Math Help - Circumference of a circle

1. Circumference of a circle

Use integrals to
a) calculate the circumference of a circle with radius=5
b) calculate the length of the curve $y=\frac{2}{3}x^\frac{3}{2}$
How do I do this?

2. Originally Posted by lynch-mob
Use integrals to calculate the circumference of a circle with radius=5

How do I do this?
Use the formula for arc length: the arc length $s$ of a continuous function $f$ over an interval $[a,\;b]$ is given by

$s = \int_a^b\sqrt{1+\left(f'(x)\right)^2}\,dx$

A circle with radius $r$ can be represented with the equation $x^2 + y^2 = r^2$.

So, you can calculate the circumference of a circle by finding the arc length of the curve $y = \sqrt{r^2 - x^2}$ from $[-r,\;r]$ (this will give you the top half of it), and then multiplying by 2 to get the whole perimeter.

In order to evaluate the integral, be sure you know your inverse trig rules.

Edit: For your second curve, are you given an interval? If not, $y = \frac23x^{3/2}$ is defined for all real $x\geq0$, and the curve will be infinitely long (as we should expect, the improper integral diverges).

3. This will be a long one

Originally Posted by Reckoner
Use the formula for arc length: the arc length $s$ of a continuous function $f$ over an interval $[a,\;b]$ is given by

$s = \int_a^b\sqrt{1+\left(f'(x)\right)^2}\,dx$

A circle with radius $r$ can be represented with the equation $x^2 + y^2 = r^2$.

So, you can calculate the circumference of a circle by finding the arc length of the curve $y = \sqrt{r^2 - x^2}$ from $[-r,\;r]$ (this will give you the top half of it), and then multiplying by 2 to get the whole perimeter.

In order to evaluate the integral, be sure you know your inverse trig rules.
Let´s see now. From the equation of a circle I get $y=\sqrt{25-x^2}$ after which i try to get the derivative of f(x) which is:
$(25-x^2)^\frac{1}{2} = \frac{1}{2}(25-x^2)^{-\frac{1}{2}} (-2x) = \frac{-x}{\sqrt{25-x^2}}$

now I´ll adjust the expression after which I´ll put it in the formula for the perimeter length. Here goes:

I´ll square (is this the correct word for it?) the expression like this:
$\frac{(-x)^2}{25-x^2}$ and then in the formula:

$\int_{-5}^{+5} \sqrt{1+\frac{x^2}{25-x^2}} = \int_{-5}^{+5} \sqrt{\frac{25-x^2}{25-x^2}+\frac{x^2}{25-x^2}}$

which is ...

$\int_{-5}^{+5} \sqrt{\frac{25}{25-x^2}} = \int_{-5}^{+5} \sqrt{\frac{25}{25(1-\frac{x^2}{25})}} =$

$\int_{-5}^{+5} \sqrt{\frac{1}{1-(\frac{x}{5})^2}}$ and after a root rule $\int_{-5}^{+5} \frac{1}{\sqrt{1-(\frac{x}{5})^2}}$

And this is the hardest part to me. We get:

$arcsin\frac{x}{5}\bigg]_{-5}^{+5}$ and after putting in the interval values we get $(arcsin1-arcsin{-1}$ which equals $\pi$

and since we got half the circle we will multiply it with 2 but we still need a 5 to multiply with. Because when I use the easy formula for calculating the perimeter of a circle with radius 5 I get 31,41...

So I have the $\pi$ and I have 2 but where can I get a five (which is the radius in this task)?????

Originally Posted by Reckoner
Edit: For your second curve, are you given an interval? If not, $y = \frac23x^{3/2}$ is defined for all real $x\geq0$, and the curve will be infinitely long (as we should expect, the improper integral diverges).
I´ll get back to this one. This Latex thing takes a lot of energy for a newbie. Thanks for your help!

4. Originally Posted by Reckoner
Edit: For your second curve, are you given an interval? If not, $y = \frac23x^{3/2}$ is defined for all real $x\geq0$, and the curve will be infinitely long (as we should expect, the improper integral diverges).
Ok I´ll try. There were interval values, sorry for not posting. They are between 0 and 8 on x-axis.

First I´ll try to differentiate the function like this

$D\frac{2}{3}x^\frac{3}{2} = x^\frac{1}{2}$ which is $\sqrt{x}$

When putting this into the formula we get

$\int_{0}^{8}\sqrt{1+x} = \int_{0}^{8}(1+x)^\frac{1}{2} = \frac{2}{3}(1+x)^\frac{3}{2}\bigg]_0^8 =$

which is

$\frac {2}{3}(1+x)\sqrt{(1+x)}$ and putting in the values we get:

$18-\frac{2}{3} = 17\frac{1}{3}$ which can´t be true?? Can it? I mean the graph is almost as straigt as a line and when looking at it (sides 8 and 8) the hypotenuse should be about 11,3. But maybe Im wrong?

