Math Vector Calculus II

Transcription

1 Math Vector Calculus II Review Notes Vectors We assume the reader is familiar with all the basic concepts regarding vectors and vector arithmetic, such as addition/subtraction of vectors in R n, scalar multiplication, the triangle and parallelogram laws, etc. So we begin our review with some other operations that are important. Recall: A unit vector (a vector of length 1) is denoted with a hat on top such as û. The coordinate unit vectors in engineering notation for R 3 (or R 2 ) are î = (1, 0, 0) (or (1, 0)) ĵ = (0, 1, 0) (or (0, 1)) ˆk = (0, 0, 1). To translate between the notations we use a, b, c = aî + bĵ + cˆk (or a, b = aî + bĵ). Dot Product The dot product of two vectors u = u 1, u 2,..., u n and v = v 1, v 2,..., v n is a scalar given by u v = n u i v i. i=1 Homework for the reader: Compute some random dot products.

2 Basic properties of the dot product (verification is left as homework to the reader): (1) u v = v u. (2) u (v + w) = u v + u w. (3) u u > 0 if and only if u 0. u u = 0 if and only if u = 0. (4) u u = u 2. Using the Law of Cosines (draw a triangle!): u v 2 = u 2 + v 2 2 u v cos θ (u v) (u v) = u u + v v 2 u v cos θ 2u v = 2 u v cos θ u v = u v cos θ. Geometric interpretation of the dot product: u v = u v cos θ. Moreover, two vectors in R n are orthogonal (perpendicular) if and only if their dot product is zero! (Note: Two vectors in R n are parallel if and only if they are scalar multiples of each other.)

3 Orthogonal Projections: The orthogonal projection of u onto v 0, denoted p = proj v (u), is defined as the scalar multiple of v that is orthogonal to u p as seen here: u cos θ So proj v (u) = ( u cos θ)(ˆv) = v = v The component of u along v is ( u cos θ) = u v the projection, denoted comp v (u). u v cos θ v 2 v = v ( u v ) v. v v, the signed magnitude (aka length) of Homework for the reader: Find the projection of (2, 4) onto (5, 1). Find the projection of (1, 2, 3) onto (1, 1, 1).

4 Cross Product The dot product, takes two vectors in R n and produces a scalar. The cross product takes two vectors in R 3 and produces another vector in R 3 that is perpendicular to both input vectors and in a direction determined by the right-hand rule. Let u = (u 1, u 2, u 3 ) and v = (v 1, v 2, v 3 ). Then the cross product given by: u v = (u 2 v 3 u 3 v 2, u 3 v 1 u 1 v 3, u 1 v 2 u 2 v 1 ). There is a method using a matrix determinant:. u v = det î ĵ ˆk u 1 u 2 u 3 v 1 v 2 v 3 Really, the most efficient way to calculate a cross product is to use the following template: (u 1, u 2, u 3 ) x (v 1, v 2, v 3 ) = (u 2 v 3 u 3 v 2, u 3 v 1 u 1 v 3, u 1 v 2 u 2 v 1 ). Homework for the reader: Perform some random cross products! vectors are embedded in 3-space. Do one where 2-dim.

5 The following derivation is shown in the notes, but we will not spend the time in class to verify the steps (that s homework for the reader). u v 2 = (u 2 v 3 u 3 v 2 ) 2 + (u 3 v 1 u 1 v 3 ) 2 + (u 1 v 2 u 2 v 1 ) 2 u v 2 = u 2 2v 2 3 2u 2 v 2 u 3 v 3 + u 2 3v u 2 3v 2 1 2u 1 v 1 u 3 v 3 + u 2 1v u 2 1v 2 2 2u 1 v 1 u 2 v 2 + u 2 2v 2 1 u v 2 = u 2 1v u 2 2v u 2 3v u 2 1v u 2 2v u 2 3v u 2 1v u 2 2v u 2 3v 2 3 (u v) 2 u v 2 = (u u u 2 3)(v v v 2 3) (u v) 2 u v 2 = u 2 v 2 u 2 v 2 cos 2 θ u v 2 = u 2 v 2 (1 cos 2 θ) u v 2 = u 2 v 2 sin 2 θ u v = u v sin θ. Properties of the cross product (homework for the reader): (1) u v = v u, (2) (c 1 u) (c 2 v) = c 1 c 2 (u v), (3) u (v + w) = u v + u w. Geometrically, the cross product is perpendicular to both inputs. Its direction is given by the right-hand rule. Also, u v = u v sin θ gives the area of the parallelogram spanned by u, v. Furthermore, the volume of the parallelepiped spanned by u, v, w is u (v w) = u v w (sin θ) cos α where θ is the angle between v, w and α is the angle between u and v w. Homework for the reader: Find the area of the parallelogram with vertices (1, 2), (5, 4), (3, 9), (7, 11). Find the volume of the parallelepiped spanned by the vectors (0, 1, 2), (3, 1, 5), (1, 1, 1).

