Vector review. MAT Multi Variable Calculus. Class Exercises. The Inner product, or Scalar Product

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1 Vector review MAT Multi Variable Calculus W. Stefan Arizona State University October 31, 2008 Let u =< u1, u2,, un > and v =< v1, v2,, vn > be two vectors in Vn and c a scalar. Operations with vectors Addition: u + v =< u1 + v1, u2 + v2,, un + v2 > scalar multiplication: cu =< cu1, cu2,, cun >. Subtraction: u v = u + ( 1v) Note: The result of addition and scalar multiplication is a vector. Length or Magnitude u = u = n ui 2 i=1 Note: The result of the magnitude operator is a scalar (number). The Inner product, or Scalar Product The Scalar product is the product of two vectors u Vn and v Vn and results in a scalar (i.e. u v R). Definition u v = u1v1 + u2v2 + + unvn n = uivi i=1 u =< 4, 3, 1 >, v =< 2, 4, 1 >. Class Exercises Show the following properties of the scalar product If u and v are in V2 and c is a scalar, then 1. u u = u 2 2. u v = v u 3. u (v + w) = u v + u w 4. (cu) v = c(u v) = a (cv) 5. 0 u = 0 u v = ( 3) 4 + ( 1) ( 1) = = 3

2 1 2 u =< 4, 6 >, v =< 4, 1 > u = = 21 u =< 2, 8 >, v =< 4, 1 > u = 8 8 = 0 Scalar Product and Angles Scalar Product and Projections Theorem If Φ is the angle between the vectors u and v then u v = u v cos Φ Note Two vectors u and v are perpendicular or orthogonal if and only if the angle is Φ = π/2 and. u v = 0 The projection proj a b is called vector projection of b onto a. The length of the projected vector is called scalar projection comp a b.

3 Scalar Product and Projections cont. Exercise Scalar and vector projection Scalar Projection of b onto a: comp a b = a b a Vector Projection of b onto a: ( ) a b a proj a b = a a The cross product Ways to remember the formula The cross product between vectors a, b V3 is a vector in V3. Definition Determinant of a two-by-two matrix a b =< a2b3 a3b2, a3b1 a1b3, a1b2 a2b1 > Properties a b is perpendicular to both a and b. a b is perpendicular to the plane generated by a and b. a b = a b sin Φ, where Φ is the angle between a and b. Vectors a and b are parallel if and only of a b = 0 Determinant of a three-by-three matrix

4 Ways to remember the formula, cont. 1st way i j k a1 a2 a3 = i(a2b3 b2a3) j(a1b3 a3b1) + k(a1b2 b1a2) b1 b2 b3 Properties of the cross product Properties of the cross product Let a, b, c V3 and c R, then 2nd way But... and a (b c) (a b) c a b = b a Applications of the cross product Applications of the cross product Find a vector that is perpendicular to two other vectors Area of a parallelogram: A = a b Torque Triple Product a1 a2 a3 a (b c) = b1 b2 b3 c1 c2 c3 Can be used to compute the volume of a parallelepiped: V = a (b c) τ = r F

5 line and plane equations Symmetric equation r = r0 + ta Vector equation t R is the parameter each value of t corresponds to a point on the line r0 =< x0, y0, z0 >, v =< a, b, c > < x, y, z >=< x0, y0, z0 > +t < a, b, c > solve 1st equation for t : t = x x0 a solve 2nd equation for t : t = y y0 b solve 3rd equation for t : t = z z0 c Symmetric equation of a line (t =) x x0 a = y y0 = z z0 b c Line segments Parallel lines, skew lines and intersecting lines Line segments from r0 to r1 r(t) = (1 t)r0 + tr1 0 t 1 check if equation makes sense: r(0) = r0 + 0r1 = r0 r(1) = 0r0 + 1r1 = r1 1. lines can be parallel 2. equal 3. skew 4. intersecting

6 Planes Vector equation For a plane containing point r0 and normal vector n: n (r r0) = 0 Planes cont. Scalar equation for a plane r0 =< x0, y0, z0 > and n =< a, b, c >: multiply out vector equation: a(x x0) + b(y y0) + c(z z0) = 0 All vectors in the plane are perpendicular to n. ax + by + cz + d = 0 Parallel planes, intersecting planes 1. Two planes are parallel if Two planes intersect at a line: v lies in both planes: v is perpendicular to n1 and n2 Distance to a plane Review A vector a can be defined using points Distance to a plane ax + by + cz + d = 0 A point in the plane can be identified by a vector D = comp n b = = = a(x1 x0) + b(y1 y0) + c(z1 z0) a 2 + b 2 + c 2 { d }} { ax1 + by1 + cz1 (ax0 + by0 + cz0) a 2 + b 2 + c 2 ax1 + by1 + cz1 + d a 2 + b 2 + c 2 To define a line we need two things: a direction and a point on the line.

