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v.s

  • 3 years ago

Differentiate with respect to t. y = b cos t + t2 sin t

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  1. Shriya
    • 3 years ago
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    sin t (2-b) + 2t cos t

  2. v.s
    • 3 years ago
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    - sint

  3. Chlorophyll
    • 3 years ago
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    ( cost)' ? (sint)' ?

  4. Chlorophyll
    • 3 years ago
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    -> b ( cos t ) ' = - bsint Okie, so far?

  5. v.s
    • 3 years ago
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    okayy

  6. v.s
    • 3 years ago
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    y' = -bsin+2tcos

  7. Shriya
    • 3 years ago
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    y=bcost+t2sint differentiating w.r.t t, dy/dt= -bsint+ d/dt(2tsint) now, we have to differentiate 2tsint using product rule for derivative. so, the derivative of 2tsint becomes (2tcost+2tsint) therefore, dy/dt= -bsint+2tsint+2tcost taking sint common, dy/dx=sint(2-b)+2tcost

  8. v.s
    • 3 years ago
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    i got it wrong

  9. v.s
    • 3 years ago
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    but i understand what you did

  10. Shriya
    • 3 years ago
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    datz ok! :) not a big deal. derivatives is realy easy as compared to the other sections of calculus. u jus need 2 know the ways to differentiate! :)

  11. v.s
    • 3 years ago
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    yeaaa thank you :)

  12. Chlorophyll
    • 3 years ago
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    Apply product rule here ( t²* sin t ) ' with : u = t² --> u' = ... v = sint --> v' = ....

  13. Chlorophyll
    • 3 years ago
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    @v.s Do you know product rule?

  14. v.s
    • 3 years ago
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    yes

  15. v.s
    • 3 years ago
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    2t*sint + cost*t^2

  16. Chlorophyll
    • 3 years ago
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    Yep, now put the 2 results together :) y' = ....

  17. Chlorophyll
    • 3 years ago
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    y = b cos t + t² sin t ->y' = ...

  18. v.s
    • 3 years ago
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    y"= -bcost+2t*sint+cost*t^2

  19. Chlorophyll
    • 3 years ago
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    y = b cos t + t² sin t -> b ( cos t ) ' = - bsint -> ( t² sin t )' = 2t sint+ t² cost

  20. v.s
    • 3 years ago
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    you just put them together

  21. v.s
    • 3 years ago
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    - bsint + 2t sint+ t² cost

  22. Chlorophyll
    • 3 years ago
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    What's my instruction above?

  23. v.s
    • 3 years ago
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    product rule?

  24. Chlorophyll
    • 3 years ago
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    The product rule applied to take derivative of the second term!

  25. v.s
    • 3 years ago
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    i'm confused

  26. Chlorophyll
    • 3 years ago
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    y = b cos t + t² sin t -> b ( cos t ) ' = - bsint ( derivative of the first term) -> ( t² sin t )' = 2t sint+ t² cost ( derivative of the second term)

  27. v.s
    • 3 years ago
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    what do you do next?

  28. Chlorophyll
    • 3 years ago
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    Yep, now put the 2 results together :) y = b cos t + t² sin t y' = - bsint + 2t sint+ t² cost

  29. Chlorophyll
    • 3 years ago
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    Now you can make it neat by simplifying - bsint + 2t sint together.

  30. v.s
    • 3 years ago
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    okaay thnks you

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