# Plotting a parametric function + area

1. Jun 27, 2011

### autre

Given the parametric function defined by x = a cos ^3t, y = a sin^3t, plot the curve.

So I converted the above to (x/a)^(2/3) + (y/a)^(2/3) = 1, and from that got y +-a(1-(x/a)^(2/3))^(3/2). However, I have no idea of how to actually plot a function of this form. Is my only choice to make a table?

I also have to find the area of the region enclosed by this function. To do this, I need to find the integral of y(t)x'(t)dt from alpha to beta. However, y(t)x'(t)dt = sin^4tcos^2tdt and I can't find a proper u-substitution for integration.

Thanks for any help!

2. Jun 27, 2011

### Asphyxiated

When I work with parametric equations I find it simplest to use a graphing program. If you do not have such a program available to you then your only options are to make a table and graph it out by hand, this is of course assuming that you can not turn it back to Cartesian form such as y=f(x) and decipher its shape from there.

For the area problem you could sub:

$$cos^{2}(x)=1-sin^{2}(x)$$

which will give:

$$\int sin^{4}(t)-sin^{6}(t) dt$$

in which you can use the fact that:

$$\frac {1-cos(2x)}{2}= sin^{2}(x)$$

and:

$$(\frac {1-cos(2x)}{2})^{2} = sin^{4}(x)$$

etc....

to solve that problem from there.

Note you will have to use the cosine half angle identity because you will get a cos^4 and cos^2 when you foil out the terms if you make the substitutions shown.

3. Jun 27, 2011

### autre

The issue is that the Cartesian form is y = +-a(1-(x/a)^(2/3))^(3/2), and I have no idea about the shape of such a graph (nor of the simpler y = +-(1-x^(2/3))^(3/2)).
Thanks for the tip on the area problem!

4. Jun 27, 2011

### lanedance

have you seen wolfram alpha
http://www.wolframalpha.com/input/?i=plot+%28x%29^%282%2F3%29+%2B+%28y%29^%282%2F3%29+%3D+1

now to imagine the plot, first assume a=1 for simplicity

clearly (0,1) and (1,0) are on the curve

now imagine the point where y=x, then 2x^(2/3)=1 which gives
x=(1/2)^(3/2)

comparing a similar form x^2 + y^2 = 1, would give the unit circle, and at y=x=(1/2)^(1/2), so the circle is "squashed" away form the axes

note it is not defined for negative values of x or y

5. Jun 27, 2011

### LCKurtz

There is no reason to get y in terms of x to plot this equation. You could easily make a rough sketch by hand if you don't have a graphing calculator which plots parametric equations. Just take θ to be 0, π/6, π/4, π/3 and π/2. You can write exact values for the sine and cosine of those angles and use them to plot the (x,y) points in the first quadrant, then free-hand sketch it. If you look carefully you will see that there is a lot of symmetry you can use to get the graph in the other quadrants.