while its direction is given by the right hand rule: point fingers of the right hand in a 1 a 2 a 3 b 1 b 2 b 3 A B = det i j k

Transcription

1 I.f Tangent Planes and Normal Lines Again we begin by: Recall: (1) Given two vectors A = a 1 i + a 2 j + a 3 k, B = b 1 i + b 2 j + b 3 k then A B is a vector perpendicular to both A and B. Then length of A B is A B sin θ, while its direction is given by the right hand rule: point fingers of the right hand in the direction of A, curl them in the direction of B. Then the thumb points in the direction of A B. To compute A B, it is convenient to use the formula: A B = det i j k a 1 a 2 a 3. b 1 b 2 b 3 A B B A Observe that A B = (B A), while if A and B are parallel, A B = 0. (2) A plane is determined by a vector N = αi + βj + γk (called the normal vector), and a point P 0 = (x 0, y 0, z 0 ) on the plane. Any other point P = (x, y, z) of space is then on the plane iff the vector P 0 P is perpendicular to N, i.e., P 0 P N = 0, or (x x 0 )α + (y y 0 )β + (z z 0 )γ = 0. 50

2 N P0 P (3) Instead of a vector N and a point P 0, we may specify three points of the plane P 0, P 1, P 2 (not collinear). N P 2 P 0 P 1 We then construct the vectors P 0 P 1, P 0 P 2 and set N = P 0 P 1 P 0 P 2. For planes it does matter whether we choose N or N as the normal! (4) Curves in space are given in parametric form as: x = x(t) y = y(t) z = z(t) a t b. This is a very convenient way for us! Remember, once again, that z = f(x, y) gives a surface, not a curve. Suppose the curve C is given in the above form. Then 51

3 r(t 0 ) = x(t 0 )i + y(t 0 )j + z(t 0 )k is the vector from the origin (0, 0, 0) to the point on C at t = t 0. We recall that dr/dt = (dx/dt)i + (dy/dt)j + (dz/dt)k gives a tangent vector to the curve C. z d r d t C r ( t ) y x Suppose now we are given a surface in (x, y, z) space. It is convenient for us to write the surface S in the form F (x, y, z) = 0, so if we had the equation z = f(x, y), then we would get F (x, y, z) = z f(x, y). Let P 0 = (x 0, y 0, z 0 ) be a point of S. Then the Tangent Plane T of S at P 0 is the plane which: (1) contains P 0 and (2) if C is any curve on S through P 0, then the tangent vector to C at P 0 is on the plane. S P 0 C 1 C3 C 2 The tangent plane plays the same role for surfaces that the tangent line played for y = f(x) in the (x, y) plane. We obtain the equation of the tangent plane T at P 0 as follows: We know one point of T (namely P 0 ), so we need only find the normal vector N. To do this, let C be any given curve through P 0 such that C lies on S. 52

4 Then C is given by x = x(t) y = y(t) z = z(t) a t b for some functions x(t), y(t), z(t). On the other hand, since C is on the surface S, we must have F (x(t), y(t), z(t)) = 0 for all t! I.e., differentiating both sides with respect to t, we obtain F x dx dt + F y dy dt + F z dz dt = 0. F P 0 d r d t C We know that dr/dt = (dx/dt)i + (dy/dt)j + (dz/dt)k is tangent to C at P 0, and thus F = ( F/ x)i + ( F/ y)j + ( F/ z)k is perpendicular to dr/dt! But C was not specified! So F is perpendicular to the tangent of any curve, and we conclude that we may take F as the normal to the tangent plane! In summary the equation of the tangent plane to F (x, y, z) = 0 at P 0 = (x 0, y 0, z 0 ) is F x (x 0, y 0, z 0 )(x x 0 ) + F y (x 0, y 0, z 0 )(y y 0 ) + F z (x 0, y 0, z 0 )(z z 0 ) = 0. By definition, the line through P 0 in the direction of F (x 0, y 0, z 0 ) is called 53

5 the Normal Line. Its equation is x = x 0 + t F x (x 0, y 0, z 0 ) y = y 0 + t F y (x 0, y 0, z 0 ) z = z 0 + t F z (x 0, y 0, z 0 ) < t <. Finally, two surfaces are perpendicular (or orthogonal) at a point of intersection iff the normal vectors N 1, N 2 of their respective tangent planes are perpendicular: N 1 N 2 = 0. Remark. Observe that F is also perpendicular to the equipotential surfaces F (x, y, z) = c, with c = constant, by exactly the same calculations. Earlier we showed that F gave the direction of maximal rate of increase for the function F (x, y, z). We now also know that F is perpendicular to the surface F (x, y, z) = 0. F F = c We thus have that the maximal rate of increase of F (x, y, z) can be obtained by going in a perpendicular direction to the equipotential surfaces: F (x, y, z) = c. We illustrate the above results with the following examples. Example 1. Find the directional derivative of f = x sin(e z ) + ye x at the point (0, 1, 1) 54

