Sec. 1.1: Basics of Vectors

Transcription

1 Sec. 1.1: Basics of Vectors Notation for Euclidean space R n : all points (x 1, x 2,..., x n ) in n-dimensional space. Examples: 1. R 1 : all points on the real number line. 2. R 2 : all points (x 1, x 2 ) or (x, y) in the plane. 3. R 3 : all points (x 1, x 2, x 3 ) or (x, y, z) in space. Notation for vectors Books: v (bold symbol). Handwritten: v (underlined) or v (right arrow). NEVER JUST v which is the scalar (number) v.

2 Sec. 1.1: Basics of Vectors Definition of vector: line segments having a direction and magnitude. Note: Line segments obtained by parallel translation represent the same vector. Standard basis vectors 1. In R 2 : i = (1, 0) and j = (0, 1). 2. In R 3 : i = (1, 0, 0), j = (0, 1, 0), and k = (0, 0, 1). Note: vector notation for i, j, and k. Alternative notation: î, ĵ, and ˆk. Notation for vectors in component form (Example in R 3 ) In the book: v = (1, 2, 3) or v = i + 2j + 3k. Alternative: v = 1, 2, 3.

3 Sec. 1.1: Basics of Vectors Scalar multiplication: Multiply each component α a = α(a 1, a 2,..., a n ) = (αa 1, αa 2,..., αa n ) a 2a Example: 2(1, 2, 3) = (2, 4, 6) 1 2 a Vector addition: Add each component a + b = (a 1, a 2,..., a n ) + (b 1, b 2,..., b n ) = (a 1 + b 1, a 2 + b 2,..., a n + b n ) b a+b b translated Example: (1, 2, 3) + (4, 5, 6) = (5, 7, 9) a

4 Sec. 1.1: Basics of Vectors Vector subtraction: Subtract each component b a b = (a 1, a 2,..., a n ) (b 1, b 2,..., b n ) = (a 1 b 1, a 2 b 2,..., a n b n ) a a b (translated) Example: (1, 2, 3) (4, 5, 6) = ( 3, 3, 3) b a+( b) b translated Vectors between two points (a 1 b 1, a 2 b 2, a 3 b 3 ) is the vector from point (b 1, b 2, b 3 ) to (a 1, a 2, a 3 ) Example: Compute the vector v from point (1, 2, 3) to (6, 5, 4). v = (6, 5, 4) (1, 2, 3) = (5, 3, 1)

5 Sec. 1.1: Equation of a Line Vector equation of a line: x(t) = x 0 + tv Needed: a point x 0 and a direction vector v < t < Example: Equation of the line through (4, 5, 6) in the direction of (1, 2, 3) Point x 0 = (4, 5, 6) and direction v = (1, 2, 3) Using the formula: x(t) = x 0 + tv = (4, 5, 6) + t(1, 2, 3) Geometrical interpretation Any point on the line can be reached when you start at a position x 0 on the line and go in the direction of the direction vector v of the line

6 Sec. 1.2: Inner Product Definition: inner product in R n (dot product) a b = (a 1, a 2,..., a n ) (b 1, b 2,..., b n ) = a 1 b 1 + a 2 b a n b n Example: (1, 2, 3) (4, 5, 6) = (1)(4) + (2)(5) + (3)(6) = 32. Definition: length of a vector a = a a = a a a2 n. Example: (1, 2, 3) = = 14. Definition of unit vector: vector with length 1. Normalization: Find unit vector u = v v in direction of a vector v 0. Example: Unit vector u in direction of vector v = (1, 2, 3). u = (1, 2, 3) (1, 2, 3) = 1 14 (1, 2, 3)

7 Sec. 1.2: Inner Product Angle between two vectors: a b = a b cos θ with θ [0, π] Example: Angle between a = (0, 1, 1) and b = (0, 2, 0). (0, 1, 1) (0, 2, 0) = ( 1) cos θ cos θ = 0(0) + ( 1)2 + 1(0) 2 2 = 1 2 = θ = 3π/4 Perpendicular (orthogonal) Two vectors a and b are perpendicular if a b = 0. Example: (1, 1) and (1, 1) are perpendicular since (1, 1) (1, 1) = 0. Work: W = F d Force times displacement Example: Find the work done by the force field F = (1, 1, 3) that moves an object along a straight line from the point (1, 0, 2) to the point (6, 2, 4). W = F d = (1, 1, 3) (6 1, 2 0, 4 + 2) = 21

