Plotting a parametric function + area

In summary, the conversation discusses plotting a parametric function and finding the area enclosed by it, as well as suggestions for solving the problem. It is recommended to use a graphing program or make a table and graph by hand. For the area problem, suggested substitutions are provided. The shape of the graph is discussed and it is noted that it is not defined for negative values of x or y. An alternative method for plotting the graph is suggested using symmetry and specific values of theta.
  • #1
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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!
 
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  • #2
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:

[tex] cos^{2}(x)=1-sin^{2}(x) [/tex]

which will give:

[tex] \int sin^{4}(t)-sin^{6}(t) dt [/tex]

in which you can use the fact that:

[tex] \frac {1-cos(2x)}{2}= sin^{2}(x) [/tex]

and:


[tex] (\frac {1-cos(2x)}{2})^{2} = sin^{4}(x) [/tex]

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
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.

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
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
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.
 

FAQ: Plotting a parametric function + area

1. What is a parametric function?

A parametric function is a mathematical function that describes the relationship between two variables using a third variable, known as a parameter. The variables are usually represented as x and y, while the parameter is represented as t.

2. How do you plot a parametric function?

To plot a parametric function, you first need to substitute different values for the parameter t to get corresponding values for x and y. These values can then be plotted on a graph to create a curve or a set of points.

3. What is the difference between a parametric function and a standard function?

In a standard function, the variables x and y are directly related to each other. In a parametric function, x and y are related through a third variable, the parameter t. This allows for more complex and dynamic relationships between x and y.

4. How do you find the area of a parametric function?

The area under a parametric function can be found by using the definite integral of the function, with the limits of integration being the starting and ending values of the parameter t. The integral will give the area between the curve and the x-axis.

5. Can a parametric function have more than one parameter?

Yes, a parametric function can have multiple parameters, each representing a different variable. This allows for even more complex relationships between variables. However, the number of parameters should match the number of equations in the function.

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