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baldymcgee6 Group Title

Derivatives of parametric curves: http://screencast.com/t/x5c5hBqr

  • one year ago
  • one year ago

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  1. baldymcgee6 Group Title
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    @hartnn, @klimenkov

    • one year ago
  2. baldymcgee6 Group Title
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    |dw:1351711571629:dw|

    • one year ago
  3. hartnn Group Title
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    ok, d2y/dx^2 = d/dx (-cot t) , right ? d/dx (-cot t) = d/dt(-cot t) * dt/dx and u know dt/dx from dx/dt

    • one year ago
  4. klimenkov Group Title
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    By what parameter you want to differentiate cot?

    • one year ago
  5. hartnn Group Title
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    i think x, isn't it ?

    • one year ago
  6. baldymcgee6 Group Title
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    |dw:1351712169898:dw|

    • one year ago
  7. hartnn Group Title
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    u got what i did ?

    • one year ago
  8. baldymcgee6 Group Title
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    when t = pi/4

    • one year ago
  9. baldymcgee6 Group Title
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    I'm still confused how you got d/dt(-cot t) * dt/dx

    • one year ago
  10. hartnn Group Title
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    |dw:1351712487011:dw|

    • one year ago
  11. hartnn Group Title
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    \(\huge \frac{d}{dx}f(t)=\frac{d}{dt}f(t) \frac{dt}{dx}\)

    • one year ago
  12. baldymcgee6 Group Title
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    oohhh.. okay... so we can write:|dw:1351712793765:dw|

    • one year ago
  13. baldymcgee6 Group Title
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    |dw:1351712929460:dw|

    • one year ago
  14. hartnn Group Title
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    seems correct, now u can put t=pi/4

    • one year ago
  15. baldymcgee6 Group Title
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    |dw:1351713083407:dw|

    • one year ago
  16. hartnn Group Title
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    yup.

    • one year ago
  17. baldymcgee6 Group Title
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    hmmmm... okay thanks so much for the help... you're the best!

    • one year ago
  18. hartnn Group Title
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    not really... welcome ^_^

    • one year ago
  19. baldymcgee6 Group Title
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    i am still a little bit confused how this works: d/dx (-cot t) = d/dt(-cot t) * dt/dx but I will just have to study a bit more

    • one year ago
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spraguer (Moderator)
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is replying to Can someone tell me what button the professor is hitting...

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