6 Lines and Curves A function from one variable to many is a vector-valued function. Such a function has a form v(t) = (x 1 (t), x 2 (t),..., x n (t)). The graph of the outputs gives a 1-dimensional curve in n-dimensional space. The curve is a line if each component function x i (t) is linear. For example, graph r(t) = cos t, sin t. It s the unit circle traversed counterclockwise. What about the graph of u(t) = t, t 2? It s the parabola y = x 2 traversed left-to-right. What about the graph of l(t) = 1+t, 2t, 1+3t? It s the line through the point (1, 0, 1) in the direction of the vector v = 1, 2, 3. Unit Tangent Vector: When v (t) exists and is nonzero, it is a tangent vector to the graph of v(t) and we can define the unit tangent vector: T(t) = v (t) v (t). Rules for derivatives (proofs are left to the reader): Let u, v be vector-valued functions of t, let f be a scalar-valued differentiable function of t, and let c be a constant vector. (1) (2) (3) (4) (5) d (c) = 0. dt d dt (u(t) + v(t)) = u (t) + v (t). d dt (f(t)u(t)) = f (t)u(t) + f(t)u (t). d dt (u(t) v(t)) = u (t) v(t) + u(t) v (t). d dt (u(t) v(t)) = u (t) v(t) + u(t) v (t).

7 Arc Length The arc-length is the integral of the speed with respect to time along the path (treating the path as the nice trajectory of an object). Suppose that r(t) = (x 1 (t), x 2 (t),..., x n (t)) represents a trajectory and each component function is differentiable. The the arc-length L of the curve between r(a) and r(b) is given by the following integral: L = b r (t) dt = b a a (x 1 (t)) 2 + (x 2(t)) (x n(t)) 2 dt. Homework for the reader: Calculate the arc length of r(t) = (cos (2πt), sin (2πt), t) from (1, 0, 0) to (1, 0, 1). Polar coordinates: One can use coordinates (r, θ) to describe points in the xy-plane, where r is the distance to the origin and θ is an angle measured counter-clockwise from the positive x-axis. In terms of the standard, rectangular coordinates (x, y), and x = r cos θ and y = r sin θ. r 2 = x 2 + y 2 and tan (θ) = y x or θ = ±π matching the sign of y. 2

8 Cylindrical and Spherical Coordinates This picture helps visualize cylindrical and spherical coordinates: Cylindrical coordinates: (r, θ, z) where (r, θ) are polar coordinates for the xy-plane. Note: Whenever convenient, one can permute the role of z with either x, y. From rectangular to cylindrical: r 2 = x 2 + y 2, tan (θ) = y x or ±π, and z = z. 2 From cylindrical to rectangular: x = r cos θ, y = r sin θ, z = z.

9 Spherical coordinates: (ρ, θ, φ) where ρ is the distance to the origin, θ is the polar angle made in the xy-plane, and φ is the angle the point regarded as a vector makes with the positive z axis. From rectangular to spherical: ρ 2 = x 2 +y 2 +z 2, tan (θ) = y x or ±π, and φ = cos 1 z 2 x. 2 +y 2 +z 2 From spherical to rectangular: x = ρ sin φ cos θ, y = ρ sin φ sin θ, and z = ρ cos φ. From cylindrical to spherical: ρ 2 = r 2 + z 2, θ = θ, and φ = cos 1 From spherical to cylindrical: r = ρ sin φ, θ = θ, and z = ρ cos φ. z r. 2 +z 2

10 Parameterizations and Arc-Length (aka Unit-Speed) What is the difference between the trajectories r 1 = (cos t, sin t) and r 2 = (cos 2πt, sin 2πt)? Both parameterize the unit circle. Use the second one if you are in a hurry... As t goes from 0 to 2π, r 1 goes through the unit circle once. So t is equal to the arc length from r 1 (0) to r 1 (t). We call such a parametrization an arc-length parametrization or a unit speed parametrization. This is not the case for r 2 as it goes through 2π units of arc length as t goes from 0 to 1. Let r(t) be a path. Consider the arc-length function for the arc length from r(0) to r(t): s(t) = t 0 r (x) dx. If s(t) can be solved for the inverse function t(s), then by substitution, we get r(t(s)) and thus it can be parameterized by its own arc-length s. Homework for the reader: Find an arc-length parametrization for r(t) = (3 cos t, 3 sin t, t). Find one for r(t) = (4t, 3t, 2t 3/2 )

