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Homework Help: Simple parametric equations

  1. Nov 30, 2012 #1
    A curve has parametric equations

    x = 2 cot t, y = 2sin² t , 0 < t ≤ pi/2

    (a) find an expression for dy/dx in terms of the parameter t
    (b) Find an equation of the tangent to the curve at the point where t = pi/4
    (c) Find a cartesian equation of the curve in the form y = f(x). State the domain on which the curve is defined.


    I've done part a),
    I've done part b),
    I've partway done part c), I have found the equation (y = 8/(4+x^2)), however I don't know how to find the domain on which the curve is defined, any help?
     
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  3. Nov 30, 2012 #2

    LCKurtz

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    If x = 2 cot t, 0 < t ≤ pi/2, what values of x can you get?
     
  4. Nov 30, 2012 #3

    Mark44

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    x as a function of t is defined for all t in (0, ##\pi/2##]. What interval constitutes the codomain for x = f(t) = 2 cot(t)?. That will be your domain for your nonparametric equation.
     
  5. Nov 30, 2012 #4
    Well drawing the cot graph as t approaches pi/2, x approaches 0 ∴ domain will approach infinity? Similar when t is really small the domain will approach 0? so 0<x<∞ or x>0?
     
  6. Nov 30, 2012 #5

    Mark44

    Staff: Mentor

    The domain doesn't "approach" anything - it's just an interval that you specify by an inequality or interval notation. (I'm taking shortcuts here - a domain might be the union of several disjoint intervals.)
    Well, almost. It's actually 0 ≤ x < ∞, or x ≥ 0. Both of those represent the same interval, which you can also write as [0, ∞).

    I don't know if you noticed, but the domain of f(x) = 8/(4 + x2) is all real numbers. The restricted domain you found comes from x(t) = cot(t).
     
  7. Dec 1, 2012 #6
    okay, thank you

    just one question - why does it include 0?
     
  8. Dec 1, 2012 #7

    HallsofIvy

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    The "the domain on which the y= f(x) is defined" is a set of x values, not t.
    t cannot be 0 but x can. [itex]x(\pi/2)= cot(\pi/2)= 0[/itex]
     
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