Tangent Plane to two parametric curves which intersect

Click For Summary
To find the tangent plane to surface S at point P(2,1,3), the curves r1(t) and r2(u) must be evaluated at their intersection, where t=0 and u=1. The derivatives of both curves, denoted as v1(t) and v2(u), should be calculated at these respective values to determine the direction vectors. The equation of the tangent plane can then be formulated using the point and the direction vectors derived from these derivatives. The confusion arises from the need to correctly apply the tangent plane formula, which requires the partial derivatives of the surface at the given point. Understanding the relationship between the curves and their derivatives is crucial for solving the problem.
Vadermort
Messages
11
Reaction score
0

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
 
Last edited:
Physics news on Phys.org


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\ .
 
Truly only this talking dog can express my love for you, man.
 
Last edited by a moderator:
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

Similar threads

  • · Replies 12 ·
Replies
12
Views
2K
Replies
1
Views
1K
Replies
1
Views
3K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 11 ·
Replies
11
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K