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Obeserve the following curve:

$$ x^{2/n}+y^{2/n}=a^{2/n} $$

where a is a positive, real number.

Sketch the curve when

$$n=1, n=3, n=5$$

I don't know how to go forward. It looks a little like an ellipse, but how would I show how I thought when I outlined the curve? I've read about sketching the graph in R. Adams - Calculus A complete course, but it seems that I can't relate things I've learned from there to this particular question.

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up vote 1 down vote accepted

There is the little preliminary question what $x^{2/{\rm odd number}}$ means when $x$ is negative. Putting this aside we can restrict to $\bigl({\mathbb R}_{\geq0}\bigr)^2$.

So we are told to "observe" the three curves $$x^2+y^2=a^2\ ,\quad x^{2/3}+y^{2/3}=a^{2/3}\ ,\quad x^{2/5}+y^{2/5}=a^{2/5}\qquad(*)$$ in the first quadrant. Note that we can solve for $y$ in the form $$y=\bigl(a^{2/n}-x^{2/n}\bigr)^{n/2}\qquad(0\leq x\leq a)$$ in each of the cases $n=1,\ 3,\ 5$, so that each of these curves appears as a graph. Looking at the equations $(*)$ we see that all three graphs will be symmetric with respect to the line $x=y$ and connect the point $(0,a)$ monotonically decreasing with $(a,0)$. See the following figure.

enter image description here

The case $n=1$ is obvious: The curve is a quarter circle. Its tangent at $(0,a)$ is horizontal and at $(a,0)$ vertical. This is different from the cases $n=3$ and $n=5$: For $$f_n(x):=\bigl(a^{2/n}-x^{2/n}\bigr)^{n/2}$$ we have $$f_n'(x)=-{2\over n}\bigl(a^{2/n}-x^{2/n}\bigr)^{{n\over2}-1}\ x^{{2\over n}-1}\qquad (0<x\leq a)$$ and therefore $f_n'(a)=0$ when $n>2$. It follows that in these cases the tangent is horizontal at $(a,0)$ and, by symmetry, vertical at $(0,a)$.

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Nice sketch you included, And also, nice use of the derivative! – amWhy Nov 26 '12 at 15:00
Christian, what do you use for sketching curves? Just curious, as I've used an assorted mix of tools for doing so. – amWhy Nov 26 '12 at 15:16
Thank you Christian! – Curtain Nov 26 '12 at 15:33
@amWhy: For plots (as here) I'm using Mathematica, for "technical figures" the software "Canvas", which is not available, nor supported, anymore. – Christian Blatter Nov 26 '12 at 15:36


When $n = 1$, you have a circle centered at the origin with radius $a$. So essentially, you have a family of circles centered at the origin, where the radius of the circles correspond to different values of $a$. (The larger the value of $a$, the larger the circle.)

Experiment with graphing curves for $n = 3$ and pick $a$ to be some positive real. Likewise, for $n = 5$. Can you make any generalizations about the resulting graphs, and how the value of $a$ impacts the graph?

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Thanks! After reading Christian's post, I can see that your hint makes sense. :) – Curtain Nov 26 '12 at 15:34
Julian: Sometimes pictures speak louder than words!$\quad$:-) – amWhy Nov 26 '12 at 15:44
@amWhy: that is a nice hint indeed! +1 – Amzoti May 16 '13 at 2:11

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