7 Extra Homework due 09/12/08 Cylinders and quadratic surfaces You know the line equation in 2-D: y = mx + b An alternative way to describe the line equation in 2-D is using vectors: r = r0 + tv with r, r0, v V2 1. write down the parametric equation in 2-D 2. plot the line 3. find m and b, if r0 =< 2, 1 > and v =< 1, 3 > Cylinders and quadratic surfaces cont. Cross section at y=2 Traces (or cross-section) Curves of intersection with the x y, x z, z y plane.

8 Vector functions and space curves Vector functions and space curves Vector-valued functions Function with input is a number and output is a vector. domain is a set of numbers and its range is a set of vectors. Its r(t) =< f (t), g(t), h(t) >= f (t)i + g(t)j + h(t)k r(t) =< cos4t, t, sin4t > Vector functions and space curves cont Vector functions and space curves exercises Find a vector valued function that represents the intersection of the two surfaces: x 2 + y 2 = 4 and z = xy.

9 Derivative of a vector valued function Derivative of a vector valued function Vector-valued functions Derivative r dt = r(t + h) r(t) r (t) = lim h 0 h The vector function is smooth in the interval I if r (t) 0 for all t in I. Properties Integral of a vector valued function Arc Length Integral b b b b r(t)dt =< f (t)dt, g(t)dt, h(t)dt > a a a a Arc Length b L = a [f (t)] 2 + [g (t)] 2 + [h (x)] 2 dt = b a r (t) dt

10 Christmas lighting business I m trying to figure out how long of a line of lights I would need to wrap around a tree 110 tall and 6 diameter at the base (assume 0 diameter at the top), if it only wraps around 6 times? I m thinking the equation of a helix but not sure. Any ideas?! Arc length function Arc length function s(t) s(t) = t r (u) du a Used to parametrize a curve wrt. the arc length: r(s) = r(t(s)) Curve can be defined independent of the parameter t. Curvature Curvature Or more conveniently: κ = dt ds = dt dt dt ds κ(t) = T (t) r (t) κ(t) = r (t) r (t) r (t) 3 The Normal and Binormal Vectors Unit Tangent vector T(t) = r (t) r (t) Observe that T(t) T (t) = 0: Principal Unit Normal Vector N(t) (or simply unit normal) Binormal Vector N(t) = T (t) T (t) B(t) = T(t) N(t)

11 Motion in Space Velocity Vector v(t) r(t + h) r(t) v(t) = lim = r (t) h 0 h The speed of the particle at time t is the magnitude of the velocity vector v(t) = r (t). Acceleration Vector a(t) Class Exercises a(t) = v (t) = r (t) Find the velocity and position vectors of a particle that has the given initial velocity and position: a(t) =< 1, 2, 0 >, v(0) =< 0, 0, 1 > and r(0) =< 1, 0, 0 >. What is the velocity vector as the particle passes through the line: r2(t) =< 1, 2, 3 > +t < 2/3, 1/3, 2/3 > What is the velocity vector as the particle passes through the plane: x = 4 Tangential and normal components of acceleration Acceleration With v = v, note that v = vt, i.e. v = v T + vt. With T = κv and T = T N we have: a(t) = v (t)t + κv 2 N = att + ann Note that v a = vt (v (t)t + κv 2 N) = vv T T + κv 3 T N = vv (because T T = 1 and T N = 0). Tangential and normal components Partial Derivatives at = v = v a = r (t) r (t) v r (t) an = κv 2 = r (t) r (t) r (t) 3 r (t) 2 = r (t) r (t) r (t) Partial Derivatives If f (x, y) is a function of two variables, its partial derivatives are the functions fx(x, y) and fy (x, y), defined by and Notations fx(x, y) = fx(x, y) = lim h 0 f (x + h, y) f (x, y) h fy (x, y) = lim h 0 f (x, y + h) f (x, y) h ( ) f (x, y) = Dxf (x, y) = = f1(x, y) = D1f (x, y) x x

