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Tangent Plane to two parametric curves which intersect

  1. Nov 7, 2011 #1
    1. The problem statement, all variables and given/known data
    Suppose you need to know an equation of the tangent plane to a surface S at the point P(2,1,3). You don't know the equation for S but you know that the curves

    r1(t)=<2+3t,1-t^2,3-4t+t^2>

    r2(u)=<1+u^2,2u^3-1,2u+1>

    both lie on S. Find an equation of the tangent plane at P.

    2. Relevant equations
    z-z0=a(x-x0)+b(y-y0) The equation for a plane where a=fx(x0,y0) and b=fy(x0,y0)


    3. The attempt at a solution
    I know that r1 and r2 intercept at P when u=1 and t=0, so I think x0=2, y0=1, and z0=3 but I'm not sure how to find the partial derivatives with respect to x and y of the plane. I tried taking dz/dt divided by dy/dt and comparing it with dz/du divided by dy/du but I got different answers for the two intercepting curves so I dunno what to do. Do I need to find an orthogonal vector- if so, how do I do that :P
     
    Last edited: Nov 7, 2011
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  3. Nov 7, 2011 #2

    SammyS

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    Re: Tangent Plane to two 2-independent variable parametric curves which intersect

    What you have just given looks to me like the procedure for finding the plane which is tangent to the surface given by z = f(x,y) at the point (x0, y0, z0).

    That's not what you have in this problem.
    What you need for this problem is to take the derivatives of r1 and r2: [itex]\displaystyle \vec{v}_1(t)=\frac{d}{dt}\vec{r}_1(t)\,,\text{ and }\vec{v}_2(u)=\frac{d}{du}\vec{r}_2(u)\,,[/itex] evaluated at t = 0 and u = 1 respectively.

    The plane may be described parametrically by, [itex]<x,\,y,\,z>=<x_0,\,y_0,\,z_0> +\ \vec{v}_1(t_0)t+\vec{v}_2(u_0)u\ .[/itex]
     
  4. Nov 7, 2011 #3
    Truly only this talking dog can express my love for you, man.
     
    Last edited by a moderator: Sep 25, 2014
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