Mth 234 Multivariable Calculus 1

Transcription

1 Mth 234 Multivariable Calculus 1 Cartesian coordinates in space (12.1) Overview of vector calculus. Cartesian coordinates in space. Right-handed, left-handed Cartesian coordinates. Distance formula between two points in space. Equation of a sphere.

2 Mth 234 Multivariable Calculus 2 Overview of Multivariable calculus Mth 132 Calculus I: f : R R, f(), differential calculus. Mth 133 Calculus II: f : R R, f(), integral calculus. Mth 234 Multivariable Calculus: f : R 2 R, f(, ) f : R 3 R, f(,, z) } scalar-valued. r : R R 3, r(t) = (t), (t), z(t) } vector-valued. We stud how to differentiate and integrate such functions.

4 Mth 234 Multivariable Calculus 4 Review: Cartesian coordinates on the plane (R 2 ) Ever point on a plane is labeled b an ordered pair (, ). 0 (, ) 0 0 z z 0 (,,z ) z Cartesian coordinates in space (R 3 ) Ever point in space is labeled b an ordered triple (,, z).

5 Mth 234 Multivariable Calculus 5 Eample: Find the set S = { 0, 0, z = 0} R 3. Solution: z z = 0 > 0 > 0 S

6 Mth 234 Multivariable Calculus 6 Eample: Find the set S R 3 given b S = {0 1, 1 2, z = 1}. Solution: z S 1 1 2

8 Mth 234 Multivariable Calculus 8 There are two tpes of Cartesian coordinate sstems ecept b rotations: Right-handed (RH) and Left-handed (LH) z (,,z ) z (,,z ) z 0 z Right Handed 0 0 Left Handed No rotation transforms one into the other

9 Mth 234 Multivariable Calculus 9 This coordinate sstem is right handed z z z This coordinate sstem is left handed z z z

10 Mth 234 Multivariable Calculus 10 Remark: The same classification occurs in Cartesian coordinates on the plane. Right Handed Left Handed In R 3 we will define the cross product of vectors. This product has different results in RH or LH Cartesian coordinates. There is no cross product in R 2. In class we use RH Cartesian coordinate sstems

12 Mth 234 Multivariable Calculus 12 Distance formula between two points in space Theorem 1 The distance P 1 P 2 between the points P 1 = ( 1, 1, z 1 ) and P 2 = ( 2, 2, z 2 ) is given b P1 P 2 = (2 1 ) 2 + ( 2 1 ) 2 + (z 2 z 1 ) 2. The distance between points in space is crucial to define the idea of limit to functions in space

13 Mth 234 Multivariable Calculus 13 Proof: Pthagoras Theorem. z P 1 P 2 (z z ) 2 1 ( ) 2 1 a ( ) 2 1 P1 P 2 2 = a 2 + (z 2 z 1 ) 2, a 2 = ( 2 1 ) 2 + ( 2 1 ) 2.

14 Mth 234 Multivariable Calculus 14 Eample: Find the distance between P 1 = (1, 2, 3) and P 2 = (3, 2, 1). Solution: P1 P 2 = (3 1)2 + (2 2) 2 + (1 3) 2 = = 8 P 1 P 2 = 2 2.

15 Mth 234 Multivariable Calculus 15 Eample: Use the distance formula to determine whether three points in space are collinear. Solution: d 21 P 2 d 32 P 2 d 32 P 3 P 3 d 21 P 1 d 31 P 1 d 31 d 21 + d 32 > d 31 d 21 + d 32 = d 31 Not collinear, collinear.

17 Mth 234 Multivariable Calculus 17 A sphere is a set of points at fied distance from a center Definition 1 A sphere centered at P 0 = ( 0, 0, z 0 ) of radius R is S = { P = (,, z) : P0 P } = R. z R That is, (,, z) S iff (if and onl if) ( 0 ) 2 + ( 0 ) 2 + (z z 0 ) 2 = R 2.

18 Mth 234 Multivariable Calculus 18 A Ball is a set of points contained in a sphere Definition 2 A ball centered at P 0 = ( 0, 0, z 0 ) of radius R is B = { P = (,, z) : P0 P } < R. That is, (,, z) B iff ( 0 ) 2 + ( 0 ) 2 + (z z 0 ) 2 < R 2.

19 Mth 234 Multivariable Calculus 19 Eample: Plot a sphere centered at P 0 = (0, 0, 0) of radius R > 0. Solution: z R

20 Mth 234 Multivariable Calculus 20 Eample: Plot the sphere z = 0 Solution: Technique: Complete the square. 0 = z 2 ( 4 ( 4 2 ] = [ ) 2) ( = z 2) 2 4. ( 4 2 ) 2 + z z 2 = ( + 2) 2 + z 2 = 2 2.

