1. Not finding help here? Sign up for a free 30min 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!

X intercept of Line Tangent to Curve

  1. May 20, 2014 #1
    1. The problem statement, all variables and given/known data

    What is the x-intercept of the line tangent to the curve x(t) = 3 + cos(∏t), y(t) = t^2 + t + 1, when t = 1?

    2. Relevant equations

    Derivative, y=mx+b

    3. The attempt at a solution

    To find the line tangent to the curve:

    d/dt = <-∏sin(∏t), 2t+1>

    at t=1 <-∏, 3>

    dy/dx = dy/du * du/dx = -3/∏

    y=(-3/∏)x+b

    at t=1, <x,y>=<3,3>

    y-3 = (-3/∏)(x-3)

    y = -3x /∏ + 9/∏ +3

    To find x-int. set y=0 so 0=-3x /∏ + 9/∏ +3

    (-3 - 9/∏)(-∏/3)=x

    The answer is x=2. I am not sure where I went wrong.
     
    Last edited: May 20, 2014
  2. jcsd
  3. May 20, 2014 #2

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    ##\sin(\pi)=0##

    What is ##u##? I think you mean ##\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}##. But you could just use the parametric equation of the tangent line instead of using ##y=mx+b## form.

     
  4. May 20, 2014 #3
    I see sin(∏)=0. Then that would mean dy/dx = 0.

    By parametric equation of the tangent line do you mean r'(t)=x' + y'? To find int. I would set r'(1)=x' + y' and x'=-∏sin(t∏). t=1 so x'=-∏sin(∏)=0. y'=3

    x'=sin(∏)=0 Inv sin (0)=0

    r'(t)=3
     
  5. May 20, 2014 #4

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    No. The 0 is in the denominator, so the slope is undefined. All the more reason not to try using ##y=mx+b##.

    That is really poor notation. ##r(t) = \langle x(t),y(t)\rangle## which is a vector, not a sum, and ##r'(t) = \langle x'(t),y'(t)\rangle## so ##r'(1) = \langle -\pi\sin\pi,3\rangle=\langle 0,3\rangle##. Since ##r(1) = \langle 2,3\rangle##, you have a point and direction vector for the tangent line when ##t=1##. What is the parametric equation of that tangent line?
     
  6. May 20, 2014 #5
    Would it be r(t)=<0, 3> + t<2,3> for the tangent? I am not sure how I can say the intercept is two from this equation. If I just look at x then x=2t which at t=1 is 2. Is that how it is found?
     
  7. May 20, 2014 #6

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Close but you have it backwards. <2,3> is the point and <0,3> is the direction vector. And it's best to use a different parameter and a different name so ##L(s) = \langle<x(s),y(s)\rangle=\langle 2,3\rangle + s\langle 0,3\rangle ##, where ##L## is the tangent line.

    What value of ##s## makes ##y(s)=0##? What is ##x(s)## for that value.

    Also, try not to lose sight of the forest for the trees. In the xy plane you have a line through (2,3) and its direction vector is <0,3>. What direction is that? What does the line look like?
     
  8. May 20, 2014 #7
    s would need to be 0. Would I necessarily need to find y in order to find the x-intercept?

    The point (2,3) is going towards (0,3) so it is a line headed downwards along x?
     
  9. May 20, 2014 #8

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Why?

    The x intercept on any graph is the x value when y = 0.

    I don't know what "(2,3) going towards (0,3)" means nor what "headed downwards along x" means. The graph is a straight line. Surely you can describe it better than that. Plot a few points for various s.
     
  10. May 20, 2014 #9
    If I separate them into x=2+s(0) and y=3+s(3) then the only thing that makes y=0 is s=0?
     
  11. May 20, 2014 #10

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Are you serious?
     
  12. May 20, 2014 #11
    If I weren't, I wouldn't still be here eight hours later, sadly. Going off your equation L(s) =<x(s), y(s)> I assumed that I can simply look at x(s) since it is the x-intercept. s=0 so x=2. If I'm missing something, I don't know what it is.
     
  13. May 20, 2014 #12

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    You have y = 3+3s and you are saying y = 0 when s = 0???

    In this problem x happens to be constant so x=2 is the correct intercept even though you have the incorrect value of s for y=0. Can you describe what the line looks like?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted