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

[kingwinner]

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!

 

[D H]

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 let's 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, $\cos^{2/3}\theta + \sin^{2/3}\theta$. This in turn will tell you somthing qualitative about $r$.[/tex]

[tex][/tex]

 

[kingwinner]

But this is not a function, so do I have to implicitly differenatiate to find dy/dx?

Also, how can I locate the cusps?

Thanks!

 

[D H]

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: $y=(1-x^{2/3})^{3/2}$. Doing so cuts off half of the curve, but that is easily fixed by taking advantage of the symmetries.

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