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trusev1 Group Title

Find all points (i.e. all x-values) between 0 and 2pi where the line tangent to the graph of y =sinx/2+cosx is horizontal

  • one year ago
  • one year ago

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  1. tkhunny Group Title
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    Have you considered the 1st Derivative?

    • one year ago
  2. trusev1 Group Title
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    1st derivative would be -(cot x) (cosx/-sinx) how would I get the points from that?

    • one year ago
  3. tkhunny Group Title
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    If that is the correct 1st derivative, which I didn't check, you determine where it is zero.

    • one year ago
  4. trusev1 Group Title
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    well the derivative of sinx is cosx and the derivative of any constant is always 0 and the derivative of cosx is -sinx and we know that cotx =cosx/sinx

    • one year ago
  5. trusev1 Group Title
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    I am not sure how I would determine where it would be zero..

    • one year ago
  6. RaphaelFilgueiras Group Title
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    horizontal tg = 0(no variation)

    • one year ago
  7. tkhunny Group Title
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    No. That will not do. You need the Quotient Rule. I get the 1st Derivative as \(\dfrac{2\cos(x)+1}{(2+\cos(x))^{2}}\). After that, simply concern yourself with the numerator being zero, unless the denominator happens to be zero in the same place.

    • one year ago
  8. tkhunny Group Title
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    This assumes your original expression was \(\dfrac{\sin(x)}{2+\cos(x)}\). Was it?

    • one year ago
  9. RaphaelFilgueiras Group Title
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    |dw:1372217350005:dw|

    • one year ago
  10. trusev1 Group Title
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    yes tkhunny that was the original equation

    • one year ago
  11. tkhunny Group Title
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    Then I gave the correct 1st Derivative and you should use parentheses to clarify your meaning. Remember your Order of Operations.

    • one year ago
  12. tkhunny Group Title
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    gtg sorry.

    • one year ago
  13. trusev1 Group Title
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    rapahael can you still help?

    • one year ago
  14. RaphaelFilgueiras Group Title
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    yes

    • one year ago
  15. trusev1 Group Title
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    so do I need to set the numerator =0

    • one year ago
  16. RaphaelFilgueiras Group Title
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    yes, because if the tangent line is horizontal, his angular coefficient is zero

    • one year ago
  17. trusev1 Group Title
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    so then I end up with cosx=-1/2

    • one year ago
  18. RaphaelFilgueiras Group Title
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    yes

    • one year ago
  19. trusev1 Group Title
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    so is that my only x value , -1? ( my reasoning behind that is that cos of 60degrees=1/2) so then y x value is -1?

    • one year ago
  20. RaphaelFilgueiras Group Title
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    |dw:1372217864898:dw|

    • one year ago
  21. RaphaelFilgueiras Group Title
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    pi-pi/3,pi+pi/3

    • one year ago
  22. RaphaelFilgueiras Group Title
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    @trusev1 get it?

    • one year ago
  23. trusev1 Group Title
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    I get that but it is asking for the x-values that are at the line tangent is horizontal. so we got the derivative and even from your picture both those points would have the same x-value.

    • one year ago
  24. RaphaelFilgueiras Group Title
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    no. my figure isnt the graph of your question,it's just the trigonometric circle

    • one year ago
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