8 Sec. 1.3: Cross Product Definition: cross product in R 3 (outer product) i j k a b = a 1 a 2 a 3 b 1 b 2 b 3 = i a 2 a 3 b 2 b 3 j a 1 a 3 b 1 b 3 + k a 1 a 2 b 1 b 2 Example: Cross product of a = (1, 2, 3) and b = (4, 5, 6): i j k a b = = i j k = ( )i ( )j + ( )k = ( 3, 6, 3) Geometrical interpretation: a b is perpendicular to both a and b a b is oriented according to the right-hand rule Area of parallelogram: A = a b Example: Area of parallelogram spanned by a = (1, 2, 3) and b = (4, 5, 6). A = ( 3, 6, 3) = ( 3) ( 3) 2

9 Sec. 1.3: Equation of a Plane Equation of a plane: n [x x 0 ] = 0 Needed: a point x 0 and a vector n normal to the plane Example: Plane through (4, 5, 6) and perpendicular to the vector (1, 2, 3) Point x 0 = (4, 5, 6) and normal direction n = (1, 2, 3) Using the formula: (1, 2, 3) [(x, y, z) (4, 5, 6)] = 0 Evaluating the dot product: 1(x 4) + 2(y 5) + 3(z 6) = 0 Geometrical interpretation Normal vector n is perpendicular to any vector in the plane, x x 0 Thus the dot product is zero: n (x x 0 ) = 0

10 Sec. 2.3: Partial Derivatives Notation for partial derivatives of f = f(x 1,..., x n ) 1. f x, f y, etc. 2. f x 1, f x 2, etc. 3. f x, f y, etc. Notation for regular derivatives of f = f(x) 1. df dx 2. f

11 Sec. 2.3: Partial Derivatives Computing partial derivatives of f = f(x 1,..., x n ) f x 1 : derivative of f w.r.t. x 1, keeping other variables x 2,..., x n constant etc. Example: f(x, y) = x y + y f x = 1 y f y = x y Gradient vector of f = f(x 1,..., x n ): f = Example: f(x, y) = x y + y ( f x 1,..., f x n ) f = ( f x, f ) y = ( 1 y, x ) y

12 Sec. 2.4: Paths and Curves For a path c(t) = (x(t), y(t), z(t)) that is sufficiently smooth Velocity: v(t) = c (t) Speed: v(t) = c (t) Acceleration: a(t) = c (t) Newton s 2nd law: F = ma Tangent vector: c (t) Tangent line: l(t) = c(t 0 ) + t c (t 0 ) For c (t) 0 For c (t 0 ) 0 Example: Path c(t) = (t 3, t 2, t). Velocity: c (t) = (3t 2, 2t, 1) Acceleration: c (t) = (6t, 2, 0) Speed: c (t) = (3t 2 ) 2 + (2t) Tangent vector at (8, 4, 2): c(t 0 ) = (t 3 0, t2 0, t 0) = (8, 4, 2) gives t 0 = 2. Tangent vector: c (2) = (12, 4, 1) Tangent line at (8, 4, 2): l(t) = c(2) + t c (2) = (8, 4, 2) + t(12, 4, 1)

13 Sec. 2.5: Differentiation Rules For vector functions of several variables of class C 1 Constant multiple rule for f : R n R m and c R: D(cf) = cdf Example: D(2x 2, 2x + 2y) = 2D(x 2, x + y) Sum rule for f, g : R n R m : D(f + g) = Df + Dg Example: D [ (x 2, y) + (1, x 2 ) ] = D(x 2, y) + D(1, x 2 ) Note: Df etc. are m n matrices.

14 Sec. 2.5: Differentiation Rules For scalar functions of several variables of class C 1 Product rule for f, g : R n R: D(fg) = fdg + gdf Example: D [ x 2 (xy) 3] = x 2 D(xy) 3 + (xy) 3 Dx 2 Quotient rule for f, g : R n R and g 0: ( ) f D g = gdf fdg g 2 ( x 2 ) + y Example: D = exy D(x 2 + y) (x 2 + y)de xy e xy e 2xy Note: Df etc. are 1 n matrices.

15 Sec. 2.5: Chain Rule Chain rule for scalar-valued functions Example 1: z(u, v) = u + 2v, u(x) = x 2, and v(x) = x 3. Compute dz/dx using the chain rule. dz dx = z du u dx + z dv v dx = (1)(2x) + (2)(3x2 ) Example 2: z(u, v) = u + 2v, u(x, y) = 3x + 4y, and v(x, y) = xy. z x = z u u x + z v v x z y = z u u y + z v v y Compute z/ x and z/ y using the chain rule. = (1)(3) + (2)(y) = (1)(4) + (2)(x)

