- #1

- 11

- 0

## Homework Statement

Parametrize the curve x^(2/3) + y^(2/3) = a^(2/3) in the standard counterclockwise sense.

## Homework Equations

x^(2/3) + y^(2/3) = a^(2/3)

Any trig identity... I was thinking cos(x)^2 + sin(x)^2 = 1

## The Attempt at a Solution

Because it has to be parameterized in the standard counterclockwise sense, I know I have to use a cosine.

The shape of the graph is a diamond-like shape. (NOT a diamond, though.) At first I let

x = a(cos[t])^(3/2), so that

y = a(sin[t])^(3/2)

*if my substitutions and simplifications are correct... Maybe my problem lies in here?

My results are

__t__

x

y

0

a

0

pi/2

pi/2

0

a

pi

pi

nonreal answer

0

3*pi /2

3*pi /2

0

nonreal answer

Instead of the nonreal answers, it should be -a. Does any one know how I can get around it?

Thank you in appreciation for your help, and I'd just like to say to the people that help everyone else, You guys are awesome. I've seen your answers before, and you're really helping a lot of people out! (Hopefully me!)

Instead of the nonreal answers, it should be -a. Does any one know how I can get around it?

Thank you in appreciation for your help, and I'd just like to say to the people that help everyone else, You guys are awesome. I've seen your answers before, and you're really helping a lot of people out! (Hopefully me!)