Math 241, Exam 1 Information. 9/24/12, LC 310, 11:15-12:05. Exam 1 will be based on: Sections 12.1-12.5, 14.1-14.3. The corresponding assigned homework problems (see http://www.math.sc.edu/ boylan/sccourses/241fa12/241.html) At minimum, you need to understand how to do the homework problems. Topic List (not necessarily comprehensive): You will need to know: theorems, results, and definitions from class. 12.1: 3-dimensional space. Distance formula: d(p 1, P 2 ) = (x 2 x 1 ) 2 + (y 2 y 1 ) 2 + (z 2 z 1 ) 2. Sphere with center (x 0, y 0, z 0 ) and radius r: 12.2: Vectors. {(x, y, z) : (x x 0 ) 2 + (y y 0 ) 2 + (z z 0 ) 2 = r 2 }. Vectors have length (magnitude) and direction. Operations: 1. Addition. (a) Algebraically: u 1, u 2, u 3 + v 1, v 2, v 3 = u 1 + v 1, u 2 + v 2, u 3 + v 3. (b) Geometrically: Parallelogram Law. (c) Properties: closure, zero element, associativity, inverses ( u = ( 1) u), commutativity. 2. Scalar multiplication. (a) Algebraically: k u 1, u 2, u 3 = ku 1, ku 2, ku 3. (b) Geometrically: k u scales u by a factor of k ; if k < 0, then k u points in the direction opposite u. (c) Properties: closure, associativity, distributivity.
Length. v 1, v 2, v 3 = v1 2 + v2 2 + v3 2 Notes. 1. For all v R 3, we have v 0 and v = 0 v = 0. 2. For k R, we have k v = k v. Unit vectors. A vector v is a unit vector if and only if v = 1. 1. Standard basis unit vectors: i = 1, 0, 0, j = 0, 1, 0, k = 0, 0, 1. 2. The normalization of v R 3 is the unit vector v v. Parallel vectors. Two non-zero vectors u and v are parallel ( u v) if and only if there exists k 0 in R with k u = v. 12.3: The dot product. Definition: u 1, u 2, u 3 v 1, v 2, v 3 = u 1 v 1 + u 2 v 2 + u 3 v 3. The input consists of two vectors; the output is a scalar. Properties. 1. Commutative, distributive. 2. u u = u 2. 3. Let 0 θ π be the angle between vectors u and v. Then we have cos θ = u v u v. Orthogonality. Two vectors u and v are orthogonal if and only if u v = 0; geometrically, this holds if and only if the vectors are perpendicular: the angle between them is π/2. ( 1. The orthogonal projection of v on v b is proj b v = b ) b. b 2 2. The component of v on b is comp b v = proj b v = v b b.
12.4: The cross product. Definition: Let u = u 1, u 2, u 3, v = v 1, v 2, v 3. Then we have i j k u v = u 1 u 2 u 3 v 1 v 2 v 3 = (u 2v 3 v 2 u 3 ) i (u 1 v 3 v 1 u 3 ) j + (u 1 v 2 v 1 u 2 ) k Two vectors are input; the output is a vector. Facts. 1. Distributive. 2. Not associative in general. 3. u u = 0. 4. Anti-commutative: u v = v u. 5. i j = k, j k = i, k i = j. The other products of these vectors are obtainable from anti-commutativity. Geometry. 1. u v is orthogonal to both u and v. Of two possible directions, the correct one is obtained via the right-hand rule. 2. Let 0 θ π be the angle between u and v. Then we have sin θ = 3. The area of the parallelogram determined by u and v is u v. 4. The vectors u and v are parallel ( u v) if and only if u v = 0. u v u v. Triple product: Let u = u 1, u 2, u 3, v = v 1, v 2, v 3, w = w 1, w 2, w 3. Then we have u 1 u 2 u 3 u ( v w) = v 1 v 2 v 3 w 1 w 2 w 3. Moreover: 1. This value gives the volume of the parallelepiped determined by u, v, and w. 2. This value is zero if and only if the three vectors are coplanar.
