Results 1 to 5 of 5

Math Help - Parametric Curves

  1. #1
    Newbie iebi's Avatar
    Joined
    May 2009
    Posts
    4

    Parametric Curves

    I'm currently doing Parametric Curves where we have to fine the length of the curve using L = the integral of sqrt((dx/dt)^2+(dy/dt)^2) dt, but I'm having a lot of trouble simplifying the equations to make it easier to take the integral of them. Can I get some help on these few question?



    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member redsoxfan325's Avatar
    Joined
    Feb 2009
    From
    Swampscott, MA
    Posts
    943
    Quote Originally Posted by iebi View Post
    I'm currently doing Parametric Curves where we have to fine the length of the curve using L = the integral of sqrt((dx/dt)^2+(dy/dt)^2) dt, but I'm having a lot of trouble simplifying the equations to make it easier to take the integral of them. Can I get some help on these few question?



    x=\frac{t}{4+t} \implies \frac{dx}{dt}=\frac{4+t-t}{(t+4)^2} = \frac{4}{(t+4)^2}
    y=\ln(4+t) \implies \frac{dy}{dt}=\frac{1}{t+4}
    0\leq t\leq 5

    \sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}  {dt}\right)^2} = \sqrt{\frac{16}{(t+4)^4}+\frac{1}{(t+4)^2}} = \sqrt{\frac{16}{(t+4)^4}+\frac{(t+4)^2}{(t+4)^4}} = \frac{\sqrt{(t+4)^2+16}}{(t+4)^4}

    \int_0^5\frac{\sqrt{(t+4)^2+16}}{(t+4)^2}\,dt

    This is a really ugly integral. Let u=t+4 and du=dt

    Now you have \int_0^5\frac{\sqrt{u^2+16}}{u^2}\,du

    Let u=4\tan(v) and du = 4\sec^2(v)\,dv

    Now you have \int_0^5\frac{\sqrt{16\tan^2(v)+16}}{16\tan^2(v)}\  cdot4\sec^2(v)\,dv = \int_0^5\frac{\sec^3(v)}{\tan^2(v)}\,dv

    I have to go. Can you finish from here? If not, I'm sure someone else will help you. Hint: Split \sec^3(v) into \sec(v)(1+\tan^2(v)).
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie iebi's Avatar
    Joined
    May 2009
    Posts
    4
    Thanks. I got to the part where I combine the equations, but didn't know what to do after that. I can finish that problem. Can anyone start me off with the other 3 problems?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member redsoxfan325's Avatar
    Joined
    Feb 2009
    From
    Swampscott, MA
    Posts
    943
    Quote Originally Posted by iebi View Post
    Thanks. I got to the part where I combine the equations, but didn't know what to do after that. I can finish that problem. Can anyone start me off with the other 3 problems?
    8.) \frac{dx}{dy} = \frac{y^4}{2}-\frac{1}{2y^4} = \frac{y^8-1}{2y^4}

    Square it: \frac{(y^8-1)^2}{4y^8} = \frac{y^{16}-2y^8+1}{4y^8}

    Add 1: \frac{y^{16}-2y^8+1}{4y^8}+1 = \frac{y^{16}-2y^8+1+4y^8}{4y^8} = \frac{y^{16}+2y^8+1}{4y^8} = \frac{(y^8+1)^2}{4y^8}

    Take the square root: \sqrt{\frac{(y^8+1)^2}{4y^8}} = \frac{y^8+1}{2y^4}

    Integrate: \int_1^4\frac{y^8+1}{2y^4}\,dy = \int_1^4\frac{y^4}{2}+\frac{1}{2y^4}\,dy = \left[\frac{y^5}{10}-\frac{1}{6y^3}\right]_1^4 = \boxed{\frac{65577}{640} \approx 102.464}
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie iebi's Avatar
    Joined
    May 2009
    Posts
    4
    Can anyone show me how to start the last two problems?
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Help on Parametric Curves please
    Posted in the Calculus Forum
    Replies: 4
    Last Post: December 10th 2010, 12:02 PM
  2. Parametric curves
    Posted in the Calculus Forum
    Replies: 1
    Last Post: April 14th 2010, 01:56 PM
  3. Parametric Curves
    Posted in the Calculus Forum
    Replies: 9
    Last Post: October 20th 2009, 06:14 PM
  4. parametric curves
    Posted in the Calculus Forum
    Replies: 3
    Last Post: August 6th 2008, 10:35 PM
  5. Parametric curves
    Posted in the Calculus Forum
    Replies: 1
    Last Post: April 27th 2008, 10:11 PM

Search Tags


/mathhelpforum @mathhelpforum