TANGENT PLANESMATH 200 WEEK 6 - MONDAY

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.

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

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

MATH 200

▸ We’ll use the gradient vector as our normal vector for the tangent plane…

���F (x, y, z) = ��2x, 2y, �1� ���F (1, 2, 3) = ��2, 4, �1�

�2(x � 1) + 4(y � 2) � (z � 3) = 0

�2x + 4y � z = 3

MATH 200

YOU TRY▸ Find the tangent plane to the ellipsoid E: x2 + y2 + 2z2 = 4

at A(1,-1,1)

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���F (x, y, z) = �2x, 2y, 4z����F (1, �1, 1) = �2, �2, 4�

▸ Tangent Plane:

▸ 2(x-1) - 2(y+1) + 4(z-1) = 0 or 2x - 2y + 4z = 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

MATH 200

▸ E.g. For S: z = y2 - x2 at A(1,2,3) we had

���F (x, y, z) = ��2x, 2y, �1�

���F (1, 2, 3) = ��2, 4, �1�

F (x, y, z) = y2 � x2 � z

▸ So, using <-2,4,-1> as the direction vector for the normal line…

l :

⎧⎪⎨

⎪⎩

x = 1− 2t

y = 2 + 4t

z = 3− t

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)

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 (x, y, z) = ⟨2y − y2, 2x− 2xy,−1⟩

▸ f(-1,-1) = 3 so the point we care about it (-1,-1,3)

−→∇F (−1,−1, 3) = ⟨−3,−4,−1⟩

▸ Plane:

▸ Normal line:

−3(x+ 1)− 4(y + 1)− (z − 3) = 0

3x+ 4y + z = −4

l :

⎧⎪⎨

⎪⎩

x = −1− 3t

y = −1− 4t

z = 3− t

MATH 200

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!

−→∇F = ⟨fx, fy,−1⟩

MATH 200

▸ We can use this to write a general formula for the tangent plane to f(x,y) at (x0,y0):

fx(x0, y0)(x− x0) + fy(x0, y0)(y − y0)− (z − z0) = 0

z = fx(x0, y0)(x− x0) + fy(x0, y0)(y − y0) + z0

▸ Solve for z:

▸ Since z0 = f(x0,y0),

z = fx(x0, y0)(x− x0) + fy(x0, y0)(y − y0) + f(x0, y0)

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

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?

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.

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

−→∇F = ⟨0, 2y, 2z⟩

−→∇G = ⟨1,−1, 0⟩

−→∇F (2, 1, 2) = ⟨0, 2, 4⟩

−→∇G(2, 1, 2) = ⟨1,−1, 0⟩

▸ Cross the normals:

▸ The line tangent to the intersection is

�v = �0, 2, 4� � �1, �1, 0�

=

������

i j k0 2 41 �1 0

������

= �4, 4, �2�

l :

���

��

x = 2 + 4t

y = 1 + 4t

z = 2 � 2t

MATH 200

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

� = arccos

��0, 2, 4� · �1, �1, 0�

||�0, 2, 4�|| ||�1, �1, 0�||

�

= arccos

��2�20

�2

�

= arccos

��2�40

�

= arccos

��2

2�

10

�

= arccos

��1�10

�

▸ If we want the acute angle…

�acute = � � arccos

��1�10

�

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

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:

���

��

4x = 4k

2y = 4k

6z = �12k

���F = k���G

�4x, 2y, 6z� = k�4, 4, �12�

���

��

x = k

y = 2k

z = �2kSETTING THE X, Y, AND Z

COMPONENTS EQUAL

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)

MATH 200

▸ Testing our answers:���F (x, y, z)) = �4x, 2y, 6z�

���F (�2, �4, 4) = ��8, �8, 24����F (2, 4, �4) = �8, 8, �24�

▸ Planes:

▸ Or (better yet):

P1 : x + y � 3z = 18

P2 : x + y � 3z = �18

P1 : �8(x + 2) � 8(y + 4) + 24(z � 4) = 0

P2 : 8(x � 2) + 8(y � 4) � 24(z + 4) = 0