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MathSofiya

  • 3 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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  1. MathSofiya
    • 3 years ago
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    oh I wrote it wrong

  2. MathSofiya
    • 3 years ago
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    \[\hat r t=<t+sin\frac12 \pi t,cos \frac12 \pi t>\]

  3. MathSofiya
    • 3 years ago
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    much better!

  4. MathSofiya
    • 3 years ago
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    got it

  5. mukushla
    • 3 years ago
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    still something wrong !!

  6. MathSofiya
    • 3 years ago
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    yeah still wrong \[\hat r t=<t+sin\frac12 \pi t,t+cos \frac12 \pi t>\]

  7. MathSofiya
    • 3 years ago
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    how about now?

  8. TuringTest
    • 3 years ago
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    \[t\vec r=<t+\sin\frac12 \pi t,t+\cos \frac12 \pi t>\]makes more sense

  9. TuringTest
    • 3 years ago
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    but wait, brb

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

  11. TuringTest
    • 3 years ago
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    ?

  12. TuringTest
    • 3 years ago
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    and is t being multiplied by r on the left? this notation is confuzzling me

  13. MathSofiya
    • 3 years ago
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    no it r(t), my bad

  14. TuringTest
    • 3 years ago
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    \[\vec r(t)=\langle t+\sin(\frac\pi2 t),t+\cos(\frac\pi2t)\rangle\]correct?

  15. MathSofiya
    • 3 years ago
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    yes

  16. TuringTest
    • 3 years ago
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    \[\vec r(t)=\langle x(t),y(t)\rangle\]so that implies that\[\vec r(t)=\langle t+\sin(\frac\pi2 t),t+\cos(\frac\pi2t)\rangle\]leads to\[x(t)=t+\sin(\frac\pi2t)\]\[y(t)=t+\cos(\frac\pi2t)\]

  17. TuringTest
    • 3 years ago
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    so it is just a matter of identifying the components

  18. MathSofiya
    • 3 years ago
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    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.

  19. TuringTest
    • 3 years ago
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    no prob, this is not one of the harder parts of vector calculus fortunately :)

  20. MathSofiya
    • 3 years ago
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    :P

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