Arc Length Formulas Use the arc length formula to find the arc length of the upper half of the circle with center at $(0,0)$ and radius $3$. Also, find the arc length of the curve in the first question by using elementary geometry.

If you are in the "arclength" section of a calculus course, you will have seen that the arclength of the curve $y=f(x)$, from $x=a$ to $x=b$, is given by
$$\int_a^b \sqrt{1+(f'(x))^2}\,dx=\int_a^b \sqrt{1+\left(\frac{dy}{dx}\right)^2}\,dx.$$
Our circle has equation $x^2+y^2=9$, so the upper half of the circle has equation $y=\sqrt{9-x^2}$. Differentiate. We get 
$$\frac{dy}{dx}=-\frac{x}{\sqrt{9-x^2}}.$$
Thus 
$$1+\left(\frac{dy}{dx}\right)^2=1+\frac{x^2}{9-x^2}=\frac{9}{9-x^2},$$
and therefore for the arclength we need to find
$$\int_{x=-3}^3 \frac{3\,dx}{\sqrt{9-x^2}}.$$
To evaluate the integral, either make the substitution $x=3\sin t$, or make the substitution $x=3u$. We do the second. Then $dx=3\,du$. After we make the substitution, we arrive at
$$\int_{u=-1}^1 \frac{3\,du}{\sqrt{1-u^2}}.$$
Now, using the fact that $\int\frac{du}{\sqrt{1-u^2}}=\arcsin u+C$, we can complete the calculation. 
