1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Find the curve given the tangent

  1. Jun 5, 2010 #1
    1. The problem statement, all variables and given/known data

    Given that the tangent to the curve [itex] c(t) [/itex] at any point on the curve is [itex] T(t) = (-sin(t), cos(t) )[/itex], find [itex] c(t) [/itex] if the curve passes through the point [itex] (0,0) [/itex].


    3. The attempt at a solution

    I try to let
    [itex] c(t) = ( x(t), y(t) ) [/itex]
    Then
    [itex] c'(t) = ( x'(t), y'(t) ) [/itex]
    [itex]| c'(t) | = \sqrt{[x'(t)]^2 + [y'(t)]^2 } [/itex]
    And
    [itex] T(t) = \frac{c'(t)}{|c'(t)|} [/itex]

    However this is complicated and consequently I am not sure how to solve it. I am also not sure how to "use" the point given since (0, 0) correspond the the values x and y respectively rather than t.

    Thanks.
     
  2. jcsd
  3. Jun 5, 2010 #2

    rock.freak667

    User Avatar
    Homework Helper

    If you had c'(t) =<2t,1> then c(t) = <t2+A,t+B> where A,B=constant.


    But you have

    c'(t)/|c'(t)| = <-sint,cost>

    so what is c(t) ? (note that |c'(t)| is just a constant)

    At (0,0), t=0. So what is c(t) now?
     
  4. Jun 5, 2010 #3
    c(t) = (cos(t) + C, sin(t) + K)
    t= 0, point is (0,0)

    So x(t) = 1 + C = 0, C = -1
    And y(t) = 0 + K, K=0

    So c(t) = (cos(t) -1 , sin(t) ) ?
     
  5. Jun 5, 2010 #4

    rock.freak667

    User Avatar
    Homework Helper

    Yes but you forgot out |c'(t)| which is the distance from the center to any tangent. In this case it would just be the same as |T(t)|
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook