Finding the Tangent Plane to a Surface at a Given Point

In summary, the equation of the tangent plane to the surface S at the point P(2,1,3) can be found by taking the cross product of the tangent vectors of the curves r_1(t) and r_2(u) and plugging them into the general form of a plane at a point. The resulting equation is 12(x - 2) - 7(y - 1) + 9(z - 3) = 0.
  • #1
tylerc1991
166
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 have an equation for S but you know that the curves:

r_1(t) = <2 + 3t, 1 - t^2, 3 - 4t + t^2>
r_2(u) = <1 + u^2, 2u^3 - 1, 2u + 1>

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

Homework Equations



The general form of a plane at a point P(x_0, y_0, z_0) with normal vector n = <a, b, c> is

a(x - x_0) + b(y - y_0) + c(z - z_0) = 0

The Attempt at a Solution



t = 0 for P to lie on r_1(t), and u = 1 for P to lie on r_2(u)
taking the first derivatives of r_1(t) and r_2(u) and evaluating them at the specified t and u values gives:
r'_1(0) = <3, 0, -4> = A
r'_2(u) = <2, 6, 2> = B

Since both of these vectors are tangent to the surface at P, I took the cross product of them, A x B and got <24, 14, 18>, which simplifies to <12, 7, 9>.

so the normal vector to the tangent plane to the surface is <12, 7, 9> and the point is P.
plugging this into the general form gives:

12(x - 2) + 7(y - 1) + 9(z - 3) = 0.

This can be simplified but I wanted to know if this was the correct answer. Thank you very much anyone for your help!
 
Physics news on Phys.org
  • #2
Certainly nothing wrong with the approach. Are you sure <3, 0, -4>X<2, 6, 2>=<24, 14, 18>. I get something a little different. Can you check it?
 
  • #3
Dick said:
Certainly nothing wrong with the approach. Are you sure <3, 0, -4>X<2, 6, 2>=<24, 14, 18>. I get something a little different. Can you check it?

Whoops, forgot the minus in the determinant, I am getting <24, -14, 18> which reduces to <12, -7, 9>. Thank you.
 

1. What is a tangent plane in Calculus III?

A tangent plane in Calculus III is a plane that touches a three-dimensional surface at a single point and is parallel to the surface's local direction at that point. It is used to approximate the behavior of a function at a specific point.

2. How do you find the equation of a tangent plane?

To find the equation of a tangent plane at a given point on a surface in Calculus III, you need to calculate the partial derivatives of the function at that point and use them to find the normal vector of the tangent plane. Then, you can use the point-normal form of a plane to write the equation of the tangent plane.

3. What is the significance of tangent planes in Calculus III?

Tangent planes are significant in Calculus III because they allow us to approximate the behavior of a function at a specific point on a three-dimensional surface. They are also important in understanding the concept of differentiability and the gradient of a function.

4. How do you determine if a point lies on a tangent plane?

A point lies on a tangent plane if the gradient of the function at that point is perpendicular to the plane. In other words, the dot product of the gradient vector and the normal vector of the plane should be equal to zero.

5. Can a function have multiple tangent planes at a single point?

Yes, a function can have multiple tangent planes at a single point if the function is not differentiable at that point. This is because the tangent planes are used to approximate the behavior of the function, and if the function is not smooth at that point, there can be multiple possible tangent planes.

Similar threads

  • Calculus and Beyond Homework Help
Replies
1
Views
1K
  • Calculus and Beyond Homework Help
Replies
6
Views
706
  • Calculus and Beyond Homework Help
Replies
6
Views
474
  • Calculus and Beyond Homework Help
Replies
7
Views
2K
  • Calculus and Beyond Homework Help
Replies
9
Views
1K
  • Calculus and Beyond Homework Help
Replies
4
Views
776
  • Calculus and Beyond Homework Help
Replies
7
Views
339
  • Calculus and Beyond Homework Help
Replies
7
Views
135
  • Calculus and Beyond Homework Help
Replies
2
Views
564
  • Calculus and Beyond Homework Help
Replies
4
Views
1K
Back
Top