Sketch the curve given parametrically $x=a \cos^3t, y=a \sin^3t$

Question

A question in my maths book starts by asking "sketch for $0\le t\le2\pi$ the curve given parametrically by $x=a \cos^3t, y=a\sin^3 t$."

I can see that $\cos t = (\frac{x}{a})^\frac{1}{3}$ and $\sin t = (\frac{y}{a})^\frac{1}{3}$ but beyond that I cannot think how to proceed. I know from internet research that the graph is an astroid.

I also realise that $\cos^2t + \sin^2t = (\frac{x}{a})^\frac{2}{3} + (\frac{y}{a})^\frac{2}{3} = 1$ but am not sure how/if this is relevant.

Answer 1

Let ignore the scaling factor $a$, the resulting curve is just homothetic to the unitary one (with $a=1$).

$$\begin{cases}x=\cos(t)^3\\y=\sin(t)^3\end{cases}$$

• First let study the symmetries of this figure.

$\Big(x(t+\frac{\pi}2),y(t+\frac{\pi}2)\Big)=\Big(-y(t),x(t)\Big)$

$\Big(x(t+\pi),y(t+\pi)\Big)=\Big(-x(t),-y(t)\Big)$

$\Big(x(t+\frac{3\pi}2),y(t+\frac{3\pi}2)\Big)=\Big(y(t),-x(t)\Big)$

And of course it is $2\pi$-periodic.

Therefore we can limit the study to $t\in[0,\frac{\pi}2]$ and complete the remaining by symmetry.

Notice that for this range of $t$ we get that $x\ge 0,y\ge 0$ so the curve will fit entirely in the first quadrant.

• Then let study the variations.

The derivatives are very pleasant for sign study (we can just ignore the squared terms) $$\begin{cases}x'=-3\cos(t)^2\sin(t)&\le 0\\y'=3\sin(t)^2\cos(t)&\ge 0\end{cases}$$

Thus we get $x\searrow$ and $y\nearrow$.

From the origin point $A=(x(0),y(0))=(1,0)$ to the end point $B=(x(\frac{\pi}2),y(\frac{\pi}2))=(0,1)$.

The slope of the tangent is given by $$\dfrac{y'(t)}{x'(t)}=-\tan(t)$$

so in $A$ we have an horizontal tangent (zero value), and in $B$ a vertical one (infinite value).

• Critical points

On interval $[0,\frac{\pi}2]$ only the extremities annulate both $x'(t)$ and $y'(t)$ so only $A$ and $B$ are critical points.

From the already known tangents and symmetries, it is easy to figure out that they are both cusp points.

• Additional points

Since $x,y$ are monotonic in this quadrant, we already have a pretty good idea of the sketch of this curve, but for a more precise drawing you can calculate a few more points in easy trigonometric lines like $t=\frac{\pi}6,\frac{\pi}4,\frac{\pi}3$ for instance.

I calculate just one in $\frac{\pi}4$ it is $C=(\frac{\sqrt{2}}4,\frac{\sqrt{2}}4)$ (better to figure out it is at distance $\frac 12$ from origin on the diagonal $y=x$) and tangent there is slope $-1$, so perpendicular to this diagonal.

• Let draw it

And finish the diamond shape by completing the symmetric parts in the $3$ other quadrants.

Note: the astroid is a pretty simple curve, for other parametric curves, there might be other steps, more specifically finding multiple points or determining the exact nature of critical points.