16 Sec. 2.6/2.3: Tangent Planes Tangent plane to a level surface F (x, y, z) = constant at (x 0, y 0, z 0 ) n (x x 0 ) = 0 with n = F (x 0 ) if F (x 0 ) 0 Example: Tangent plane to x + y 2 + z 3 = 8 at (3, 2, 1). F (x, y, z) = x + y 2 + z 3 and F = (1, 2y, 3z 2 ). Normal n = F (3, 2, 1) = (1, 4, 3). Tangent plane: (1, 4, 3) (x 3, y 2, z 1) = 0 1(x 3) + 4(y 2) + 3(z 1) = 0. Special case: Tangent plane to a graph z = f(x, y) at (x 0, y 0 ) F (x, y, z) = f(x, y) z = F = (f x, f y, 1) and z 0 = f(x 0, y 0 ) z = f(x 0, y 0 ) + f x (x 0, y 0 )[x x 0 ] + f y (x 0, y 0 )[y y 0 ]

17 Sec. 3.1: Higher-Order Partial Derivatives Notation for second partial derivatives of f = f(x 1,..., x n ) 1. R 2 : f xx = 2 f x = 2 x f yy = 2 f y 2 = y ( ) f x ( ) f y ( ) f f xy = (f x ) y = y x f yx = (f y ) x = ( ) f x y = 2 f y x = 2 f x y 2. R n : 2 f, x f x 1 x n etc. Notation for third partial derivatives of f = f(x 1,..., x n ) R 2 : f xxy, 3 f y x 2, 3 f. x 2 x 2 1 Th. 1: Mixed derivative theorem If f is of class C 2, then the mixed 2nd partial derivatives are equal. Examples f xy = f yx f xz = f zx

18 Sec. 3.2: Taylor Polynomials n-th order Taylor polynomial at x 0 for f(x) of class C n+1 n f (k) (x 0 ) T n (x) = (x x 0 ) k k! Note k=0 = f(x 0 ) + f (x 0 )[x x 0 ] + f (x 0 ) 2! [x x 0 ] f (n) (x 0 ) [x x 0 ] n n! 1. A Taylor polynomial of degree n consists of the first n + 1 terms of the Taylor series. 2. T 1 (x) = L(x), the linearization of f. For n 2 you get a higher-order approximation of f. Example: Compute T 2 (x) at x 0 = 2 for f(x) = e x. f (x) = e x and f (x) = e x. T 2 (x) = f(x 0 ) + f (x 0 )[x x 0 ] + f (x 0 ) 2! [x x 0 ] 2 = e 2 + e 2 [x 2] + e2 [x 2]2 2

19 Sec. 3.2: Taylor Polynomials n-th order Taylor polynomial at (x 0, y 0 ) for f(x, y) of class C n+1 ( f T n (x, y) = f(x 0, y 0 ) + x (x 0, y 0 ) [x x 0 ] + f ) y (x 0, y 0 ) [y y 0 ] + 1 ( 2 ) f 2! x 2(x 0, y 0 ) [x x 0 ] f x y (x 0, y 0 ) [x x 0 ][y y 0 ] + 2 f y 2(x 0, y 0 ) [y y 0 ] n! n ( ) n n f j x n j y (x 0, y j 0 ) [x x 0 ] n j [y y 0 ] j j=0 Remarks ( ) n = j n! j! (n j)! accounts for equal mixed partial derivatives The order of the derivative and power of the corresponding variable are equal

20 Sec. 4.1: Differentiation Rules for Paths For paths b(t) and c(t) of class C 1 and scalar functions q(t) of class C 1 Sum rule: d dt [b + c] = db dt + dc dt Product rule: d dt dq [q c] = dt c + q dc dt Chain rule: d dt [c(q(t))] = D (c q) = dc dt (q(t)) dq dt (t) Note: Specific case of chain rule in Sec Dot product rule: d dt db [b c] = dt c + b dc dt Cross product rule: d dt db [b c] = dt c + b dc dt

21 Sec. 4.4: Identities of Vector Analysis For f and g scalar functions of class C 1 and c constant 1. (f + g) = f + g 2. (cf) = c f 3. (fg) = f g + g f ( ) f 4. g = g f f g g 2 at points where g(x) 0.

22 Sec. 4.4: Identities of Vector Analysis For F and G vector fields of class C 1 and f a scalar function of class C 1 5. (F + G) = F + G 6. (F + G) = F + G 7. (ff ) = f F + F f 8. (F G) = G ( F ) F ( G) 10. (ff ) = f F + ( f) F

23 Sec. 4.4: Identities of Vector Analysis For f and g scalar functions of class C 2 and F a vector field of class C 2 9. ( F ) = 0 Compare v (v w) = ( f) = 0 Compare v v = (fg) = f 2 g + g 2 f + 2 f g 13. ( f g) = (f g g f) = f 2 g g 2 f