5. Originally Posted by lynch-mob
Use integrals to
a) calculate the circumference of a circle with radius=5
The arc length formula is actually $\int \sqrt{(dx)^2+(dy)^2}$.

We can calculate it easily by using this formula. Firstly, parameterize the circle.

$x = 5 \cos \theta$

$y=5\sin \theta$

And the integral will be from 0 to 2π.

Find the differentials:

$dx = -5 \sin \theta~d\theta$

$dy = 5 \cos \theta~d\theta$

$\int_0^{2\pi} \sqrt{25 \sin^2 \theta~(d\theta)^2 + 25 \cos^2 \theta (d\theta)^2}$

$\int_0^{2\pi} 5~d\theta = 10\pi$

6. hmm

Originally Posted by wingless
The arc length formula is actually $\int \sqrt{(dx)^2+(dy)^2}$.

We can calculate it easily by using this formula. Firstly, parameterize the circle.

$x = 5 \cos \theta$

$y=5\sin \theta$

And the integral will be from 0 to 2π.

Find the differentials:

$dx = -5 \sin \theta~d\theta$

$dy = 5 \cos \theta~d\theta$

$\int_0^{2\pi} \sqrt{25 \sin^2 \theta~(d\theta)^2 + 25 \cos^2 \theta (d\theta)^2}$

$\int_0^{2\pi} 5~d\theta = 10\pi$
I read in a book that if we parameterize x itself we get the formula $\int_{-5}^{5}\sqrt{1+f'(x)^2}$

Now I would like to try it with this formula cuz I really don´t understand this parameter stuff. Maybe someday? But there must be a way to calculate the task with the formula. What did I do wrong?

7. Originally Posted by lynch-mob
I read in a book that if we parameterize x itself we get the formula $\int_{-5}^{5}\sqrt{1+f'(x)^2}$

Now I would like to try it with this formula cuz I really don´t understand this parameter stuff. Maybe someday? But there must be a way to calculate the task with the formula. What did I do wrong?
$\int \sqrt{(dx)^2+(dy)^2}$

$\int \sqrt{(dx)^2 \left [1+\left ( \frac{dy}{dx} \right )^2 \right ]}$

$\int \sqrt{1+\left ( \frac{dy}{dx} \right )^2}~dx$

$\int \sqrt{1+ [f'(x)]^2}~dx$

So your formula can be derived from mine.

Ok, let's solve it with yours.

Equation of the circle is: $x^2 + y^2 = 25$.

$y^2 = 25 -x^2$

$y = +\sqrt{25 -x^2}~\text{and}~-\sqrt{25 -x^2}$

The circle can't be written as a function of x. But it can be written as two functions.

$y = \sqrt{25 -x^2}$ is the top semi circle and $-\sqrt{25 -x^2}$ is the bottom semi circle.

Just take one of them and calculate it's length. Let's use the top semi circle, $y = \sqrt{25 -x^2}$.

$f(x) = \sqrt{25 -x^2}$

$f'(x) = \frac{1}{2\sqrt{25 -x^2}}\cdot (25-x^2)'$

$f'(x) = \frac{-2x}{2\sqrt{25 -x^2}}$

$f'(x) = \frac{-x}{\sqrt{25 -x^2}}$

Plug it in the formula:

$\int_{-5}^5 \sqrt{ 1+ \left ( \frac{-x}{\sqrt{25 -x^2}} \right )^2 }~dx$

$\int_{-5}^5 \sqrt{1+ \frac{x^2}{25 -x^2}}~dx$

$\int_{-5}^5 \sqrt{\frac{25}{25 -x^2}}~dx$

Let $x=5u$, then $dx = 5~du$ and the bounds will change too,

$5\int_{-1}^1 \sqrt{\frac{25}{25 -25u^2}}~du$

$5\int_{-1}^1 \sqrt{\frac{1}{1 -u^2}}~du$

$5 \arcsin x \bigg |_{-1}^1$

$5\pi$

Length of the bottom semi circle is $5\pi$. Then circumference of the circle will be $2\cdot 5\pi = 10\pi$.

