Arc Length Formulas

Question

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.

Answer 1

HINTS: The parametrisation of a circle of radius $r$ in $\mathbb{R}^2$ is given by

$$\gamma(t) = (r \cos (t), r \sin (t)).$$

Arclength is calculated by working out

$$\int_a^b | \gamma'(t)| dt,$$

where $a, b$ are the two ends of your interval and $|\gamma ' (t)|$ is the velocity of the cure.

EDIT: Ok, assuming you have made some kind of attempt, here is how to answer the question:

We are told that the radius of the circle is $3$, so using my first hint, we can write the circle as

$$\gamma(t) = (3 \cos (t), 3 \sin (t)).$$

Now, we want to calculate the arc length of the upper half. We know that a regular circle goes from $0$ round to $2 \pi$ and so if we want just the upper half, we take half of this and we get our interval to be from $0$ and $\pi$, i.e we have $a = 0$ and $b = \pi$.

Next, we want to calculate $| \gamma ' (t)|$:

$$\gamma(t) = (3 \cos (t), 3 \sin (t)),$$ $$\gamma'(t) = (-3 \sin (t), 3 \cos (t)),$$ $$|\gamma'(t)| = |(-3 \sin (t), 3 \cos (t)| = \sqrt{(-3 \sin (t))^2 + (3\cos (t))^2} = \sqrt{9\sin ^2(t) + 9 \cos ^2(t)} = \sqrt{9(\cos ^2(t) + \sin ^2(t))} = 3,$$

which means to find the arc length, we now have to solve the equation

$$\int_{0}^{\pi} 3 dt$$ $$= [3t]_{0}^{\pi}$$ $$= 3(\pi) - 3(0) = 3\pi.$$

Now, if you wanted to check to see if your answer is correct, you can use a different method, i.e like one Andre Nicolas used, and you will see that you get the same answer.

Answer 2

The preexisting (good) answer addresses the first half of your question, on calculus. As to the second half —

find the arc length of the curve in the first question by using elementary geometry

— note that the arclength of the upper half of a circle is half the perimeter of the circle, and that a formula for a circle's perimeter is (likely) known to you from a previous course in geometry.

Answer 3

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.