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How to sketch x^(2/3) + y(2/3) = 1 ?

  1. Nov 29, 2007 #1
    Let f(t)=( (cos t)^3, (sin t)^3 )
    Let S = {f(t) | t E R}
    Draw a sketch of S.

    Now I know that S is x^(2/3) + y(2/3) = 1

    What is the fastest way to draw a really rough sketch of this? (bascially I need the shape) I am very lost in situations like this. I was never taught how to do quick sketch on graphs like this one...(no graphing calculators allowed, of course)

    Please help me! Thanks!
  2. jcsd
  3. Nov 29, 2007 #2

    D H

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    Some things to do:
    • The function in question looks kind of like the equation for a unit circle. Sketching a circle as reference function might help.
    • Look for symmetries. The function in question has some obvious and not-so-obvious symmetries. This lets you focus in on a part of the graph. The rest you can fill in by reflections.
    • Look at the derivative of y with respect to x, x with respect to y. For a qualitative sketch you are looking for qualitative anomalies, like cusps.
    • Convert to another form, such as polar. [tex]r^{2/3}(\cos^{2/3}\theta + \sin^{2/3}\theta) = 1[/itex]. Try to derive some qualitative relation regarding the angular term, [itex]\cos^{2/3}\theta + \sin^{2/3}\theta[/itex]. This in turn will tell you somthing qualitative about [itex]r[/itex].
  4. Nov 29, 2007 #3
    But this is not a function, so do I have to implicitly differenatiate to find dy/dx?

    Also, how can I locate the cusps?

    Last edited: Nov 29, 2007
  5. Nov 29, 2007 #4

    D H

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    I was using the term function a bit too loosely. It is an equation which becomes a function upon restricting the sqaure root to the principal value: [itex]y=(1-x^{2/3})^{3/2}[/itex]. Doing so cuts off half of the curve, but that is easily fixed by taking advantage of the symmetries.

    IMO, implicit differentiation of [itex]x^{2/3}+y^{2/3}-1=0[/itex] is easier than explicit differentiation of [itex]y=(1-x^{2/3})^{3/2}[/itex]
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