Results 1 to 11 of 11

Math Help - Circumference of a circle

  1. #1
    Junior Member
    Joined
    Jun 2008
    Posts
    44

    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?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Reckoner's Avatar
    Joined
    May 2008
    From
    Baltimore, MD (USA)
    Posts
    1,024
    Thanks
    75
    Awards
    1

    Smile

    Quote Originally Posted by lynch-mob View Post
    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).
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Jun 2008
    Posts
    44

    This will be a long one

    Quote Originally Posted by Reckoner View Post
    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.
    Letīs 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 Iīll adjust the expression after which Iīll put it in the formula for the perimeter length. Here goes:

    Iīll 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)?????

    Quote Originally Posted by Reckoner View Post
    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).
    Iīll get back to this one. This Latex thing takes a lot of energy for a newbie. Thanks for your help!
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Jun 2008
    Posts
    44
    Quote Originally Posted by Reckoner View Post
    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 Iīll try. There were interval values, sorry for not posting. They are between 0 and 8 on x-axis.

    First Iīll 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 canīt 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?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member wingless's Avatar
    Joined
    Dec 2007
    From
    Istanbul
    Posts
    585
    Quote Originally Posted by lynch-mob View Post
    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
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Junior Member
    Joined
    Jun 2008
    Posts
    44

    hmm

    Quote Originally Posted by wingless View Post
    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 donīt understand this parameter stuff. Maybe someday? But there must be a way to calculate the task with the formula. What did I do wrong?
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Super Member wingless's Avatar
    Joined
    Dec 2007
    From
    Istanbul
    Posts
    585
    Quote Originally Posted by lynch-mob View Post
    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 donīt 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.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor Reckoner's Avatar
    Joined
    May 2008
    From
    Baltimore, MD (USA)
    Posts
    1,024
    Thanks
    75
    Awards
    1

    Smile

    Nice work! Just one problem:

    Quote Originally Posted by lynch-mob View Post
    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:

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

    18-\frac{2}{3} = 17\frac{1}{3} which canīt 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).
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Junior Member
    Joined
    Jun 2008
    Posts
    44
    Quote Originally Posted by Reckoner View Post
    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...Iīve never used that arcsin rule before. Great.


    Quote Originally Posted by Reckoner View Post
    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 Iīll have to overcome in order to learn new stuff.


    Quote Originally Posted by Reckoner View Post
    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 didnīt 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?
    Follow Math Help Forum on Facebook and Google+

  10. #10
    MHF Contributor Reckoner's Avatar
    Joined
    May 2008
    From
    Baltimore, MD (USA)
    Posts
    1,024
    Thanks
    75
    Awards
    1

    Smile

    Quote Originally Posted by lynch-mob View Post
    I didnīt 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.
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Junior Member
    Joined
    Jun 2008
    Posts
    44
    Quote Originally Posted by Reckoner View Post
    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!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. circumference of the circle
    Posted in the Geometry Forum
    Replies: 1
    Last Post: November 3rd 2010, 11:33 PM
  2. pdf of area and circumference of a circle
    Posted in the Advanced Statistics Forum
    Replies: 6
    Last Post: May 10th 2010, 09:56 PM
  3. Spherical Circle Circumference
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: October 13th 2009, 11:33 PM
  4. Circumference of a circle HELP!!!!
    Posted in the Math Topics Forum
    Replies: 4
    Last Post: April 22nd 2008, 01:19 PM
  5. circumference of a circle
    Posted in the Geometry Forum
    Replies: 1
    Last Post: December 7th 2006, 07:39 PM

Search Tags


/mathhelpforum @mathhelpforum