12 Partial Derivatives cont. Partial Derivatives cont. Rules for finding partial derivatives 1. To find fx, regard y as a constant and differentiate f (x, y) with respect to x. 2. To find fy, regard x as a constant and differentiate f (x, y) with respect to y. f (x, y) = xe 3y fx(x, y) = e 3y fy (x, y) = 3xe 3y Geometric Interpretation fx(a, b) slope of the tangents to curves of the intersection of the graph with plane y = b f (x, y) = 4 x 2 2y 2 fx(x, y) = 2x fy (x, y) = 4y fx(1, 1) = 2 fy (1, 1) = 4 f (1, 1) = 1 Higher derivatives Second order derivatives 1. (fx)x (fx)x = fxx = ( f ) x x = 2 f x x 2. (fx)y (fx)y = fxy = ( f ) y x = 2 f y x 3. (fy )x = fyx = x 4. (fy )y = fyy = y Clairaut s Theorem ( ) f y ( f y = 2 f x y ) = 2 f y y fxy = fyx Ideal gas The gas law for a fixed mass m of an ideal gas at absoute temperature T, preassure P, and volume V is PV = mrt. Show that T = PV mr P = mrt V V = mrt P T P V T T = mr PV mr mr mr V P = mr P T = mr V V T = mr P

13 Tangent-planes Equation of the tangent plane Plane through a point P(a, b, c), with position vector r0 =< a, b, c > width vectors in the plane v =< 1, 0, fx(a, b) > and u =< 0, 1, fy (a, b) > equation of plane n(r r0) = 0 Tangent plane through the point P(x0, y0, z0), i.e. z0 = f (x0, y0), of the function z = f (x0, y0). z z0 = fx(x0, y0)(x x0) + fy (x0, y0)(y y0) plot Zoom in

14 Linear Approximation Linear approximation may not exist Linear Approximation f (x, y) L(x, y) = f (a, b) + fx(a, b)(x a) + fy (a, b)(y a) is the linear approximation or linearization of f (x, y) at (a, b) Differentability If the partial derivatives fx and fy exist and are continuous at (a, b), then f is differentable at (a, b). Differentials 2 Total differential dz = fx(x, y)dx + fy (x, y)dy = x dx + x dy = fx(x, y) x + fy (x, y) y Resistors A model of the surface of the human body is given by S = w h 0.725, where w is weight in pounds and h is the height in inches. S is measured in square feet. Use differentials to estimate the maximum error in percent in the calculated surface if the error in measurement in w and h is at most 2%.

15 The Chain Rule The Cain Rule (Case 1) Suppose z = f (x, y) with x = g(t) and y = h(t): dz dt = f dx x dt + f dy y dt z = sin(x) cos(y), x = πt, y = t dz dt = cos(πt) cos( sin(πt) sin( t) t)π 2 t The Chain Rule cont. The Cain Rule (Case 2) Suppose z = f (x, y) with x = g(s, t) and y = h(s, t): s = f x x s + f y y s t = f x x t + f y y t z = x 2 + xy + y 2, x = s + t, y = st = 2(s + t) + st + (2st + s + t)s t = 2(s + t) + st + (2st + s + t)t s The Chain Rule cont. Implicit Differentiation z x y u v w u v w Class Exercise Using a rule similar to the chain rule (case 2) to find all partial derivatives of z = x 2 + xy 3, x = uv 2 + w 3, y = u + ve w. Formulate a version of the chain rule for f (x1, x2, xn) with xi = xi(t1, tm) Let F (x, y) = 0 and y = f (x) differentable, where F (x, f (x)) = 0. We want to compute the derivative of f (x): Use chain-rule (case I) to differentate both sides: with d dx x = x = 1 now if F y 0: F x d dx F (x, y) = d dx 0 dx dx + F dy y dx = 0 F dx x dy = F y dy dx = F x = Fx Fy y

16 Implicit Differentiation cont. Implicit Differentiation cont. Implicit Differentiation Let F (x, y) = 0 and y = f (x) differentable, where F (x, f (x)) = 0. Then f (x) = dy F dx = x = Fx F y Fy Note Let F (x, y) = 0 then y can be expressed as a function of x, i.e. y = f (x) whenever Fy (x, y) 0 Find dy dx for x 2 + y 2 = 4 F (x, y) = x 2 + y 2 4 dy dx = 2x 2y = x y.

