# Proof of Parametrized Functions

1. Sep 21, 2009

### H2instinct

1. The problem statement, all variables and given/known data

2. Relevant equations

The hint.

3. The attempt at a solution

So using the hint I took the derivative of each of the parametrized functions.

$$\frac{dx}{d\theta} = -3sin(\theta)*cos^{2}(\theta)*a$$

$$\frac{dy}{d\theta} = 3cos(\theta)*sin^{2}(\theta)*a$$

Then I plugged them into this:

$$ds = \sqrt{dx^{2}+dy^{2}}d\theta$$

Giving:

$$ds = 3*\left|a*sin(\theta)*cos(\theta)\right|d\theta$$

$$\intds = \int3*\left|a*sin(\theta)*cos(\theta)\right|d\theta$$

I know ds is infinitesimally small point on the original graph to measure slope, but I am still confused slightly about integrating both sides. Also assuming the absolute value goes away because sin and sos can only go from -1 to 1, so I took out the abs. value because I am assuming for positive number. Don't know if that's actually possible to do though.

$$s = \frac{-3*(cos(\theta))^{2}*a}{2}$$

This is where I need some pointing in the right direction.

Have been trying while waiting for some help:
Plugging into the other part of the hint:
$$ds = \sqrt{1 + \frac{-3sin(\theta)*cos^{2}(\theta)*a}{3cos(\theta)*sin^{2}(\theta)*a}}\left|-3sin(\theta)*cos^{2}(\theta)*a\right|$$

Gives:
$$ds = 3*\left|a\right|(cos(\theta))^{2}*d\theta$$

Tried integrating:
$$\intds = \int3*\left|a\right|(cos(\theta))^{2}d\theta$$

Gives:
$$S = \frac{3*(sin(\theta)*cos(\theta)+\theta)*\left|a\right|}{2}$$

Still not totally sure where to go with this.

Last edited: Sep 22, 2009
2. Sep 22, 2009

### LCKurtz

$$ds = 3|a||\cos(\theta)\sin(\theta)| d\theta$$

looks correct. You can't just ignore the absolute value signs. Since the curve traced when theta goes from 0 to 2pi and is symmetric, you can integrate from 0 to pi/2 and multiply by four. On that domain for theta the sine and cosine are positive and you can drop the absolute value signs. So calculate

$$4\int_0^{\pi/2} 3|a||\cos(\theta)\sin(\theta)| d\theta$$

and see if it works.

3. Sep 22, 2009

### H2instinct

Yes it did work, thanks for the help, I will remember to change it to the $$(0 , \frac{\pi}{2})$$ quadrant when it calls for abs. value.

4. Sep 22, 2009

### LCKurtz

Just remember that you can only use that shortcut when the curve is symmetric so that the total perimeter is 4 times the arc length in the first quadrant.

You might find it instructive to do it the long way so you learn how to handle the absolute values if you have to. Break the integral from 0 to 2pi up into four pieces, one for each quadrant. Then, for example, in the second quadrant where cosine is negative and sine is positive you would use

| cos(t) | = - cos(t)
| sin(t) | = sin(t)

to calculate the absolute values. Similarly for the other quadrants. Try it and see that you get the same answer.

5. Sep 22, 2009

### H2instinct

Ah Ok, I am getting how you got the boundaries then. And right, it obviously wouldn't work for a parametrized function that is only in the 1st and 3rd quadrant, etc.