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Determine the equation for the the tangent

  1. Sep 1, 2012 #1
    1. The problem statement, all variables and given/known data
    Determine the equation for the the tangent-to the curve:
    y=3 sin 2x - cos 2x
    if x=3∏/4


    2. Relevant equations
    So I thought I might get the y?

    y=3 sin 2(3∏/4) - cos 2(3∏/4)
    y= ~0.25 - ~1 ≈ 0.75

    k(?) =
    m(?) =
    Then what?
    Please help :(

    3. The attempt at a solution

    The answer is y = -2x + 3∏/2 - 3
     
    Last edited: Sep 1, 2012
  2. jcsd
  3. Sep 1, 2012 #2

    rock.freak667

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    For y=3 sin2x - cos2x, how would you find the gradient function i.e. the function which gives the gradient at any point x?
     
  4. Sep 1, 2012 #3

    eumyang

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    This isn't correct. If you are using a calculator, are you sure you are in radian mode?

    Once you get the correct y coordinate in the point, find the derivative of y. Plug x = 3π/4 into the derivative to get the slope of the tangent line. Finally, plug in the point and slope into the point-slope form of the equation of a line.
     
  5. Sep 1, 2012 #4
    I don't know really. Should I deriving?
     
  6. Sep 1, 2012 #5
    No I think its adjustments are set to angles. Is it possible to do it without setting it to radians?
     
  7. Sep 1, 2012 #6

    Bacle2

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    Do you know the definition of the derivative in terms of the tangent line?

    This is what I think roc.freak was getting at.
     
  8. Sep 1, 2012 #7

    eumyang

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    You could also memorize the sine and cosine of special angles.
    [itex]y = 3 \sin \left(2 \cdot \frac{3\pi}{4}\right) - \cos \left(2 \cdot \frac{3\pi}{4}\right)[/itex]
    [itex]y = 3 \sin \left(\frac{3\pi}{2}\right) - \cos \left(\frac{3\pi}{2}\right)[/itex]
    3π/2 is one such special angle. What is
    [itex]\sin \left(\frac{3\pi}{2}\right)[/itex]
    and
    [itex]\cos \left(\frac{3\pi}{2}\right)[/itex]?
     
  9. Sep 1, 2012 #8
    When I put that on the calculator I get:
    sin to be -0.99... and cos to be -2.38 × 10-3 :cry:
    And it's set to radians.
     
  10. Sep 1, 2012 #9
    Should I take 3 × -0.9999 = -3. Is that the gradient?
     
  11. Sep 1, 2012 #10
    No. Sorry. I know how to derivate sin, cos etc. though :redface:
     
  12. Sep 1, 2012 #11

    eumyang

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    Something is wrong with your calculator. Because
    sin (3π/2) = -1 and cos (3π/2) = 0.

    No, it's not the gradient! It's the y-coordinate of the point where the tangent line intersects the curve. Please reread my post.
     
  13. Sep 1, 2012 #12

    Oh sorry!:redface: Hm. Okay so is y= -1 then? Sigh.. I don't know?
    Should I derivate the function and put -1 = (the derivation)? (And include x)?
     
  14. Sep 1, 2012 #13

    eumyang

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    No, y does not equal -1. There was a 3 in front of the sine.

    No again. You should take the derivative of
    [itex]y = 3 \sin 2x - \cos 2x[/itex]
    [itex]y' = ...[/itex]
    and then plug in x = 3π/4. y' is the slope of the tangent line at x = 3π/4.
     
  15. Sep 1, 2012 #14
    So
    y' = 3cos2x + 2sin2x right?
    y' = 3 cos 2(3∏/4) + 2 sin 2(3∏/4) = - 4 ?
    Could you lead me through please? I can't do this on my own. :confused:
     
  16. Sep 1, 2012 #15

    eumyang

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    No. You are forgetting the chain rule.
     
  17. Sep 1, 2012 #16
    I give up... Thanks very much for the help though.. :)
     
  18. Sep 1, 2012 #17

    HallsofIvy

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    The problem appears to be that you are trying to do a Calculus problem involving trig functions but have never learned Calculus or Trigonometry.
     
  19. Sep 2, 2012 #18

    Simon Bridge

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    It's in "homework" ... so it looks like OP is doing a course that requires calculus and trigonometry without having learned the prerequisites.

    Actually, the question would be a good entry test for any situation where a basic understanding of calc and trig are required but not taught. Like a job interview.
     
  20. Sep 2, 2012 #19
    Yes. It's in "Homework" which means that I'm no pro on this one. I just started with the concept of derivations in trigonometry. I get that there're a lot of rules when one derivate functions including trigonometry. What have caused me to fail the calulations is because my calculator seem to not be set to the primary settings. I hoped that someone could more or less tell me the finnish. That would be helpful, not just telling me what's wrong all the time. :smile:
     
  21. Sep 2, 2012 #20

    Simon Bridge

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    You mean, "derivatives of trigonometric functions"?
    Or, that your course is on the concepts of differentiating trigonometric functions?
    "when one differentiates functions"
    You don't need a calculator at all so this is not true.
    That would be against the forum rules.

    The idea is that you do your homework - we can help out where you get stuck but we will not do the work for you. Thing is, you have been stuck on everything. You do not know how a sine function varies with angle, and you don't know the chain rule. Both of these things were in an earlier part of your coursework.

    Therefor: you have to go back over your course notes.
     
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