Arc length of vector function with trigonometric components

[Subdot]

Homework Statement

Find the length of the path traced out by a particle moving on a curve according to the given equation during the time interval specified in each case.
r(t) = (c²/a)cos³t i + (c²/b)sin³t j

where i and j are the usual unit vectors, 0 \leq t \leq 2\pi, c² = a² - b², and 0 < b < a

Also, r(t) is the position vector-valued function of t. Scalar functions will not be bolded.

Homework Equations

I know that the arc length in this case is the integral of ||r'(t)|| = v(t) from 0 to 2\pi.

The Attempt at a Solution

Differentiating gives: r'(t) = (-3c²/a)cos²t sin t i + (3c²/b)sin²t cos t j.

Therefore, v(t) = √{(9c⁴/a²)cos⁴t sin²t + (9c⁴/b²)sin⁴t cos²t)} = √{(9c⁴/a²)cos²t(cos²t sin²t) + (9c⁴/b²)sin²t(sin²t cos²t)} =

√{(9c⁴/a²)cos²t(0.25sin²2t) + (9c⁴/b²)sin²t(0.25sin²2t)} = √{(9c⁴/a²)cos²t(0.25sin²2t) + (9c⁴/b²)sin²t(0.25sin²2t)} =

(3c²/2)(sin2t)*√{(1/a²)cos²t + (1/b²)sin²t} = (3c²/(2ab))(sin2t)*√{(b²)cos²t + (a²)sin²t}

And that is where I get stuck. I'm not sure what the integral of that expression is, nor do I know what I can do to simplify it further. I would say that it has no simple solution, since the argument under the square root looks an awfully lot like an integral for an ellipse, and I have heard one needs something called an "elliptic integral" to solve. However, the answer for this problem is quite simple: (4a³-4b³)/(ab). Will someone please help me with this?

 

[tiny-tim]

Hi Subdot!

(have a square-root: √ )

(3c²/2)(sin2t)*Sqrt {(1/a²)cos²t + (1/b²)sin²t} = (3c²/(2ab))(sin2t)*Sqrt {(b²)cos²t + (a²)sin²t}
…
I'm not sure what the integral of that expression is …

Use standard trigonometric identities to write it as sin2t f(cos2t)

 

[Subdot]

Hello! Oh! I get it now. I just use the identities: sin² = .5 - .5cos2t and cos² = .5 + .5cos2t after which I do a simple substitution. Thanks for that! On another related note, now it's obvious that I'm going to need to integrate it a step over the interval at a time.

[(1/(ab)) (.5b² + (.5b² - .5a²)cos(2t) + .5a²)^1.5] from 0 to 2\pi. If I try to integrate it from 0 to .5\pi then from .5\pi to \pi and so on I come up with (2a³ - 2b³)/(ab) + (2b³ - 2a³)/(ab). I'm guessing from the answer that I am justified in flipping that last fraction around to (2a^3 - 2b^3)/(ab). However, I'd like to know why, so I'll be able to do it in the future without knowing the answer. I think it's because the arc length must be positive and so I must flip (b³ - a³)/(ab) to (a³ - b³)/(ab) because b < a? So then do you take the absolute value of the integral as you move up the interval, not the integrand, when you find arc length?

 

[Dick]

The reason is that you factored cos(t)^2*sin(t)^2 outside of the sqrt as cos(t)*sin(t). It really should be |cos(t)*sin(t)|. So you have to break the integral where cos(t)*sin(t) changes sign.

 

[Subdot]

Okay, I get it all now. I can't believe I missed that... Thanks both!