# length of astroid curve

• Dec 7th 2008, 12:50 PM
littlejodo
length of astroid curve
I'm stuck because the answer I keep arriving at is negative under a square root sign. In the original question the variable is theta but for ease of typing I'm going to make it x.

If f(x) = 8cos^3(x) and g(x) = 8sin^3(x) , Find the total length of the astroid described by f(x) and g(x) (f(x), g(x)) as x ranges from x=0 to x=2pi

So I found that f'(x) = -24x^2(sin(x^3))
and g'(x) = 24x^2(cos(x^3))

I multiplied through to get:

so int_0^2pi sqrt[ 576x^4(sin(x^6)) + 576x^4(cos(x^6))]
simplified:
24 int_0^2pi x^2 sqrt[sin(x^6) + cos(x^6)]
integrated:
24 int_0^2pi 1/3x^3 sqrt[-cos(x^6) + sin(x^6)]

when I plug in 2pi for x, I get -.0436499839 under the square root sign.
where did I go wrong?

Thanks!
• Dec 7th 2008, 03:42 PM
whipflip15
I'm not sure what you have done with you derivatives because
$\frac{d}{dx}(8\cos^3x)=-24\sin x \cos^2 x$
but
$\frac{d}{dx}(8\cos (x^3))=-24x^2 \sin (x^3)$.
• Dec 7th 2008, 04:19 PM
littlejodo
I made the mistake of thinking sin^2(x) was the same thing as sin(x^2). oops.

Let me try this again --

int_0^2pi sqrt[(-24sin(x)cos^2(x))^2 + (24cos(x)sin^2(x))^2]

I will multiply through (I'm not sure if I'm distributing the exponent correctly)

int_0^2pi sqrt[576sin^2(x)cos^4(x) + 576cos^2(x)sin^4(x)]

I can factor out the 576sin^2cos^2 and move out of the square root to get:

24 int_0^2pi sin(x)cos(x) sqrt[cos^2(x) + sin^2(x)]

I integrated to get:

24 [-cos(x)sin(x) sqrt(sin^2(x) - cos^2(x))

I plugged in 2pi and got the answer 0 (because of the sin(2pi)) and this is still the wrong answer. Is my problem with integrating cos^2(x) and sin^2(x)? I have very little idea how to integrate these. If anybody wants to help me out with that I'd be very grateful!!
• Dec 7th 2008, 04:41 PM
skeeter
using symmetry ...

$S = 96 \int_0^{\frac{\pi}{2}} \sqrt{\cos^4{t}\sin^2{t} + \sin^4{t}\cos^2{t}} \, dt$

note the following ...

$\cos^4{t}\sin^2{t} + \sin^4{t}\cos^2{t} =$

$\cos^2{t}\sin^2{t}(cos^2{t} + \sin^2{t}) = \cos^2{t}\sin^2{t}$

$S = 96 \int_0^{\frac{\pi}{2}} \sqrt{\cos^2{t}\sin^2{t}} \, dt$

$S = 48 \int_0^{\frac{\pi}{2}} 2\cos{t}\sin{t} \, dt$

$S = 48 \int_0^{\frac{\pi}{2}} \sin(2t) \, dt$

$S = 48 \left[-\frac{\cos(2t)}{2}\right]_0^{\frac{\pi}{2}}$

$S = 48 \left[\frac{1}{2} - \left(-\frac{1}{2}\right)\right] = 48$
• Dec 7th 2008, 05:20 PM
littlejodo
Thank you very much for the answer!

I haven't seen this use of symmetry before. Is there a specific formula? where did the 96 and pi/2 come from?

Is this something that you just have memorized? or how is it found?
http://www.mathhelpforum.com/math-he...cf0454d3-1.gif

http://www.mathhelpforum.com/math-he...e4e168e6-1.gif
• Dec 7th 2008, 05:27 PM
skeeter
look at the graph of the astroid ... it has the same shape and length in each quadrant.

0 to pi/2 finds the arc length in quadrant I ... I just multiplied by 4 to get the entire length.

as far as your 2nd question, I just factored and used the Pythagorean identity.