Equation of tangent plane: for explicitly defined surfaces Suppose that the surface z = f(x,y) has a non-vertical tangent plane at a point (a, b, f(a,b)). The plane y = b intersects the surface at a curve C. The plane y = b intersects the tangent plane at a line, and this line is tangent to the curve C at the point (a, b, f(a,b)). Call this line L 1. The slope of the line L 1 with respect to the system xz is equal to f x (a,b). So, L 1 is parallel to the vector T 1 = i + f x (a,b)k. Similarly, the tangent plane intersects the plane x = a in a line, and this line has the slope f y (a,b) and is parallel to the vector T 2 = j + f y (a,b)k. The lines L 1 and L 2 are both on the tangent plane, therefore the vectors T 1 and T 2 are parallel to the tangen plane, therefore the cross product T 1 T 2 can be taken as a normal vector of the tangent plane. But: i j k 0 1 f x (a,b) 1 0 f y (a,b) = f x (a,b), f y (a,b), 1 Then the equation of tangent plane is: f x (a,b)(x a) + f y (a,b)(y b) (z f(a,b)) = 0 tangent plane at (a,b,f(a,b)) z = f(a,b) + f x (a,b)(x a) + f y (a,b)(y b) The line through the point (a, b, f(a,b)) with the vector f x (a,b), f y (a,b), 1 as its direction vector is perpendicular to the plane at the point (a, b, f(a,b)), because this direction vector of the line is perpendicular to the plane. The equation of this normal line to the surface is: normal line at (a,b,f(a,b)) x a f x (a,b) = y b f y (a,b) = z f(a,b) 1 Example. Find the equation of the tangent plane and normal line to the surface z = e x2 y 2 at the point 1
(1, 1, 1). z x = 2xe x2 y 2 z y = 2ye x2 y 2 z x (1, 1) = 2 z y (1, 1) = 2 Then the normal vector to the tangent plane is the vector 2, 2, 1. The equation of the tangent plane will be: 2(x 1) + 2(y + 1) (z 1) = 0 z = 1 + 2x + 2y The equation of the normal line will be: x 1 2 = y + 1 2 = z 1 1 2
Equation of tangent plane: for implicitly defined surfaces Some surfaces are defined implicitly, such as the sphere x 2 + y 2 + z 2 = 1. In general an implicitly defined surface has the equation F(x,y,z) = k where k is a fixed number. Supposing that z is a function of (x,y) around the point (a,b,c), the equation of the tangent line will be: z x (a,b)(x a) + z y (a,b)(y b) (z c) = 0 But as we have seen before, for implicitly defined functions we have: z x = F x F z z y = F y F z cmat the point (a,b,c) Substituting these into the equation of the tangent plane we get: F x(a,b,c) F z (a,b,c) (x a) F y(a,b,c) (y b) (z c) = 0 F z (a,b,c) By multiplying both sides by F z (a,b,c) we get tangent plane at (a,b,c) F x (a,b,c)(x a) + F y (a,b,c)(y b) + F z (a,b,c)(z c) = 0 This equation shows that the vector F z, F y, F z is the normal vector of the tqngent plane. This vector is then parallel to the line that is perpendicular to the surface at (a,b,c). So, the equation of the normal line to the plane at (a,b,c) is: normal line to the surface at (a,b,c) x a F x (a,b,c) = y b F y (a,b,c) = z c F z (a,b,c) Note. If f(x,y) = k is the equation of a curve in the plane xy, then similarly one can show that the equation of the tangent line at (a,b) is: 3
f x (a,b)(x a) + f y (a,b)(y b) = 0 Example. Find the equation of the the tangent line to the ellipse x 2 + 2y 2 = 6 at the point (2,1). Solution. Set f(x,y) = x 2 + 2y 2. Then: f x = 2x f y = 4y f x (2,1) = 4 f y (2,1) = 4 Then the equation of the tangent plane will be: 4(x 2) + 4(y 1) = 0 x + y = 3 Example (from the textbook). Find the equation of the the tangent line to the ellipsoid x 2 4 + y2 + z2 9 = 3 at the point ( 2,1, 3). Solution. Set F(x,y,z) = x2 4 + y2 + z2 9. So, the surface is implicitly written as F(x,y,z) = 3 F x = x 2 F y = 2y F z = 2z 9 These derivatives at the point ( 2,1, 3) become: F x = 1 F y = 2 F z = 2 3 Then the equation of the tangent plane: (x + 2) + 2(y 1) 2 (z + 3) = 0 3x 6y + 2z + 18 = 0 3 And the equation of the normal line: 4
