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anonymous

  • 4 years ago

Find an equation of the tangent line to the curve y=arccos(x/2) at the point (1,4pi)

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  1. anonymous
    • 4 years ago
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  2. anonymous
    • 4 years ago
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    Any ideas?

  3. anonymous
    • 4 years ago
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    well, the point (1,4pi) is not on the line

  4. anonymous
    • 4 years ago
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    Oh sorry, it was y=3arccos(x/2)

  5. amistre64
    • 4 years ago
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    the slope of the line at any given point is the derivative of the function

  6. amistre64
    • 4 years ago
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    is the equation spose to be in point slope or slope intercept?

  7. anonymous
    • 4 years ago
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    either way, the point is not on the line, the derivative of y=3arccos(x/2) is y'=-3/(sqrt(4-x^2)) subbing x=1 we get the slope we get the slope of -sqrt(3)

  8. anonymous
    • 4 years ago
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    so it narrows down to 2 answers

  9. amistre64
    • 4 years ago
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    cos(60)=1/2 so (1,pi/3) is on it i believe

  10. anonymous
    • 4 years ago
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    -sqrt(3)=(y-1,x-4pi)

  11. anonymous
    • 4 years ago
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    oops, i mean -sqrt(3)=(y-1/x-4pi)

  12. anonymous
    • 4 years ago
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    oops, i mean -sqrt(3)=(y-1)/(x-4pi)

  13. anonymous
    • 4 years ago
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    So the answer would be?

  14. anonymous
    • 4 years ago
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    funny thing is none of them match, so i must have made a mistake somewhere

  15. anonymous
    • 4 years ago
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    Got any ideas amistre?

  16. anonymous
    • 4 years ago
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    i got it

  17. anonymous
    • 4 years ago
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    -sqrt(3)=(y-4pi)/(x-1

  18. anonymous
    • 4 years ago
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    mixed up the coordinates

  19. anonymous
    • 4 years ago
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    answer is the 3rd one

  20. anonymous
    • 4 years ago
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    -sqrt(3)=(y-4pi)/(x-1) isolate y and you'll get the answer

  21. anonymous
    • 4 years ago
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    Thanks man, I never would have got that one, I guess I need more studying :P

  22. anonymous
    • 4 years ago
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    no prob, by math skills are getting rusty

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spraguer (Moderator)
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