6 in the direction of the vector v = i j + k. Answer. f = (sin(e z ) + ye x )i + e x j + x cos(e z )e z k, so f = (sin e + 1)i + j + 0k (0,1,1) while a unit vector u in the direction of v is u = i j + k Finally, df dx = f u = v (sin e + 1) 1 3 = sin(e) 3. Example 2. Let the electrical potential V be given by V = 5x 2 + 3y 3 z + xz 4. You are at the point (1, 1, 0). (a) In which direction should you head for the maximal rate of change of V with respect to distance? What is the maximal rate? (b) Do the same for the minimal (i.e., most negative) rate. Answer. V = (10x + z 4 )i + 9y 2 zj + (3y 3 + 4z 3 x)k. Therefore at (1, 1, 0), V = 10i + 0j + 3k. (1,1,0) (a) We have the direction of maximal rate of change = direction of V = 10i+3k. The rate = V = 109. (b) The direction of the most negative rate of change = direction of V = 10i 3k. This rate = V =

7 Example 3. Let (u, v) be differentiable functions. Show: (uv) = v u + u v. Answer. (uv) = x (uv)i + y (uv)j + z (uv)k [ u = x v + u v ] [ u i + x y v + u v y [ u = v x i + u y j + u ] z k + u = v u + u v. ] [ u j + z v + u v z ] [ v x i + v y j + v z k ] i Example 4. Find the tangent plane and normal line to the surface z = x 2 + y 2 at the point x = 1, y = 1. Answer. Write the surface as F (x, y, z) = 0. Here F (x, y, z) = x 2 + y 2 z. Then the normal vector to the tangent plane is F = 2xi + 2yj k = 2i 2j k. (1, 1,2) So the equation of the tangent plane is (2i 2j k) ((x 1)i + (y ( 1))j + (z 2)k) = 0 or 2(x 1) 2(y + 1) (z 2) = 0, while the normal line is: x = 1 + 2t y = 1 2t < t <. z = 2 t 56

8 Example 5. Find the point(s) on the sphere z 2 + x 2 + y 2 = 1 where the tangent plane is parallel to the plane x + y + z = 17. Answer. The normal vector N 1, to the given plane is N 1 = i + j + k. We now calculate the normal N 2 to the sphere x 2 + y 2 + z 2 = 1. Here F (x, y, z) = x 2 + y 2 + z 2 1 and so N 2 = F = 2xi + 2yj + 2zk. We wish N 2 and N 1 to be parallel, i.e., N 1 N 2 = 0 or det i j k = 0. 2x 2y 2z Equivalently, i[2z 2y] j[2z 2x] + k[2y 2x] = 0. We have z = y, z = x, y = x. Note that only two of these three equations are useful, since they imply the third. We choose: y = z and x = z. Then, we keep in mind that (x, y, z) is on the sphere! I.e., x 2 + y 2 + z 2 = 1, and so we must have z 2 + z 2 + z 2 = 1 or z 2 = 1 3. We thus find z = 1/ 3, z = 1/ 3 and the two points on the sphere are (1/ 3, 1/ 3, 1/ 3) and ( 1/ 3, 1/ 3, 1/ 3). 57

9 Example 6. Find parametric equations for the tangent line to the curve of intersection of the paraboloid z = 4(x 2 + y 2 ) and the ellipsoid x 2 + 4y 2 + z 2 = 21 at the point (1, 1, 4). Answer. Note that the intersection curve(s) satisfy x 2 + (z 4x 2 ) + z 2 = 21 or 3x 2 + z 2 + z = 21 and 4y 2 = 21 z 2 x 2 = 21 z z2 z. 3 These last two equations give x and y as functions of z by taking square roots. Clearly these equations imply that the intersection is complicated! Proceeding directly does not seem too wise. Instead, note that the curve of intersection must lie on both surfaces simultaneously. It follows that its tangent line must have a direction vector V which is perpendicular to the normals N 1 and N 2 of the two surfaces. We may therefore take V = N 1 N 2. Now z = 4(x 2 + y 2 ) can be written as 4(x 2 + y 2 ) z = 0. Its normal N 1 is thus N 1 = 8xi + 8yj 1k and N 1 (1,1,4) = 8i + 8j k. 58

10 In the same way, the ellipsoid x 2 + 4y 2 + z 2 = 21 has normal N 2 given by N 2 = 2xi + 8yj + 2zk and N 2 (1,1,4) = 2i + 8j + 8k. We then have V = det i j k = 72i 66j + 48k and the equation of the tangent line becomes: x = t y = 1 66t < t <. z = t 59

11 Further Exercises: 1) Find the tangent plane and normal line to z 2 = x 2 + y 2 at (1, 0, 1). 2) Find the tangent plane and normal line to x 2 + y2 4 + z2 9 = 1 at (1, 0, 0). 3) Find the tangent plane and normal line to the cylinder x 2 +y 2 = 8 at (2, 2, 1). 4) Find all points of the surface x = y 2 + z 2 where the normal line intersects the plane x + y + z = 1 at right angles. Find the points of the plane where the normal lines intersect the plane. 5) Find all points of the surface z = 4x 2 +9y 2 where the tangent plane is parallel to the plane 16x + 18y + 2z = 1. 6) Find a tangent vector at the point (1, 0, 1) to the curve of intersection of the surfaces z 2 = x 2 + y 2 and z = x + 4y. 7) Show that the normal line at any point on the surface (x 1) 2 + y 2 + z 2 = 1 passes through the point (1, 0, 0). 8) Show that the normal line to the surface z = x 2 + y 2 at the point (x 0, y 0, z 0 ) must contain the point (0, 0, 1 + z 0 ). 60