already chosen the best response.1\[\hat r t=<t+sin\frac12 \pi t,cos \frac12 \pi t>\]

mukushla
 2 years ago
Best ResponseYou've already chosen the best response.0still something wrong !!

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

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

TuringTest
 2 years ago
Best 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

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.3and is t being multiplied by r on the left? this notation is confuzzling me

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

TuringTest
 2 years ago
Best 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\[x(t)=t+\sin(\frac\pi2t)\]\[y(t)=t+\cos(\frac\pi2t)\]

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.3so it is just a matter of identifying the components

MathSofiya
 2 years ago
Best ResponseYou've already chosen the best response.1yep. 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.

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