x + 2 1 = y 1 = z + 3 2 2 3 Example. The ellipsoid 4x 2 + 2y 2 + z 2 = 15 intersects the plane y = 2 at an ellipse. Find the parametric equations of the tangent line to the ellipse at the point (1,2,2). Solution. F x = 8x F x (1,2,2) = 8 F y = 4y F z = 2z F y (1,2,2) = 8 F z (1,2,2) = 4 Then the equation of the tangent plane: 8(x 1) + 8(y 2) + 4(z 2) = 0 2(x 1) + 2(y 2) + (z 2) = 0 We then set y = 2 in the equation: 2(x 1) + (z 2) = 0 z = 4 2x x = x y = 2 z = 4 2x < x < We note further that the vector 1, 0, 2 is the direction vector of this line. Example (section 14.6 exercise 54). At what points of the paraboloid y = x 2 + y 2 the tangent plane is parallel to plane x + 2y + 3z = 1? Solution. Suppose that the point we are looking for is the point (a,b,c). We write the surface in the implicit form F(x,y,z) = x 2 + y 2 y = 0. Then: 5
F x = 2x F y = 1 F z = 2z F x (a,b,c) = 2a F y (a,b,c) = 1 F z (a,b,c) = 2c The gradient vector F(a,b,c) is perpendicular to the tangent plane at (a,b,c), therefore it must be parallel to the vector 1,2,3, therefore it must satisfy: 2a 1 = 1 2 = 2c 3 4a = 1 4c = 3 a = 1 4 c = 3 4 As the point (a,b,c) must be on the surface, we have: b = a 2 + c 2 b = 1 16 + 9 16 = 5 8 So (a,b,c) = ( 1 4, 5 8, 3 4 ) Example (section 14.6 exercise 56). Show that the ellipsoid 3x 2 + 2y 2 + z 2 = 9 and the sphere x 2 + y 2 + z 2 8x 6y 8z + 24 = 0 are tangent to each other at the point (1,1,2). (This means that they have a common tangent plane at the point.) Solution. We consider the surfaces in the implicit form: F(x,y,z) = 3x 2 + 2y 2 + z 2 = 9 G(x,y,z) = x 2 + y 2 + z 2 8x 6y 8z + 24 = 0 We must show that the surfaces have a common tangent plane at the point (1,1,2). Equivalently, we show that the vectors F(1,1,2) and G(1,1,2) are parallel (because these two vectors are the normal vectors of those planes). 6
F x = 6x F y = 4y F z = 2z F x (1,1,2) = 6 F y (1,1,2) = 4 F z (1,1,2) = 4 G x = 2x 8 G x (1,1,2) = 6 G y = 2y 6 G z = 2z 8 G y (1,1,2) = 4 G z (1,1,2) = 4 F(1,1,2) = 6, 4, 4 So we have G(1,1,2) = 6, 4, 4 and obviously these two vectors are parallel. Example (section 14.6 exercise 63). Find parametric equations for the tangent line to the curve of intersection of the paraboloid z = x 2 + y 2 and the ellipsoid 4x 2 + y 2 + z 2 = 9 at the point ( 1, 1, 2). Solution. We write the two surfaces in the implicit form: F(x,y,z) = x 2 + y 2 z = 0 G(x,y,z) = 4x 2 + y 2 + z 2 = 9 The tangent line we are looking for in the intersection of the tangent planes of the two surfaces. The vectors F( 1, 1, 2) and G( 1, 1, 2) are perpendicular to the surfaces at the common point ( 1, 1, 2) of the two surfaces. Therefore, the vector F( 1, 1, 2) G( 1, 1, 2) is parallel to the tangent line, so we will use this vector to write the equation of the tangent line. F x = 2x F x ( 1,1,2) = 2 F y = 2y F z = 1 F y ( 1,1,2) = 2 F z ( 1,1,2) = 1 G x = 8x G y = 2y G z = 2z G x ( 1,1,2) = 8 G y ( 1,1,2) = 2 G z ( 1,1,2) = 4 7
F( 1, 1, 2) G( 1, 1, 2) = i j k 2 2 1 8 2 4 = 10, 16, 12 The vector 10, 16, 12 is the direction vector of the tangent line. Therefore its parametric equations are: x = 5t 1 y = 8t + 1 z = 6t + 2 < t < 8
