1. Math 200
Week 6 - Monday
Tangent Planes

2. Math 200
Goals
Be able to compute an equation of the tangent plane at a point on the surface z = f(x,y).
Given an implicitly defined level surface F(x,y,z) = k, be able to compute an equation of the tangent plane at a point on the surface.
Know how to compute the parametric equations (or vector equation) for the normal line to a surface at a specified point.
Be able to use gradients to find tangent lines to the intersection curve of two surfaces.
And, be able to find (acute) angles between tangent planes and other planes.

3. Math 200
What we know
In the last section we saw that the gradient at a point is normal to the level curve through that point
By extension, the gradient of a function of three variables, F(x,y,z), is normal to the level surface through a given point
We need two things for a plane: (1) a point on the plane and (2) a vector normal to the plane

4. Math 200
Example
Consider the surface S: z = y2 - x2 (saddle). Let’s find the tangent plane to S at A(1,2,3).
We can treat the surface as the level surface of a function of three variables
If F(x,y,z) = y2 - x2 - z, then S is the level surface F = 0:
Check: 0 = y2 - x2 - z —> z = y2 - x2
Now find the gradient of F at A

5. Math 200
We’ll use the gradient vector as our normal vector for the tangent plane…

6. Math 200
You try
Find the tangent plane to the ellipsoid E: x2 + y2 + 2z2 = 4 at A(1,-1,1)

7. Then the surface E is the level surface F = 4
Math 200
First we treat the surface as the level surface of some function of three variables:
x2 + y2 + 2z2 = 4
Let F(x,y,z) = x2 + y2 + 2z2
Then the surface E is the level surface F = 4
Find the gradient of F at A
Tangent Plane:
2(x-1) - 2(y+1) + 4(z-1) = 0 or 2x - 2y + 4z = 8

8. Math 200
Normal Lines
We can easily find normal lines to surfaces using the same basic steps:
Treat the surface as a level surface
Compute the gradient of F(x,y,z) at the desired point
Use the gradient as the direction vector

9. E.g. For S: z = y2 - x2 at A(1,2,3) we had
Math 200
E.g. For S: z = y2 - x2 at A(1,2,3) we had
So, using <-2,4,-1> as the direction vector for the normal line…

10. One more example before moving on
Math 200
One more example before moving on
Consider the function f(x,y) = 2xy - xy2
Find the tangent plane to the surface at (-1,-1)
Find the normal line to the surface at (-1,-1)

11. Now we can compute the gradient of F
Math 200
First, we rewrite the surface as the level surface of some function of three variables:
f(x,y) = 2xy - xy2
z = 2xy - xy2
0 = 2xy - xy2 - z
F(x,y,z) = 2xy - xy2 - z
Now we can compute the gradient of F
f(-1,-1) = 3 so the point we care about it (-1,-1,3)
Plane:
Normal line:

12. Math 200

13. Generalizing some Consider any function of two variables, f(x,y).
Math 200
Generalizing some
Consider any function of two variables, f(x,y).
To find the tangent plane at (x0,y0), we should treat the surface z = f(x,y) as a level surface of some function of three variables:
z = f(x,y) can be written as 0 = f(x,y) - z
F(x,y,z) = f(x,y) - z
Notice that Fx = fx, Fy = fy, and Fz = -1
So,
And this will be the case for any function of two variables!

14. Math 200
We can use this to write a general formula for the tangent plane to f(x,y) at (x0,y0):
Solve for z:
Since z0 = f(x0,y0),

15. For example We looked at f(x,y) = 2xy - xy2 at (-1,-1)
Math 200
For example
We looked at f(x,y) = 2xy - xy2 at (-1,-1)
Let’s use this newly derived formula:
fx = 2y - y2; fy = 2x - 2xy
fx(-1,-1) = -3 and fy(-1,-1) = -4
f(-1,-1) = 3
So the plane is
z = -3(x+1) - 4(y+1) + 3
z = -3x - 4y - 4

16. An application Consider two surfaces:
Math 200
An application
Consider two surfaces:
The cylinder y2 + z2 = 5 and the plane y = x - 1
The point (2,1,2) is on the intersection of these two surfaces
Q: How can we find the line tangent to the intersection of these two surfaces?

17. We’ve already done this type of problem!
Math 200
We know how to find the tangent planes to the surfaces at the given point
The line tangent to the intersection is the line of intersection of the two tangent planes
We’ve already done this type of problem!
To get a vector parallel to the intersection of the two planes, we just need to cross their normal vectors
Let’s find the normal vectors to the surfaces at the point A(2,1,2) and cross them.

18. For y2 + z2 = 5, let’s define F(x,y,z) = y2 + z2
Math 200
For y2 + z2 = 5, let’s define F(x,y,z) = y2 + z2
For y = x - 1, let’s write it as x - y = 1 and then define G(x,y,z) = x - y
Cross the normals:
The line tangent to the intersection is

19. Math 200

20. The angle between the planes is the angle between their normal vectors
Math 200
We can also find the angle between the tangent planes to these two surfaces
The angle between the planes is the angle between their normal vectors
The normal vectors are the gradients
If we want the acute angle…

21. A tricky problem Consider the ellipsoid E: 2x2 + y2 + 3z2 = 72
Math 200
A tricky problem
Consider the ellipsoid E: 2x2 + y2 + 3z2 = 72
And also the plane P: 4x+4y-12z=-100
Find all of the points on the ellipsoid where the tangent plane to E is parallel to the plane P

22. Setting the x, y, and z components equal
Math 200
Treat both the plane and the ellipsoid as level surfaces:
F(x,y,z) = 2x2 + y2 + 3z2
G(x,y,z) = 4x+4y-12z
If the tangent planes are parallel (the plane is its own tangent plane of course), then their gradients are scalar multiples:
Setting the x, y, and z components equal

23. Math 200
What we’re looking for is any point on the ellipsoid of the form (k, 2k, -2k)
2x2 + y2 + 3z2 = 72
2(k)2 + (2k)2 + 3(-2k)2 = 72
2k2 + 4k2 + 12k2 = 72
18k2 = 72
k2 = 4
k = -2 or +2
So there are two points on the ellipsoid where the tangent plane is parallel to 4x+4y-12z=-100: (-2, -4, 4) and (2, 4, -4)

24. Math 200
Testing our answers:
Planes:
Or (better yet):