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Antiderivative of y

  1. Sep 23, 2016 #1
    1. The problem statement, all variables and given/known data
    x=sec(t),
    y=tan(t),
    -π/2 ≤ t ≤ π/2
    Try to find y in terms of x

    2. Relevant equations


    3. The attempt at a solution
    1.

    ∂y/∂x = sec(t)/tan(t)
    y=∫sec(t)/tan(t)∂x
    =∫x/y∂x
    =(1/y)*∫x∂x
    =x2/2y + C
    2y2=x2 + C
    When t=π/4, x=√2, y=1
    2(1)2 = (√2)2 + C
    C=0
    So y2 = x2/2

    2.
    y/x = sin(t)
    1/x = cos(t)
    sin2(t) + cos2(t) = 1
    y2/x2 + 1/x2 = 1
    y2 = x2 + 1

    Why couldn't I get the same equations in both (1) and (2)? It turns out only the equation 2 works well with other value of t. What did I do wrong in (1)?
     
  2. jcsd
  3. Sep 23, 2016 #2

    Mark44

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    What does this have to do with antiderivatives, as stated in your thread title?
    No -- this is just y/x, not the partial of y with respect to x. Finding y in terms of x can be done using only some trig identities and some algebra.
    This part makes no sense.
    This part does make sense, but you have some errors.
    OK to here, but you have a mistake in the work below.
     
    Last edited: Sep 23, 2016
  4. Sep 23, 2016 #3

    fresh_42

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    You've treated ##y## as a constant with respect to ##dx##. But as both are functions of ##t## and your are looking for ##y(x)##. How can this be?
     
  5. Sep 24, 2016 #4
    1. Ooops. Sorry. I actually don't quite understand the meaning of antiderivative....><
    2. I thought since ∂y/∂x = (∂y/∂t)/(∂x/∂t), and ∂y/∂t = sec2 (t), ∂x/∂t = sec(t)tan(t), then ∂y/∂x = sec(t)/tan(t) ??
    3. Ooops. It's y2 = x2 - 1

    Thanks for the help!
     
  6. Sep 24, 2016 #5
    I thought when we integrate a function with respect to only one of its variables, we treat the rest of the variables as constants?
     
  7. Sep 24, 2016 #6

    PeroK

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    I can't see the relevance of that here.
     
  8. Sep 24, 2016 #7

    haruspex

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    There is no need to use any calculus in answering the question.
     
  9. Sep 24, 2016 #8

    Mark44

    Staff: Mentor

    In your first post, you had this
    All of your derivatives you show as being partial derivatives (e.g.,∂y/∂x) . Since both x and y are functions of a single variable, t, ordinary derivatives are appropriate here. With x = sec(t), dx/dt = sec(t) tan(t), and similar for y and dy/dt.

    In any case, as already stated by haruspex, you don't need any calculus for this problem.
     
  10. Sep 25, 2016 #9
    Thanks for pointing it out. I thought they are interchangeable until now....Paradigm shift! Now I finally understand why my lecturer uses them .
     
  11. Sep 25, 2016 #10
    So I can't solve it using Calculus????
     
  12. Sep 25, 2016 #11

    PeroK

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    You could if you wanted, but it's unnecessarily complicated. In this case, if you know your trig identiies, you are essentially given:

    ##y^2 = x^2 - 1##

    So, why use calculus to solve that?
     
  13. Sep 25, 2016 #12

    Mark44

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    The problem as stated:
    To reiterate what has already been stated (by @PeroK), this problem has nothing to do with calculus.
     
  14. Sep 26, 2016 #13
    I know I can do it using trig identities but I was just curious whether the question can be solved using calculus. I tried and failed and don't know where did I get wrong.
     
    Last edited by a moderator: Sep 26, 2016
  15. Sep 26, 2016 #14

    fresh_42

    Staff: Mentor

    Can you tell which formulas you used?

    Why is ##\frac{dy}{dx}=\frac{\sec t}{\tan t} ##? (I think it's correct but don't have it in mind.)

    And why do you think you can treat ##y## as a constant with respect to the integration over ##x##?
    Isn't ##y=y(x)=\tan (arc\sec x))## dependent on ##x##?
     
  16. Sep 27, 2016 #15

    Mark44

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    Really, the question is "What's the right tool for the job?"

    The original problem is to eliminate the parameter t in two equations with trig functions, so all that is required is a bit of algebra and a trig identity or two. The problem doesn't have anything to do with rates of change or summing infinitesimally small quantities, so calculus methods are neither called for or needed.
     
  17. Sep 27, 2016 #16

    haruspex

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    I don't think it can be solved using Galois Theory either.
     
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