8. Nice work! Just one problem:

Originally Posted by lynch-mob
And this is the hardest part to me. We get:

$arcsin\frac{x}{5}\bigg]_{-5}^{+5}$ and after putting in the interval values we get $(arcsin1-arcsin{-1}$ which equals $\pi$

and since we got half the circle we will multiply it with 2 but we still need a 5 to multiply with. Because when I use the easy formula for calculating the perimeter of a circle with radius 5 I get 31,41...
Your antiderivative is a little off. The general rule is

$\int\frac1{\sqrt{a^2 - x^2}}\,dx = \arcsin\frac xa + C,\;a > 0$

So, for your integral, we have

$\int_{-5}^5\sqrt{\frac{25}{25 - x^2}}\,dx$

$=\int_{-5}^5\frac5{\sqrt{5^2 - x^2}}\,dx$

$=5\int_{-5}^5\frac1{\sqrt{5^2 - x^2}}\,dx$

$=5\arcsin\frac x5\bigg|_{-5}^5 = 5\pi$

Multiplying by two (since we only did the upper-half), we get a circumference of $2\cdot5\pi = 10\pi$. This agrees with our existing formula, $2\pi r = 2\pi\cdot5 = 10\pi$.

Regarding the parametric equations: you will probably learn this soon if you haven't already (in the US at least it's usually covered by Calc II). Parametric forms can often be a lot easier to work with.

For your other question:

Originally Posted by lynch-mob
$\frac {2}{3}(1+x)\sqrt{(1+x)}$ and putting in the values we get:

$18-\frac{2}{3} = 17\frac{1}{3}$ which can´t be true?? Can it? I mean the graph is almost as straigt as a line and when looking at it (sides 8 and 8) the hypotenuse should be about 11,3. But maybe Im wrong?
It's very good that you try to check the reasonableness of your answer like that! However, $\frac23\left(8^{3/2}\right)\approx15.084$, so your hypotenuse would be about $17.075$, which is pretty close to the arc length that you found (naturally we expect the actual length to be a little larger than this, and it is).

9. Originally Posted by Reckoner
Nice work! Just one problem:

Your antiderivative is a little off. The general rule is

$\int\frac1{\sqrt{a^2 - x^2}}\,dx = \arcsin\frac xa + C,\;a > 0$
Wow...I´ve never used that arcsin rule before. Great.

Originally Posted by Reckoner
Regarding the parametric equations: you will probably learn this soon if you haven't already (in the US at least it's usually covered by Calc II). Parametric forms can often be a lot easier to work with.
Jep, I better do that, it just feels so outer space to me. Too big of a mountain to climb. But thats an obstacle I´ll have to overcome in order to learn new stuff.

Originally Posted by Reckoner
For your other question:

It's very good that you try to check the reasonableness of your answer like that! However, $\frac23\left(8^{3/2}\right)\approx15.084$, so your hypotenuse would be about $17.075$, which is pretty close to the arc length that you found (naturally we expect the actual length to be a little larger than this, and it is).
I didn´t get this one: $\frac23\left(8^{3/2}\right)\approx15.084$ where do you get this from? And is $17\frac{1}{3}$ the correct answer?

10. Originally Posted by lynch-mob
I didn´t get this one: $\frac23\left(8^{3/2}\right)\approx15.084$ where do you get this from? And is $17\frac{1}{3}$ the correct answer?
I looked over your work, and it seems good to me. $17\frac13 = \frac{52}3$ should be correct.

About your approximation: You used an isosceles triangle with sides of length 8, but this is not correct. You want to approximate the curve $y = \frac23x^{3/2}$ over the interval $[0,\;8],$ right? Thus you draw a line through the two endpoints: when $x = 0$, $\frac23x^{3/2} = 0$ and when $x = 8$, $\frac23x^{3/2} = \frac23\cdot8^{3/2} = \frac23\cdot8\sqrt8 = \frac{32\sqrt2}3\approx15.0849$. Finding the distance between these two points we get $d = \sqrt{(x_1 - x_0)^2 + (y_1 - y_0)^2}\approx \sqrt{(8 - 0)^2 + (15.0849 - 0)^2}\approx17.075$ which is pretty close to the value you found for the length of the actual curve.

11. Originally Posted by Reckoner
I looked over your work, and it seems good to me. $17\frac13 = \frac{52}3$ should be correct.

About your approximation: You used an isosceles triangle with sides of length 8, but this is not correct. You want to approximate the curve $y = \frac23x^{3/2}$ over the interval $[0,\;8],$ right? Thus you draw a line through the two endpoints: when $x = 0$, $\frac23x^{3/2} = 0$ and when $x = 8$, $\frac23x^{3/2} = \frac23\cdot8^{3/2} = \frac23\cdot8\sqrt8 = \frac{32\sqrt2}3\approx15.0849$. Finding the distance between these two points we get $d = \sqrt{(x_1 - x_0)^2 + (y_1 - y_0)^2}\approx \sqrt{(8 - 0)^2 + (15.0849 - 0)^2}\approx17.075$ which is pretty close to the value you found for the length of the actual curve.
Ok!