12.5: Equations of lines and planes. I. Lines. A line L is determined by a direction vector v = a, b, c and a point P 0 = (x 0, y 0, z 0 ) on the line. The equation of a line can take several forms (a) Parametric form. x = x 0 + at, y = y 0 + bt, z = z 0 + ct, t R. (b) Vector form. r = r 0 + t v. Here, we have r = x, y, z and r 0 = x 0, y 0, z 0. (c) Symmetric form. Suppose that none of a, b, c is zero. Then we have newline x x 0 = y y 0 = z z 0. Suitable adjustments are required when one or more a b c of the components of v is zero. Skew lines. Two lines in R 3 are skew if and only if they are non-parallel lines lying in parallel planes. Facts. Let L 1 and L 2 be lines with direction vectors v 1 and v 2. (a) The lines L 1 and L 2 are parallel if and only if their direction vectors are; this happens if and only if there exists k R with k v 1 = v 2. (b) The lines L 1 and L 2 are perpendicular if and only if their direction vectors are orthogonal: u v = 0. (c) Two lines are skew if they are non-intersecting and non-parallel. To determine whether two non-parallel lines intersect requires one to solve a system of three equations in two variables. If the system has a solution, the lines intersect at the point given by the solution; if the system has no solution, then the lines are skew. II. Planes. A plane Q is determined by a normal vector n = a, b, c which measures the tilt of Q in R 3 and a point P 0 = (x 0, y 0, z 0 ) on Q. The normal vector is perpendicular to Q. The equation of the plane Q is (a) Q : a(x x 0 ) + b(y y 0 ) + c(z z 0 ) = 0. (b) Q : ax + by + cz = d, where d = ax 0 + by 0 + cz 0. Observations. (a) Planes Q 1 and Q 2 are parallel if and only if their normal vectors are: n 1 n 2. (b) Planes Q 1 and Q 2 are perpendicular if and only if their normal vectors are: n 1 n 2. (c) Line L is parallel to plane Q if and only if the direction vector of L is perpendicular to the normal of Q: n v. (d) Line L is perpendicular to plane Q if and only if the direction vector of L is parallel to the normal of Q: n v.
Distance. Let P 0 = (x 0, y 0, z 0 ) and let Q : ax + by + cz + d = 0. Then the distance from P 0 to Q is d(p 0, Q) = ax 0 + by 0 + cz 0 + d. a2 + b 2 + c 2 One can use this formula to find the distance between the following objects (assumed to be non-intersecting so the distance makes sense): a point and a plane; a point and a line; a line and a plane; two planes; 14.1: Functions of several variables. Given a function z = f(x, y), be able to sketch or describe the domain, the graph, and the level curves. 14.2: Limits and continuity. Limits along curves: Let C be a smooth curve in R 2 through the point P 0 = (x 0, y 0 ), and suppose that C has parametric equations C : x = x(t), y = y(t), t R, with x 0 = x(t 0 ), y 0 = y(t 0 ) for some t 0. Then we have lim f(x, y) = lim f(x(t), y(t)) (x,y) (x 0,y 0 ) along C t t0 Remarks: We have lim f(x, y) exists if and only if lim f(x, y) exists and gives a common (x,y) (x 0,y 0 ) (x,y) (x 0,y 0 ) along C value for all curves C through (x 0, y 0 ). Suppose that there exist curves C 1 and C 2 through (x 0, y 0 ) with Then lim f(x, y) does not exist. (x,y) (x 0,y 0 ) lim f(x, y) lim f(x, y) (x,y) (x 0,y 0 ) along C 1 (x,y) (x 0,y 0 ) along C 2
Definition: The function f(x, y) is continuous at (x 0, y 0 ) if and only if 1. (x 0, y 0 ) is in the domain of f and 2. lim f(x, y) = f(x 0, y 0 ). (x,y) (x 0,y 0 ) 14.3: Partial derivatives. The partial derivative of z = f(x, y) with respect to x is obtained by differentiating with respect to x while holding y constant; similarly, the partial with respect to y is obtained by differentiating with respect to y while holding x constant. Notation: Let z = f(x, y). We write f x (x, y) = f x (x, y) = x z, f y(x, y) = f y (x, y) = y z. Fact: Let z = f(x, y), and let P 0 = (x 0, y 0 ). Then f x (x 0, y 0 ) gives the slope of the line tangent to the surface at P 0 in the direction of the unit vector i; similarly, f y (x 0, y 0 ) gives the slope of the line tangent to the surface at P 0 in the direction of j. Higher order partials: 2 f x = 2 x ( ) f = f xx, x 2 f y 2 = y ( ) f = f yy. y Clairaut s Theorem implies that 2 f y x = f xy = f yx = 2 f x y