Results 1 to 5 of 5

Math Help - Curve Sketching

  1. #1
    Newbie
    Joined
    Apr 2009
    Posts
    3

    Curve Sketching

    Let x = 2t^2, y = (3/2)t^2 t = [3,5]

    I have to find to length of the curve from these equations using integration, without eliminating parameters.

    Then I need to sketch it having eliminated parameters, and finally, find the length using algebra.
    Last edited by yuropod; April 26th 2009 at 10:12 PM.
    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 yuropod View Post
    Let x = 2t^2, y = \frac{3}{2}t^2,     t = [3,5]

    I have to find to length of the curve from these equations using integration, without eliminating parameters.

    Then I need to sketch it having eliminated parameters, and finally, find the length using algebra.
    \frac{dy}{dx} =\frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{3t}{4t} = \frac{3}{4}.

    As t=[3,5], x=[18,50].

    Use the arc length formula: \int_{18}^{50}\sqrt{1+(3/4)^2}\,dx = \int_{18}^{50}\sqrt{25/16}\,dx = \frac{5}{4}\int_{18}^{50}\,dx = \frac{5}{4}\cdot 32 = 40

    ------------------------------------

    t=\sqrt{\frac{x}{2}}~~~~~t=\sqrt{\frac{2y}{3}}

    \sqrt{\frac{x}{2}}=\sqrt{\frac{2y}{3}} \implies \frac{x}{2}=\frac{2y}{3} \implies y=\frac{3x}{4} (This is the equation of the line you want to sketch.)

    Using x=[18,50] again, we have y=\left[\frac{27}{2},\frac{75}{2}\right]

    Using the Pythagorean Theorem, we have the length of this line to be \sqrt{(50-18)^2+(75/2-27/2)^2} = \sqrt{32^2+24^2} = \sqrt{1600}=40 which agrees with the above answer.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Apr 2009
    Posts
    3
    Thanks a bunch!
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,914
    Thanks
    779
    Hello, yuropod!

    Let: . \begin{array}{ccc}x &=&2t^2 \\ y &=&\frac{3}{2}t^2\end{array} \quad t \in [3,5]

    Find the length of the curve without eliminating parameters.
    Formula: . L \;=\;\int^b_a\sqrt{\left(\tfrac{dx}{dt}\right)^2 + \left(\tfrac{dy}{dt}\right)^2}\,dt

    We have: . \begin{array}{ccc}\frac{dx}{dt} &=& 4t \\ \\[-4mm]\frac{dy}{dt} &=& 3t \end{array}

    . . Then: . \left(\tfrac{dx}{dt}\right)^2 + \left(\tfrac{dy}{dt}\right)^2 \:=\:(4t)^2 + (3t)^2 \:=\:16t^2 + 9y^2 \:=\:25t^2

    . . Hence: . \sqrt{\left(\tfrac{dx}{dt}\right)^2 + \left(\tfrac{dy}{dt}\right)^2} \;=\;\sqrt{25t^2} \;=\;5t

    Therefore: . L \;=\;\int^5_3 5t\,dt \;=\;\tfrac{5}{2}\,t^2\,\bigg]^5_3 \;=\;\tfrac{5}{2}(5^2) - \tfrac{5}{2}(3^2) \;=\;40

    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    May 2007
    Posts
    237
    Quote Originally Posted by yuropod View Post
    Let x = 2t^2, y = (3/2)t^2 t = [3,5]

    I have to find to length of the curve from these equations using integration, without eliminating parameters.

    Then I need to sketch it having eliminated parameters, and finally, find the length using algebra.
    It's in fact the straight line y=3x/4 from (18,27/2) to (50,75/2) . So we can use the distance between the two pionts to find the length.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. curve sketching
    Posted in the Calculus Forum
    Replies: 1
    Last Post: June 8th 2010, 02:32 AM
  2. Curve Sketching
    Posted in the Calculus Forum
    Replies: 1
    Last Post: May 26th 2010, 07:20 PM
  3. Curve Sketching
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: April 3rd 2010, 02:48 AM
  4. Curve Sketching
    Posted in the Calculus Forum
    Replies: 3
    Last Post: March 19th 2009, 03:09 PM
  5. Curve Sketching again~
    Posted in the Calculus Forum
    Replies: 1
    Last Post: September 18th 2008, 02:55 PM

Search Tags


/mathhelpforum @mathhelpforum