MIT OpenCourseWare http://ocw.mit.edu 18.02 Multivariable CalculusFall 2007 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.� 18.02 Lecture 1. – Thu, Sept 6, 2007 Handouts: syllabus; PS1; flashcards. Goal of multivariable calculus: tools to handle problems with several parameters – functions of several variables. Vectors. A vector (notation: A�) has a direction, and a length (|A�|). It is represented by a directed line segment. In a coordinate system it’s expressed by components: in space, A�= �a1, a2, a3� = a1ˆı + a2ˆj + a3kˆ. (Recall in space x-axis points to the lower-left, y to the right, z up). Scalar multiplication Formula for length? Showed picture of �3, 2, 1� and used flashcards to ask for its length. Most students got the right answer (√14). why |A�| = a12 + a22 + a32 by reducing to the Pythagorean theorem in the plane (Draw a picture, showing A�and its projection to the xy-plane, then derived |A�| from lengthof projection + Pythagorean theorem). Vector addition: A�+B�by head-to-tail addition: in a parallelogram (s howed how the diagonals are A�+ B�and B�− A�); A�= 3ˆı + 2ˆj + kˆon the displayed example. Dot product. Definition: A�B�= a1b1 + a2b2 + a3b3 (a scalar, not a vector).· Theorem: geometrically, A�· B�= |A�||B�| cos θ. Explained the theorem as follows: first, A�A�= A�2 cos 0 = |A�2 is consistent with the definition.· | | |2Next, consider a triangle with sides A�, B�, C�= A�− B�. Then the law of cosines gives C�= |A�|2 + |B�|2 − 2|A�||B�| cos θ, while we get | |C = C C = (A�− B�) (A�− B�) = |A�2 + | �− 2A�B.| �|2 �· �· | B|2 · �Hence the theorem is a vector formulation of the law of cosines. A�B�Applications. 1) computing lengths and angles: cos θ = · . |A�||B�|Example: triangle in space with vertices P = (1, 0, 0), Q = (0, 1, 0), R = (0, 0, 2), find angle at P : P Q P R cos θ = −−→· −−→= �−1, 1, 0� · �−1, 0, 2� =1 , θ ≈ 71.5◦. |−−→P R |√2√5 √10P Q ||−−→Note the sign of dot product: positive if angle le ss than 90◦, negative if angle more than 90◦, zero if perpendicular. 2) detecting orthogonality. Example: what is the set of points w here x +2y + 3z = 0? (possible answers: empty set, a point, a line, a plane, a sphere, none of the above, I don’t know). Answer: plane; can see “by hand”, but more geometrically use dot product: call A�= �1, 2, 3�, P = (x, y, z), then A�· OP = x + 2y + 3z = 0 ⇔ | �OP | cos θ = 0 ⇔ θ = π/2 ⇔ �OP . So we−−→A||−−→A ⊥ −−→get the plane through O with normal vector A�. 1 You can explainaddition works componentwise, and it is true thatDraw a picture���� ���� ���� ���� ������ ������ ���� ���� ���� ���� ���� ���� ������ ������ ���� ���� ���� ���� ���� ���� 2 18.02 Lecture 2. – Fri, Sept 7, 2007 We’ve seen two applications of dot product: finding lengths/angles, and detecting orthogonality. A third one: finding components of a vector. If uˆis a unit vector, A�·uˆ= |A�| cos θ is the component of A�along the direction of uˆ. E.g., A�ˆı = component of A�along x-axis.· Example: pendulum making an angle with vertical, force = weight of pendulum F�pointing downwards: then the physically important quantities are the components of F�along tangential direction (causes pendulum’s motion), and along normal direction (causes string tension). Area. E.g. of a polygon in plane: break into triangles. Area of triangle = 1 base × height = 2 1 |A�||B�| sin θ (= 1/2 area of parallelogram). Could get sin θ using dot product to compute cos θ and2 sin2 + cos2 = 1, but it gives an ugly formula. Instead, reduce to complementary angle θ� = π/2 − θ by considering A�� = A�rotated 90◦ counterclo ckwise (drew a picture). Then area of parallelogram = |A�||B�| sin θ = |A��||B�| cos θ� = A�� · B�. Q: if A�= �a1, a2�, then what is A��? (showed picture, used flashcards). Answer: A�� = �−a2, a1�. (explained on picture). So area of parallelogram is �b1, b2� · �−a2, a1� = a1b2 − a2b1. a1 a2Determinant. Definition: det( �B) = A, �= a1b2 − a2b1. b1 b2 Geometrically: a1 a2 b1 b2 = ± area of parallelogram. sign of 2D determinant has to do with whether B�is counterclockwise or clockwise from A�, without details. Determinant in s pace: det( �B, �A,�C) = a1 a2 a3 b1 b2 b3 c1 c2 c3 = a1 b2 b3 c2 c3 −a2 b1 b3 c1 c3 +a3 b1 b2 c1 c2 . Geometrically: det( �B, �A,�C) = ± volume of parallelepiped. R eferred to the notes for more about determinants. Cross-product. (only for 2 vectors in space); gives a vector, not a scalar (unlike dot-product). ˆı ˆj kˆa2 a3 a1 a3 + kˆ a1 a2A�× B�= =ˆı − ˆj Definition: a1 a2 a3 . b2 b3 b1 b3 b1 b2b1 b2 b3 (the 3x3 determinant is a symbolic notation, the actual formula is the expansion). Geometrically: |A�× B�| = area of space parallelogram with s ides A�, B�; direction = normal to the plane containing A�and B�. How to decide between the two perpendicular directions = right-hand rule. 1) extend right hand in direction of A�; 2) curl fingers towards direction of B�; 3) thumb points in same direction as A�×B�. ˆı × ˆj =? (answer: kˆ, checked both by geometric description and by calculation). Triple product: volume of parallelepiped = area(base) · height = |B�× C�| (A�· nˆ), where ˆn = B�× �|B�× �|�· (B�× C�) = det( �B, �C/ C . So volume = A A,�C). The latter identity can also be checked directly using components. TheFlashcard
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