21 Mth 234 Multivariable Calculus 21 Eample: Plot the sphere z = 0 Since z 2 = ( + 2) 2 + z 2 = 2 2, we conclude that P 0 = (0, 2, 0) and R = 2, therefore, z 2

22 Mth 234 Multivariable Calculus 22 Eercise: Given constants a, b, c, and d R, show that is the equation of a sphere iff z 2 2a 2b 2c z = d d > (a 2 + b 2 + c 2 ). (1) Furthermore, show that if Eq. (1) is satisfied, then the epressions for the center P 0 and the radius R of the sphere are given b P 0 = (a, b, c), R = d + (a 2 + b 2 + c 2 ).

23 Mth 234 Multivariable Calculus 23 Vectors on a plane and in space (12.2) Vectors in R 2 and R 3. Vector components in Cartesian coordinates. Magnitude of a vector and unit vectors. Addition and scalar multiplication.

24 Mth 234 Multivariable Calculus 24 A vector in R 2 or R 3 is an oriented line segment Definition 3 A vector in R n, with n = 2, 3, is an ordered pair of points in R n, denoted as P 1 P 2, where P 1, P 2 R n. The point P 1 is called the initial point and P 2 is called the terminal point. P 1 P P 2 1 P 2 A vector is drawn b an arrow pointing to the terminal point. A vector is denoted not onl b P 1 P 2 but also b an arrow over a letter, like v, or b a boldface letter, like v.

25 Mth 234 Multivariable Calculus 25 The order of the points determines the direction. P P 2 1 P 2 P P 2 1 P 2 P 1 P 1 The vectors P 1 P 2 and P 2 P 1 have opposite directions.

27 Mth 234 Multivariable Calculus 27 Components of a vector in Cartesian coordinates Theorem 2 Given the points P 1 = ( 1, 1 ), P 2 = ( 2, 2 ) R 2, the vector P 1 P 2 determines a unique ordered pair denoted as follows, P 1 P 2 = ( 2 1 ), ( 2 1 ). Proof: Draw the vector P 1 P 2 in Cartesian coordinates. 1 2 P 1 P P 1 2 ( ) 2 1 P 2 ( )

28 Mth 234 Multivariable Calculus 28 A similar result holds for vectors in space. Theorem 3 Given the points P 1 = ( 1, 1, z 1 ), P 2 = ( 2, 2, z 2 ) R 3, the vector P 1 P 2 determines a unique ordered triple denoted as follows, P 1 P 2 = ( 2 1 ), ( 2 1 ), (z 2 z 1 ). Proof: Draw the vector P 1 P 2 in Cartesian coordinates. z P 1 P2 P 1 P 2 ( z z ) 2 1 ( ) 2 1 ( ) 2 1

29 Mth 234 Multivariable Calculus 29 Eample: Find the components of a vector with initial point P 1 = (1, 2, 3) and terminal point P 2 = (3, 1, 2). Solution: P 1 P 2 = (3 1), (1 ( 2)), (2 3) P 1 P 2 = 2, 3, 1. Eample: Find the components of a vector with initial point P 3 = (3, 1, 4) and terminal point P 4 = (5, 4, 3). Solution: P 3 P 4 = (5 3), (4 1), (3 4) P 3 P 4 = 2, 3, 1. P 1 P 2 and P 3 P 4 have the same components although the are different vectors.

30 Mth 234 Multivariable Calculus 30 The vector components do not determine a unique vector. u v v 0 0P v P v The vectors u, v and 0P have the same components but the are all different, since the have different initial and terminal points. Definition 4 Given a vector P 1 P 2 = v, v, the standard position vector is the vector 0P, where 0 = (0, 0) is the origin of the Cartesian coordinates and P = (v, v ). v

31 Mth 234 Multivariable Calculus 31 Vectors are used to describe motion of particles. z a(t) v(t) r(t) r(0) The position r(t), velocit v (t), and acceleration a(t) at the time t of a moving particle are described b vectors in space.

33 Mth 234 Multivariable Calculus 33 The length of a vector is the distance between its initial and terminal points Definition 5 The magnitude or length of a vector P 1 P 2 is the distance from the initial point to the terminal point. If the vector P 1 P 2 has components P 1 P 2 = ( 2 1 ), ( 2 1 ), (z 2 z 1 ), then its magnitude, denoted as P 1 P 2, is given b P 1 P 2 = (2 1 ) 2 + ( 2 1 ) 2 + (z 2 z 1 ) 2. If the vector v has components v = v, v, v z, then its magnitude, denoted as v is given b v = v 2 + v 2 + vz. 2

34 Mth 234 Multivariable Calculus 34 Eample: Find the length of a vector with initial point P 1 = (1, 2, 3) and terminal point P 2 = (4, 3, 2). Solution: First find the component of the vector P 1 P 2, that is, P 1 P 2 = (4 1), (3 2), (2 3) P 1 P 2 = 3, 1, 1. Therefore, its length is P 1 P 2 = ( 1) 2 P 1 P 2 = 11. Eample: If the vector v represents the velocit of a moving particle, then its length v represents the speed of the particle.

