# Sketching Parametric Curves

I am trying to sketch the curve given by the following two parametric equations.

$x=cos^3\theta$

$y=sin^3\theta$

Or the single cartesian equation:

$x^{\frac{2}{3}}+y^{\frac{2}{3}}=1$

So,

I put the graph in Desmos and got:

Now, the general advice for parametric curves is to create a table of values (by hand or with a calculator) and plot the (x,y) coordinates roughly on a graph.

Simple enough.

My problem is this:

The circle with the equation $x^2 + y^2 =1$ gives the graph: Now how do I know whether the graph curves "inwards" as in the first graph or "outwards" like a circle?

I am aware of increasing/decreasing function and implicit differentiation.

$\frac{d}{dx}[x^{\frac{2}{3}}+y^{\frac{2}{3}}]=\frac{d}{dx}$

$\frac{dy}{dx}= - \sqrt{\frac{y}{x}}$

Hence in the 1st quadrant where $x>0$ and $y>0$, and in the 3rd quadrant where $x<0$ and $y<0$ (same signs)

$\frac{dy}{dx}<0$ therefore it a decreasing function here

In 2nd quadrant where $x<0$ but $y>0$ and in the 4th quadrant where $x>0$ but $y<0$ (opposite signs)

$\frac{y}{x}<0$

$\therefore \frac{dy}{dx} >0$ Hence increasing function

which should explain the shape.

Does anyone have a less mechanical way of doing this as I feel I worked "backwards" as I knew what I was aiming for once I had seen the correct graph?

• – Emilio Novati Apr 13 '18 at 9:22

Hint:

I suppose that you know that the convexity is represented by the sign of the second derivative. In your case the first derivative is: $$\frac{dy}{dx}=\frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}=-tan \theta$$

and the second derivative is given by:

$$\frac{d^2y}{dx^2}= \frac{\frac{d}{d\theta}\frac{dy}{dx}}{\frac{dx}{d\theta}}$$

can you complete?

When plotting a parametric curve it is always a good idea to calculate the gradient as a function of the parameter using $$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$$

Thereby you can determine the slope of the curve at each point for which you calculate $(x,y)$ and find the points where the gradient is $0$ or$\infty$.