Equation of tangent plane: for the surfaces that are parametrized by two parameters We recall that some surfaces are described in parametric form and using two parameters. For example, as we have seen before, the plane that passes through the point (x 0,y 0,z 0 ) and is parallel to two vectors (or contains two vectors) a = a1, a 2, a 3 b = b1, b 2, b 3 can be described by parametric equations: x = x 0 + a 1 t + b 1 s S : y = y 0 + a 2 t + b 2 s z = z 0 + a 3 t + b 3 s For example, the parametric equations of the plane through (1, 1,2) parallel containing the vectors a = 1, 2, 3 and b = 4, 5, 6 is S : x = 1 + t + 4s y = 1 + 2t + 5s z = 2 + 3t + 6s Example. The parametric representation and the vector representation of the sphere x 2 + y 2 + z 2 = 4 is: S : x = 2sinφ cosθ y = 2sinφ sinθ z = 2cosφ 0 φ π 0 θ 2π r(θ, φ) = 2sinφ cosθ, 2sinφ sinθ, 2cosφ Example. In general, The parametric representation and the vector representation of the sphere x 2 + y 2 + z 2 = a 2 is: 9
S : x = asinφ cosθ y = asinφ sinθ z = acosφ 0 φ π 0 θ 2π r (θ, φ) = asinφ cosθ, asinφ sinθ, acosφ x 2 + y 2 = 4 Example. Find the parametric representation of the cylinder 0 z 1 Solution. x = 2cosθ y = 2sinθ 0 θ 2π z = z 0 z 1 Important example. If a surface is described explicitly by z = f(x,y) (for example, the paraboloid z = x 2 + y 2 ), then we can parametrize this surface by: parametric form x = x y = y z = f(x,y) 0 z 1 vector form r (x,y) = x, y, f(x,y) as an example, the surface z = 2 x 2 + y 2 (the upper half of a cone) is parametrized as: r(x,y) = x, y, 2 x 2 + y 2 Suppose that a surface S is parametrize as r(u,v) = x(u,v), y(u,v), z(u,v) Consider a particular point P(x(u 0,v 0 ), y(u 0,v 0 ), z(u 0,v 0 )) on the surface. The expression u = u 0 is the equation of the plane perpendicular to the u-axis all of whose points have the first component equal 10
to u 0. The intersection of this plane with the surface S is a curve on the surface. For a moment we call it the curve C 1. The vector representation for this curve is r(u 0,v) = x(u 0,v), y(u 0,v), z(u 0,v) As we learned in chapter 13, the following vector is tangent to the curve C 1 at the point, and therefore is on the tangent plane of the surface at the point P: r v (u 0,v 0 ) = x v (u 0,v 0 ), y v (u 0,v 0 ), z v (u 0,v 0 ) Now consider the plane v = v 0. This plane intersects the surface at a curve whose vector representation is: r(u,v 0 ) = x(u,v 0 ), y(u,v 0 ), z(u,v 0 ) We call this curve C 2. The following vector is tangent to C 2 at the point P and therefore is on the tangent plane of the surface at the point P: r u (u 0,v 0 ) = x u (u 0,v 0 ), y u (u 0,v 0 ), z u (u 0,v 0 ) Based on this discussion, the two vectors r u (u 0,v 0 ) and r v (u 0,v 0 ) are on the tangent plane, therefore their cross product is perpendicular to the surface at the point P and it can taken as the normal vector of the tangent plane. See the example below for an application of it. Example. Consider the surface parametrized by: S : x = u 2 y = v 2 z = u + 2v find the equation of the tangent plane at the point (1,1,3). 11
Solution. r = u 2, v 2, u + 2v r u = 2u, 0, 1 r v = 0, 2v, 2 The point P(1,1,3) corresponds to u = 1 and v = 1, and for these values of the parameters u and v the vectors r u and r v become: r u = 2, 0, 1 r v = 0, 2, 2 Then: r u r v = i j k 2 0 1 0 2 2 = 2, 4, 4 Then here is the equation of the tangent plane: 2(x 1) 4(y 1) + 4(z 3) = 0 x + 2y 2z + 3 = 0 Example. Using the parametric equations of the sphere x 2 + y 2 + z 2 = 4 x = 2sinφ cosθ 0 φ π y = 2sinφ sinθ 0 θ 2π z = 2cosφ find the equation of the tangent plane to the an arbitrary point of it. Suggested questions for this topic. The following exercises are being suggested for this lecture: section 14.4 exercises 1, 3, 5 section 14.6 exercises 41, 45, 49, 55, 57, 60 12
section 16.6 exercises 33, 35 13