# Thread: Parametric equations and the lengths of curves.

1. ## Parametric equations and the lengths of curves.

1. Find parametric equations for the semicircle: x^2 + y^2 = a^2 using as parameter the arc length s measured counterclockwise from the point (a, 0) to the point (x, y).

We were not taught this, and I don't think the book really helps. I was wondering if somebody could clearly explain this concept and problem to me.

2. Find the length of the curve: x = cos t, y = t + sin t, [0, [pi]].

Here's where I get stuck. I know that the length of a curve is the integral evaluated from a to b (or, in this case, from zero to pi) of the quadrature of dx/dt and dy/dt. However, upon doing that, I get the integral of the square root of (-sin t)^2 + (1 + cos t)^2. I cannot use u substitution (although that does not rule out it's possibility of use) and if I make (-sin t)^2 into (sin t)^2 and expand (1 + cos t)^2, I get the integral of sqrt(2 + 2cos t), which then, by pulling out the square root of 2, makes the integral of the square root of 1 + cos t. I...don't know what to do then.

I would appreciate any help.

2. Originally Posted by niyati
1. Find parametric equations for the semicircle: x^2 + y^2 = a^2 using as parameter the arc length s measured counterclockwise from the point (a, 0) to the point (x, y).

We were not taught this, and I don't think the book really helps. I was wondering if somebody could clearly explain this concept and problem to me.

...
Hello,

I've attached a drawing.

A point on the semicircle, the centre of the semicircle and the diameter form an angle $\theta$ which can be calculated in radians by

$\theta = \frac sa$

Therefore you get:

$\left| \begin{array}{l}x=a\cdot \cos\left(\frac sa\right) \\ y = a \cdot \sin\left(\frac sa\right) \end{array} \right.$

3. ## Second problem

On the second part, you said you obtained the equation $L=\sqrt{2}\int_{0}^{\pi}\sqrt{1+\cos{x}}\,dx$
To solve that, remember the trig identity $\cos{2\theta}=2\cos^2\theta-1$. Thus $1+\cos{2\theta}=2\cos^2\theta$, or using $\theta=\frac{x}{2}$, we obtain $1+\cos{x}=2\cos^2\frac{x}{2}$, so
$L=\sqrt{2}\int_{0}^{\pi}\sqrt{1+\cos{x}}\,dx$
$=\sqrt{2}\int_{0}^{\pi}\sqrt{2\cos^2\frac{x}{2}}\, dx$
$=2\int_{0}^{\pi}\left|\cos\frac{x}{2}\right|\,dx$
Note that for your region of integration, $\cos\frac{x}{2}\ge0$, so you can remove the absolute value:
$L=2\int_{0}^{\pi}\cos\frac{x}{2}\,dx$
You should be able to do that integral.

--Kevin C.