17 Implicit Differentiation for 3 variables Let F (x, y, z) = 0 and z = f (x, y) differentable, where F (x, y, f (x, y)) = 0. We want to compute the partial derivatives of f (x, y): Use chain-rule (case II) to differentate both sides with respect to x: F x x F (x, y, z) = x 0 x x + F y y x + F x = 0 with x x = x = 1 and x y = y x 1 = 0 F x x = F Implicit Differentiation for 3 variables cont. Implicit Differentiation for 3 variables Let F (x, y, z) = 0 and z = f (x, y) differentable, where F (x, y, f (x, y)) = 0. Then the partial derivatives of f (x, y) are given by x = F x = Fx y = F y F Fz = Fy Fz now if F 0: x = F x = Fx Fz Directional Derivatives Directional derivative Duf (x0, y0) = lim h 0 f (x0 + ha, y0 + hb) f (x0, y0) h Note with g(h) := f (x0 + ha, y0 + hb) we can write g(h) g(0) Duf (x0, y0) = lim = g (0) h 0 g and using the chain rule case I: g (h) = fx(x0 + ha, y0 + hb)a + fy (x0 + ha, y0 + hb)b therefore Gradient Directional derivative and gradient Directional derivative in the direction u: Duf (x0, y0, z0) = fx(x0, y0)a + fy (x0, y0)b = < fx(x0, y0), fy (x0, y0) > u = f (x0, y0) u, where f (x0, y0) is called the gradient of f (x, y) at (x0, y0). g (0) = Duf (x0, y0, z0) = fx(x0, y0)a + fy (x0, y0)b

18 Maximizing the directional derivative Useful applications of the gradient Tangent plane The tangent plane at the point P(x0, y0, z0): In what direction u is the directional derivative maximal? Duf (x0, y0, z0) = f (x0, y0) u = f (x0, y0) cosφ where φ is the angle between f (x0, y0) and u. φ = 0. The gradient is the direction of steepest assent. F (x0, y0, z0) (r r0) = 0 Gradient and level lines The gradient is perpendicular to the level lines, i.e. if F (x(t), y(t)) = k then < x (t), y (t) > F (x(t), y(t)) = 0 Local maximum Local maximum or minimum Theorem If f (x, z) has a local maximum or minimum or a critical point at (a, b) and the gradient exist then f (x, y) (x,y)=(a,b) =< 0, 0 > f (x, y) = 9 2x + 4y x 2 4y 2 Definition A function of two variables f (x, y) has a local maximum at (a, b) if f (a, b) f (x, y) when (x, y) is close to (a, b). fx(x, y) = 2 2x = 0 fy (x, y) = 4 8y = 0 x = 1 y = 1 2

19 Local maximum Second Derivative Test Let f (a, b) =< 0, 0 > and D = fxx fyz then 1. if D > 0 and fxx(a, b) > 0, then f (a, b) is a local minimum, 2. if D > 0 and fyy(a, b) < 0, then f (a, b) is a local maximum, 3. if D < 0 is not a local minimum or local maximum,it is a saddle point. Note The test gives no information if D = 0: f (a, b) could still be a local minimum or maximum. fxy fxx f (x, y) = 9 2x + 4y x 2 4y 2 plug in ( 1, 1/2) fx(x, y) = 2 2x = 0 fy (x, y) = 4 8y = 0 x = 1 y = 1 2 fxy(x, y) = 0 fyx(x, y) = 0 fxx(x, y) = 2 fyy(x, y) = 8 D = = 16 > 0 The point ( 1, 1/2) is a local maximum. Multiple Integrals Review of the definite integral Let f (x) be defined for a x b and let the interval [a, b] be divided into n subinterval [xi 1, xi] with xi = xi xi 1. Let xi be points in these subintervals. Then Vector Fields assigns a vector to each point y = F(x) = F(x, y) b n f (x)dx = lim xi 0 f (xi ) xi i=1 a

20 Vector Fields in Applications Newton s law of gravitation F(x) = mmg x 3 x Gradient Fields Consider the scalar function z = f (x, y) Note: the gradient f of f (x, y) is a vector. i.e. the gradient defines a vector field Conservative Vector Field A vector field F(x) is called conservative if a f (x) exists, such that F (x) = f (x) Flowlines The path followed by an particle whose velocity filed is the given vector field. Flowline A curve r(t) is a flow line of the vector field F(x) if r (t) = F(r(t))