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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.

Ross Millikan
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Anton
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3 Answers3

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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.

Kaish
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  • I still don't understand... – Anton Jun 11 '13 at 18:40
  • @Anton Sorry, there was a tiny mistake, corrected it now. – Kaish Jun 11 '13 at 18:45
  • I still don;t understand the hint. – Anton Jun 11 '13 at 18:47
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    @Anton, do you have a calculus book? It may be worth taking the time to read through the section on arclength, working through all the examples and making sure you understand each line before you read the next. – msh210 Jun 11 '13 at 18:48
  • No, not at this moment. – Anton Jun 11 '13 at 18:49
  • I am still lost, can you show me how to do it? – Anton Jun 11 '13 at 18:51
  • @Anton Edited my post. – Kaish Jun 11 '13 at 21:57
  • This is all manner of wrong, the parametrisation gives the circle in the interval $[0,2\pi)$, if we want the upper half we need to use $[0,\pi]$. The answer should be half the circumference of the circle so it REALLY needs a $\pi$ in it somewhere. – James Jun 11 '13 at 22:05
  • @James Ohh, yeah, oops. Sorry. I'll edit it. No real excuses for that. – Kaish Jun 11 '13 at 22:06
  • There is a typo in the 2nd line of the calculation of the integral... – James Jun 11 '13 at 22:11
  • @James Fixed it. Thanks. – Kaish Jun 11 '13 at 22:17
  • There is a fallacy in this argument which shows that the problem is bad (not the fault of the solver; I'm sure this is the intended solution). The arc length integrand relies on the fact that the derivative of sine is cosine. And that follows from the trigonometric limit $\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$. And that follows from the identity $s=r\theta$ for radian measure. And that follows from the definition of $\pi$ as the ratio of a circle's circumference to its diameter. So using an integral to “prove“ this relation is...wait for it...circular. – Matthew Leingang May 17 '20 at 19:35
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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.

Community
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msh210
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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.

André Nicolas
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