Finding length of path over an interval

**ZeroVector** · October 17th 2010, 01:49 PM

(2cost - cos2t, 2sint - sin2t) 0 <= t <= pi/2
Here's what I have so far:
x'(t)^2 = 4sin^2(t) - 8(sint)(sin2t) + 4sin^2(2t)
y'(t)^2 = 4cos^2(t) - 8(cost)(cos2t) + 4cos^2(2t)
x'(t)^2 + y'(t)^2 = 4 - 8(sint)(sin2t) + 4sin^2(2t) - 8(cost)(cos2t) +
4cos^2(2t)

What do I do next? And/or am I even doing it right?

**lvleph** · October 17th 2010, 02:53 PM

You will want to use the formula
$s = \int_{t_0}^{t_1} |v(t)|\, dt$

Where v(t) is the speed function.

**ZeroVector** · October 17th 2010, 05:58 PM

Well, I'm wondering how I can integrate it as the resultant integrand will be:
integral( SQRT(4 - 8(sint)(sin2t) + 4sin^2(2t) - 8(cost)(cos2t) + 4cos^2(2t)), t, 0, pi/2 )

I'm not sure how I can simplify it when this part is in the square root: 8(sint)(sin2t) + 4sin^2(2t) - 8(cost)(cos2t) + 4cos^2(2t)

**lvleph** · October 17th 2010, 06:02 PM

Well
$4\sin^2 2t + 4\cos^2 2t = 4$
And
$-8\sin t \sin 2t - 8 \cos t \cos 2t = -8\cos t$

EDIT: Fixed typos

**Soroban** · October 17th 2010, 06:13 PM

Hello, ZeroVector!

You're on the right track . . .

$\begin{Bmatrix}x &=& 2\cos t - \cos2t \\ y &=& 2\sin t - \sin2t\end{Bmatrix}\;\;0 \leq t \leq \frac{\pi}{2}$

Here's what I have so far:

. . $x'(t)^2 \:=\: 4\sin^2\!t - 8\sin t\sin2t + 4\sin^2\!2t$
. . $y'(t)^2 \:=\: 4\cos^2\!t - 8\cos t\cos2t + 4\cos^2\!2t$

$x'(t)^2 + y'(t)^2 \:=\: 4\underbrace{\left[\sin^2\!t + \cos^2\!t\right]}_{\text{This is 1}}$

. . . . . . . . . . . . . . . . . . $- 8\underbrace{\left[\cos t\cos2t + \sin t\sin2t\right]}_{\text{This is }\cos t}$

. . . . . . . . . . . . . . . . . . . . . . $+ 4\underbrace{\left[\sin^2\!2t + \cos^2\!2t\right]}_{\text{This is 1}}$

. . . . . . . . . . $=\;8 - 8\cos t \;=\;8(1-\cos t)$

. . . . . . . . . . $=\;16\left(\dfrac{1-\cos t}{2}\right) \;=\;16\sin^2\!\frac{t}{2}$

Hence: . $\sqrt{x'(t)^2 + y'(t)^2} \;=\;4\sin\frac{t}{2}$

Therefore: . $\displaystyle L \;=\;\int^{\frac{\pi}{2}}_0 \sin\frac{t}{2}\,dt$

Got it?

**ZeroVector** · October 17th 2010, 08:44 PM

Thanks! By the way, shouldn't the integral you wrote at the end have a 4 in front of it?
I think the main part was that I didn't think that I was supposed to use things like angle sum/double angle/half angle (which, unfortunately, my teacher pre-calculus never taught) formulas.

Also, I was wondering, how did you get the text to look "nice"? Is there some sort of web app that does it for you?

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