# Thread: length of astroid curve

1. ## 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!

2. 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
$\frac{d}{dx}(8\cos (x^3))=-24x^2 \sin (x^3)$.

3. 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!!

4. 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 ...

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

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

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

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

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

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

$S = 48 \left[\frac{1}{2} - \left(-\frac{1}{2}\right)\right] = 48$

5. 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?

6. 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.