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Point of intersection of tangent line with another line

  1. May 6, 2014 #1
    1. The problem statement, all variables and given/known data

    Find the pt. at which the tangent line to the curve x=3t^2 - t, y=2t+t^3 at t=1 intersects the line y=2-x.

    2. Relevant equations

    Possibly <6t-1, 2+3t^2> if the tangent is not already present

    3. The attempt at a solution

    I am confused about how to go about solving this. Where should I use the t=1? I thought about parametrizing y=2-x, but I am unsure about how to do this. Do I assume 0=f(x,y)=2-x-y?

    The answer is (1/2, 3/2) but I have no idea how to get there.
  2. jcsd
  3. May 6, 2014 #2


    Staff: Mentor

    Parametrizing y=2-x is not a good idea.

    You have the curve, with x and y given in terms of a parameter t. How do you find dy/dx for such a curve? Once you have dy/dx, evaluate it for t = 1. This will give you the slope of the tangent line. The point of tangency is the point on the curve where t = 1.
  4. May 6, 2014 #3

    Thanks for taking the time to reply.

    I assume <6t-1, 2+3t^2> is the dy/dx at this curve.

    Evaluated at t=1, I would get <5,5>. Is the y value 5 the slope?

    I'm guessing putting it into another y=mx+b form would be the way to go, but I'm sure that I am not correctly translating these values.
    Last edited by a moderator: May 6, 2014
  5. May 6, 2014 #4


    Staff: Mentor

    No. This is the vector <dx/dt, dy/dt>. For the tangent to the curve, you need dy/dx.

    No, see above.
  6. May 6, 2014 #5
    I guess my question would be, how would I find dy/dx when only <dx/dt, dy/dt> has been provided. Would I find the parametrization r=<dx/dt, dy/dt> + t v and take dr/dx?

    Would that look something like r=<5,5> + (1) <dx/dt, dy/dt>?
  7. May 6, 2014 #6


    Staff: Mentor

    Well, from the chain rule, dy/dt = dy/dx * dx/dt. Does that give you an idea of how to find dy/dx?
    This doesn't make much sense. You want the tangent line to be y = mx + b. No vectors.
  8. May 6, 2014 #7


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    Homework Helper

    The equation for the tangent to a parametrized curve is ##\displaystyle \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}##. You should be able to calculate this quickly in terms of t.

    You can now calculate the gradient of the tangent line at the required point. Can you figure out the equation of the tangent line? This should be elementary.
  9. May 6, 2014 #8
    From the chain rule, I got y'=3t^2 +2 / 6t-1.

    I evaluated at t=1 which is y'=1. If m=y' and I choose (1,1) as a point, y=x but this definitely not the right line, since solving for the intersection of y=x and y=2-x yields (1, 1).
  10. May 6, 2014 #9
    Gradient? This is a problem from the beginning of multivariable calc so it shouldn't figure into it yet. I found dy/dx at evaluated at t=1. From the chain rule, I got y'=3t^2 +2 / 6t-1. I chose the point (1,1) to find the line equation. From this I got y=x, which isn't correct. There are gaps in my calculus knowledge so I'm not sure what is missing.
  11. May 6, 2014 #10


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    Science Advisor
    Homework Helper
    Gold Member

    I'm a bit late to this thread and I'm going to suggest a different, and in my opinion, more appropriate approach to this problem. You don't need to mess with dy/dx at all when you have the parametric equation for the line$$
    \vec r(t) = \langle3t^2-t, 2t+t^3\rangle$$And you know the derivative$$
    \vec r'(t) = \langle6t -1, 2+3t^2\rangle$$is tangent to the curve. The tangent line to the curve when ##t=1## is$$
    \vec T(s) = \vec r(1) + s\vec r'(1)$$Figure out what value of ##s## on that line gives ##y=2-x##and use that value in ##\vec T(s)## to get the point of intersection.
  12. May 6, 2014 #11


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    Homework Helper

    Gradient is another name for slope of a line.

    You can't just "choose" a random point like (1,1). The tangent line meets the curve at exactly one point. Hence this point lies on both the tangent line *and* the curve. You don't know the equation of the tangent line (yet) but you *do* know the parametric equations determining the curve. Now you have to *calculate* the coordinates of this point using those equations, getting the x and y values corresponding to t = 1.

    Following that, you have to determine the equation of the tangent line. The standard form is ##y = mx + c## where m is the gradient (or slope) and c is the y-intercept. You already know m, and you should be able to determine c based on the coordinates you just calculated.
  13. May 6, 2014 #12
    At t=1, the coordinates are <5,5> but that still yields y=x.
  14. May 6, 2014 #13


    Staff: Mentor

    No, it doesn't.

    From post #1
    The slope of the tangent line is 1, but I don't think you have figured out what the point of tangency is on the parametric curve. To get the equation of the tangent line, you need the slope (which you have) and a point on the line (which you don't have). What are the coordinates on the parametric curve when t = 1?
  15. May 7, 2014 #14


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    You do NOT use the expression for the derivative to get that point! Go back to the original parametric equations for the curve.
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