lec_week1

lec_week1 - MIT OpenCourseWare http://ocw.mit.edu 18.02...

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Unformatted text preview: MIT OpenCourseWare http://ocw.mit.edu 18.02 Multivariable Calculus Fall 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 = ˆ . (Recall in space x-axis points to the lower-left, y to the right, z up). �a1 , a2 , a3 � = a1ˆ + a2ˆ + a3 k ı j Scalar multiplication Formula for length? Showed√ icture of �3, 2, 1� and used flashcards to ask for its length. Most p students got the right answer ( 14). � � You can explain why |A| = a2 + a2 + a2 by reducing to the Pythagorean theorem in the 1 2 3 � � plane (Draw a picture, showing A and its projection to the xy -plane, then derived |A| from length of projection + Pythagorean theorem). �� Vector addition: A + B by head-to-tail addition: Draw a picture in a parallelogram (showed how � + B and B − A); addition works componentwise, and it is true that � � the diagonals are A � � ı jˆ A = 3ˆ + 2ˆ + k on the displayed example. Dot product. �� Definition: A · B = a1 b1 + a2 b2 + a3 b3 (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. ��� � � � Next, consider a triangle with sides A, B , C = A − B . Then the law of cosines gives |C |2 = � |2 + |B |2 − 2|A||B | cos θ, while we get � �� |A � �� �� �� � � �� |C |2 = C · C = (A − B ) · (A − B ) = |A|2 + |B |2 − 2A · B. 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 : −→ −→ − − 1 PQ · PR �−1, 1, 0� · �−1, 0, 2� √√ =√ , cos θ = −→ −→ = θ ≈ 71.5◦ . − − 25 10 |P Q ||P R | Note the sign of dot product: positive if angle less than 90◦ , negative if angle more than 90◦ , zero if perpendicular. 2) detecting orthogonality. Example: what is the set of points where 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 ⇔ |A||OP | cos θ = 0 ⇔ θ = π /2 ⇔ A ⊥ OP . So we � get the plane through O with normal vector A. 1 2 18.02 Lecture 2. – Fri, Sept 7, 2007 We’ve seen two applications of dot product: finding lengths/angles, and detecting orthogonality. �ˆ � ˆ A third one: finding components of a vector. If u is a unit vector, A · u = |A| cos θ is the component � �ı � ˆ of A along the direction of u. E.g., A · ˆ = component of A along x-axis. � Example: pendulum making an angle with vertical, force = weight of pendulum F pointing � along tangential downwards: then the physically important quantities are the components of F direction (causes pendulum’s motion), and along normal direction (causes string tension). 1 Area. E.g. of a polygon in plane: break into triangles. Area of triangle = 2 base × height = 1� � 2 |A||B | sin θ (= 1/2 area of parallelogram). Could get sin θ using dot product to compute cos θ and sin2 + cos2 = 1, but it gives an ugly formula. Instead, reduce to complementary angle θ� = π /2 − θ � � by considering A� = A rotated 90◦ counterclockwise (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 � = a1 b2 − a2 b1 . � � � � � , B ) = � a1 a2 � = a1 b2 − a2 b1 . � Determinant. Definition: det(A � b1 b2 � � � � a1 a2 � � � = ± area of parallelogram. Geometrically: � b1 b2 � � � The sign of 2D determinant has to do with whether B is counterclockwise or clockwise from A, without details. � � � � � � � � � a1 a2 a3 � � � � � � � � � � B, C ) = � b1 b2 b3 � = a1 � b2 b3 �−a2 � b1 b3 �+a3 � b1 b2 �. �� Determinant in space: det(A, � c1 c2 � � c1 c3 � � � � c2 c3 � � c1 c2 c3 � ��� Geometrically: det(A, B, C ) = ± volume of parallelepiped. Referred to the notes for more about determinants. Cross-product. (only for 2 vectors in space); gives a vector, not a scalar (unlike dot-product). � � �ˆ ˆ k� � � � � � � j ˆ� �ı � a2 a3 � � a1 a3 � � � ˆ � a1 a2 �. �� � j� � Definition: A × B = � a1 a2 a3 � = ˆ � � � ı � b2 b3 � − ˆ � b1 b3 � + k � b1 b2 � � b1 b2 b3 � (the 3x3 determinant is a symbolic notation, the actual formula is the expansion). � � �� Geometrically: |A × B | = area of space parallelogram with sides 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 . ˆ Flashcard Question: ˆ × ˆ =? (answer: k, checked both by geometric description and by ıj calculation). � ��ˆ Triple product: volume of parallelepiped = area(base) · height = |B × C | (A · n), where n = ˆ � × C/|B × C |. So volume = A · (B × C ) = det(A, B, C ). The latter identity can also be checked �� � �� � ��� B directly using components. ...
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