math 200 week 6 - monday tangent planesdp399/math200/slides/... · 2018-05-09 · math 200 goals be...
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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
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