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by: Kevin Green

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MA 26100 Multivariable Calculus MA 26100

Marketplace > Purdue University > Math > MA 26100 > MA 26100 Multivariable Calculus
Kevin Green
Purdue

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Lecture Notes for lecture 1 delivered on 8/22/16
COURSE
Multivariate Calculus
PROF.
Kuan-Hua J. Chen
TYPE
Class Notes
PAGES
3
WORDS
CONCEPTS
vectors, dot product, cross product
KARMA
Free

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This 3 page Class Notes was uploaded by Kevin Green on Tuesday August 23, 2016. The Class Notes belongs to MA 26100 at Purdue University taught by Kuan-Hua J. Chen in Fall 2016. Since its upload, it has received 117 views. For similar materials see Multivariate Calculus in Math at Purdue University.

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Date Created: 08/23/16
8/22/16 MA 26100 Multivariable Calculus Introduction: Professor: Dr. Joseph Chen, Purdue University Office Hours: MWF 9:00 – 11:00 AM in MATH 848 Textbook: Calculus: Early Transcendentals – 7 Edition, Stewart Exam Dates: Exam 1 – 9/29/16, Exam 2 – 11/8/16 (100 Points Each), Final – TBD (200 Points) Class is worth 550 Points Review: 12.1, 3 Dimensional Coordinate System  Points are given in Cartesian, (x,y,z), format  Distance between 2 points [(x ,y ,z ) and (x ,y ,z )] given by: 1 1 1 2 2 2 √(x −x ) +(y −y ) +(z −z ) =d 2 o 2 1 2 1 2 1 o This is known as the distance formula  The equation for a sphere with center at (h, k, l) and a radius r is given as follows: 2 2 2 2 o (x−h) +(y−k) +(z−l) =r o This is the standard form of the equation EX. 1 Identify the center and radius of a sphere with the equation: 2 2 2 x +y +z =4x−2y+4 SOLUTION We need to first convert this equation to standard form by completing the square, which will result in the following equation x −4x+y +2 y+z =4 2  Get all the variables to one side 2 2 2  Complete the square x −4x+4+y +2y+1+z =4+4+1  Simplify 2 2 2 (x−2) +(y+1) +z =9 CENTER AT (2,-1,0) RADIUS = 3 EX. 2 What is the shape and equation of intersection with the yz plane? Shape is a circle, as the intersecting plane is vertical To find the equation, set x to zero. We can do this because there is no variation in the x plane  Set the x term to zero  Evaluate x term  Simplify (0−2) +(y+1) +z =9 4+(y+1) +z =9 (y+1) +z =5 CENTER AT (-1,0) RADIUS SQUARE ROOT OF 5 EX. 3 What is x = 6 in 3D? Plane parallel to yz plane EX. 4 What is 0≤ y≤4 ? The area between y = 0 and y = 4. This area is “free” as there are no bounds on x and z Review: 12.2, Vectors Definition of Vector: a quantity with magnitude (length) and direction a=¿2,1>¿ 2i+ j  are both examples of 2 dimensional vectors ⃗  b=¿1,2,−3>¿i+2 j−3k are both examples of 3 dimensional vectors The magnitude of a vector ( a=¿ a 1a 2 a3>¿ ) is calculated as follows: a⃗= √ +1 +a22 3 Review: 12.3, Dot Products ⃗ ⃗=¿a ,1 ,2 >3 , b=¿b ,1 ,2 >3 ⃗ Definition of Dot Product ⃗∗b=a ∗1b +a1∗b 2a 2b 3 3 THIS QUANTITY IS A SCALAR ⃗∗b=¿⃗ a∨¿b∨cos? (θ) TO BE USED IF NEEDING TO FIND THE ANGLE BETWEEEN 2 VECTORS ⃗  If a is orthogonal to b , then the result of taking the dot product is zero EX. 5 Determine if a triangle defined by P(1,0), Q(2,3), R(1,4) is a right triangle. ⃗  Establish 3 vectors ⃗,b⃗,c ⃗=¿0,4>,b=¿1,3>,c ⃗=¿−1,1>¿  Perform 3 Dot Product operations ⃗∗b= (0∗1 + 4∗3 )=12 b∗c⃗= (1∗1 + 3∗1 =2 a∗⃗c= (0∗−1 )+(4∗1 =4  Since none of these operations results in a value of zero, this triangle is not a right triangle Review: 12.4, Cross Products i j k a=¿a ,a ,a >¿ b=¿b ,b ,b >¿ a a a 1 2 3 , 1 2 3 1 2 3 b1 b2 b3 To take a cross product, take the determinant of this 3X3 matrix (prerequisite knowledge for this class) ⃗ a xb will always result in a VECTOR perpendicular to both a and b ⃗xb⃗| is equal to the area of a parallelogram formed by a and b

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