35 Mth 234 Multivariable Calculus 35 Unit vectors have length one. Definition 6 A vector v is called a unit vector iff v = Eample: Show that v =,, is a unit vector Solution: v = = v = 1. 14

36 Mth 234 Multivariable Calculus 36 The following vectors are eamples of unit vectors. i = 1, 0, 0, j = 0, 1, 0, k = 0, 0, 1. z k i j These vectors will be ver useful to write an other vector.

38 Mth 234 Multivariable Calculus 38 Vectors can be added and multiplied b scalars. Definition 7 Given the vectors v = v, v, v z, w = w, w, w z in R 3, and a number a R, then the the addition of (v + w) and the scalar multiplication (av ) are given b v + w = (v + w ), (v + w ), (v z + w z ), av = av, av, av z. The vector v = ( 1)v is called the opposite of vector v. Remark: The difference of two vectors is the addition of one vector and the opposite of the other vector, that is, v w = v + ( 1)w. In components we obtain the formula v w = (v w ), (v w ), (v z w z ).

39 Mth 234 Multivariable Calculus 39 The addition of two vectors is equivalent to the parallelogram law V V+W V v (v+w) W w (v+w) w v The vector (v + w) is the diagonal of the parallelogram formed b vectors v and w when the are in their standard position.

40 Mth 234 Multivariable Calculus 40 The addition and difference of two vectors. v v w v+w V a V a>1 a V 0 < a < 1 V a = 1 w The scalar multiplication stretches a vector if a > 1 and compresses the vector if 0 < a < 1,

41 Mth 234 Multivariable Calculus 41 Eample: Given the vectors v = 2, 3 and w = 1, 2, find the magnitude of the vectors v + w and v w. Solution: We first compute the components of v + w, that is, v + w = (2 1), (3 + 2) v + w = 1, 5. Therefore, its magnitude is v + w = v + w = 26. A similar calculation can be done for v w, that is, v w = (2 + 1), (3 2) v w = 3, 1. Therefore, its magnitude is v w = v w = 10.

42 Mth 234 Multivariable Calculus 42 Unit vectors along an direction are simple to obtain. Theorem 4 If the vector v 0, then the vector v v along the direction given b v. is a unit vector Proof: We show the case where v R 2. If we write v = v, v, then v = v 2 + v. 2 Therefore, we obtain v v = 1 v v v, v = v, v. v This is a unit vector, since v [ v ] 2 [ v ] 2 1 = + = v v v v v 2 + v 2 = v v Just divide the vector b its own length. v = 1. v

43 Mth 234 Multivariable Calculus 43 The vectors i, j, k are useful to epress an other vector Recall: i = 1, 0, 0, j = 0, 1, 0, k = 0, 0, 1. Theorem 5 Ever vector v = v, v, v z in R 3 can be epressed in a unique wa as a linear combination of vectors i, j, k as follows v = v i + v j + v z k. z v k i j

44 Mth 234 Multivariable Calculus 44 The vectors i, j, k are useful to epress an other vector Proof: Use the definitions of vector addition and scalar multiplication as follows, v = v, v, v z = v, 0, 0 + 0, v, 0 + 0, 0, v z = v 1, 0, 0 + v 0, 1, 0 + v z 0, 0, 1 = v i + v j + v z k. z v k i j

45 Mth 234 Multivariable Calculus 45 Eample: Epress the vector with initial and terminal points P 1 = (1, 0, 3), P 2 = ( 1, 4, 5) in the form v = v i + v j + v z k. Solution: We first compute the components of the vector v = P 1 P 2, that is, v = ( 1 1), (4 0), (5 3) = 2, 4, 2. We therefore conclude that v = 2i + 4j + 2k.

46 Mth 234 Multivariable Calculus 46 Eample: Epress the vector with initial and terminal points P 1 = (1, 0, 3), P 2 = ( 1, 4, 5) in the form v = v i + v j + v z k. Solution: We first compute the components of the vector v = P 1 P 2, that is, v = ( 1 1), (4 0), (5 3) = 2, 4, 2. We therefore conclude that v = 2i + 4j + 2k. Eample: Find a unit vector w opposite to v found above. Solution: Since the length of v is given b v = ( 2) = = 24, we conclude that the solution vector w is given b w = , 4, 2.