Math 200 Week 6 - Monday Tangent Planes

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…

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

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

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

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…

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)

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:

Math 200

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!

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),

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

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?

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.

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

Math 200

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…

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

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

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: Planes: Or (better yet):