Tangent Plane to two parametric curves which intersect

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SUMMARY

The discussion focuses on finding the equation of the tangent plane to a surface S at the point P(2,1,3), given two parametric curves: r1(t) = <2+3t, 1-t^2, 3-4t+t^2> and r2(u) = <1+u^2, 2u^3-1, 2u+1>. The curves intersect at P when t=0 and u=1. To derive the tangent plane, participants emphasize the need to compute the derivatives of the curves, specifically \(\vec{v}_1(t) = \frac{d}{dt}\vec{r}_1(t)\) and \(\vec{v}_2(u) = \frac{d}{du}\vec{r}_2(u)\), evaluated at the respective parameters. The correct approach involves using these derivatives to establish the plane's equation.

PREREQUISITES
  • Understanding of parametric equations
  • Knowledge of vector calculus, specifically derivatives of vector functions
  • Familiarity with the equation of a plane in three-dimensional space
  • Ability to evaluate limits and derivatives at specific points
NEXT STEPS
  • Learn how to compute derivatives of vector functions in three dimensions
  • Study the concept of tangent planes and their equations in multivariable calculus
  • Explore the application of the chain rule in vector calculus
  • Investigate the geometric interpretation of parametric curves and surfaces
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Students studying multivariable calculus, mathematicians interested in vector calculus, and educators teaching concepts related to tangent planes and parametric curves.

Vadermort
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Homework Statement


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.

Homework Equations


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

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 don't know what to do. Do I need to find an orthogonal vector- if so, how do I do that :P
 
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Vadermort said:

Homework Statement


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.

Homework Equations


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

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/dy divided by dy/dt but I got different answers for the two intercepting curves so I don't know what to do.
What you need for this problem is to take the derivatives of r1 and r2: \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)\,, evaluated at t = 0 and u = 1 respectively.

The plane may be described parametrically by, &lt;x,\,y,\,z&gt;=&lt;x_0,\,y_0,\,z_0&gt; +\ \vec{v}_1(t_0)t+\vec{v}_2(u_0)u\ .
 
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