Total length of the astroid

**operaphantom2003** · December 5th 2009, 12:27 PM

This is the last section of my class and I just don't get it. I can't even get my calculator so graph it for me (probably my stupid though). Can anyone please explain how to answer this type of question? I have 5 questions on the length of a parametric curve to do. I just can't understand what the book is saying (again probably my stupid). Any help is greatly appreciated.

If f(θ) is given by: f(θ)=18cos3θ and g(θ) is given by: g(θ)=18sin3θ
Find the total length of the astroid described by f(θ) and g(θ).
(The astroid is the curve swept out by (f(θ),g(θ)) as θ ranges from 0 to 2π. )

**galactus** · December 5th 2009, 12:59 PM

$f(t)=18cos(3t), \;\ g(t)=18sin(3t)$

$f'(t)=-54sin(3t), \;\ g'(t)=54cos(3t)$

parametric arc length is given by $\int_{0}^{2\pi}\sqrt{(-54sin(3t))^{2}+(54cos(3t))^{2}}dt=54\int_{0}^{2\pi }dt$

I suppose this is what they are getting at. To graph this in parametric, your calculator must be in parametric mode. Parametrically, this is an ellipse, not an astroid. Unless I am missing something. The convention normally used is t instead if theta. Theta is commonly used when dealing with polar coordinates. Even in polar coordinates, this is a rose and not an astroid.

Please do not tell me you mean $f(t)=18cos^{3}(t), \;\ g(t)=18sin^{3}(t)$ and did not use ^ to mean power.

**operaphantom2003** · December 5th 2009, 01:09 PM

That is what the equation reads. All greek to me

. Thanks for your help.

**galactus** · December 5th 2009, 01:20 PM

For an astroid, the equations should be $x(t)=18cos^{3}(t) \;\ and \;\ y(t)=18sin^{3}(t)$ .

$x'(t)=-54sin(t)cos^{2}(t)$

$y'(t)=54sin^{2}(t)cos(t)$

Repeating, Parametric Arc length is give by $\int_{a}^{b}\sqrt{(\frac{dx}{dt})^{2}+(\frac{dy}{d t})^{2}}dt$

Yours whittles down to

$54\int_{0}^{2\pi}|sin(t)cos(t)|dt$

Let's do it this way though. Since we have symmetry, we can integrate from 0 to Pi/2 and multiply by 4:

$216\int_{0}^{\frac{\pi}{2}}sin(t)cos(t)dt$

Can you integrate that?. See the graph?. That is an astroid. Astroid means star-shaped.

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