11 Directional Derivatives and the Gradient Suppose z = f(x, y) and you want to know how z changes at the point (x 0, y 0 ) with respect to a direction (within the domain) given by a unit vector û = (a, b) (so a 2 + b 2 = 1). Then the derivative of f with respect to the direction given by û is given by f(x 0 + ha, y 0 + hb) f(x 0, y 0 ) Dûf(x 0, y 0 ) = lim. h 0 h Definition: The gradient of f(x 1, x 2,..., x n ) is given by f = f x1, f x2,..., f xn. A nice formula for directional derivatives: Dûf(x 0, y 0 ) = f(x 0, y 0 ) (a, b). So the directional derivative is just the dot product of the gradient and a unit vector! All of this generalizes. The directional derivative of f(x 1, x 2,..., x n ) at (a 1, a 2,..., a n ) in the direction of the n-dimensional unit vector û is given by Dûf = f(a 1, a 2,..., a n ) û. Homework for the reader: Let f(x, y) = x 3 y 2 + tan 1 (xy). Find the directional derivative of f(x, y) at the point (1, 3) in the direction of the vector < 3, 4 >.

12 For what direction is Dûf(x 0, y 0 ) at a maximum? By the dot product identity, Dûf(x 0, y 0 ) = f(x 0, y 0 ) û cos θ = f(x 0, y 0 ) cos θ. So it is at a maximum when θ = 0. That occurs when û is in the direction of the gradient f(x 0, y 0 ). Moreover, this maximum directional derivative is equal to the magnitude of the gradient. So the gradient points in the direction of maximum increase (negative gradient points in the direction of maximum decrease) and its length is that rate of increase (or decrease). Homework for the reader: Find the direction of maximum increase for the function f(x, y, z) = xy 2 z 3 at the point (2, 2, 1). What is the maximum rate of increase? Furthermore, in a direction orthogonal to the gradient of a function f(x 1, x 2,..., x n ), the directional derivative is zero. Such a direction must then be along/parallel to a level set of the function (because the function does not change in that direction). Key idea regarding the gradient: The gradient at a point P is orthogonal to the level set of f at P, pointing the direction of maximum increase, with length equal to that maximum rate of increase.

13 Differential Areas and Volumes; Change of Variables Recall the following differential areas and volumes for double and triple integrals: Rectangular: da = dxdy = dydx and dv = dxdydz (in all 6 orders). Polar/Cylindrical: da = rdrdθ and dv = rdrdθdz. Spherical: dv = ρ 2 sin (φ)dρdφdθ. These are just examples of a more general idea: A change of variables in a double integral is a transformation that relates two sets of variables, (u, v) and (x, y), where the region R in the xy-plane corresponds to a region S in the uv-plane. Suppose x = g(u, v) and y = h(u, v) defines transformation where g, h have continuous first order partials over S and is one-to-one on the interior of S. The Jacobian determinant is computed in the following way: J(u, v) = (x, y) (u, v) = det x u y u x v y v = x y u v x y v u. Homework for the reader: Calculate the Jacobian determinant for x = r cos θ,y = r sin θ...

14 Theorem (the proof is beyond the scope of this course...) Under the preceding assumptions and the assumption that f is continuous on R... f(x, y)da = R S f(g(u, v), h(u, v)) (x, y) (u, v) da. Note that (x,y) (u,v) measures the magnification or reduction of a differential area in the uvplane compared to the transformed differential area in the xy-plane... Exercises left for the reader: 1) Compute (x + y) 2 da x + y 1 using a nice change of variables... [DRAW REGION!] Hint: u = x + y and v = x y. 2) Compute R y 2 da where R is the region in the first quadrant bounded by xy = 1, xy = 4, y x = 1 and y x = 3... [DRAW REGION!] Hint: u = xy and v = y x.

15 In 3 variables... Suppose x = g(u, v, w), y = h(u, v, w) and z = l(u, v, w) is a transformation of a solid B in uvw-space into a solid E in xyz-space that on the interior is one-to-one and has continuous partials. The Jacobian determinant is computed in the following way: J(u, v, w) = (x, y, z) (u, v, w) = det x u y u z u x v y v z v x w y w z w. Calculate the Jacobian determinant for x = ρ sin φ cos θ, y = ρ sin φ cos θ, and z = ρ cos φ... of course we have seen this before! Theorem (the proof is beyond the scope of this course...) Under the preceding assumptions and the assumption that f is continuous on E... f(x, y, z)dv = E B f(g(u, v, w), h(u, v, w), l(u, v, w)) (x, y, z) (u, v, w) dv. Note that (x,y,z) (u,v,w) measures the magnification or reduction of a differential volume in the uvw-space compared to the corresponding differential volume in the xyz-space... Homework for the reader: Use a change of variables to calculate the volume bounded by the ellipsoid: x 2 a + y2 2 b + z2 2 c = 1. 2 Hint: Use x = au, y = bv, and z = cw.