MSCI 300 SPRING 2016 Calculus 1 Charles Rubenstein
MSCI 300 – SPRING 2016 Calculus 1 Charles Rubenstein, Ph. D. Professor of Engineering and Information Science Week 3: Session 2: Monday 02/01/16 Mondays 6: 30 pm-8: 50 pm PMC 705 A
Instructor Contact Information Dr. Charles Rubenstein <crubenst@pratt. edu> Professor of Engineering & Information Science Office hours (by appointment *) • Mondays: 5: 00 pm-6: 00 pm Pratt Manhattan Campus Office: PMC 604 -C • Tuesdays: 12: 00 pm - 2: 00 pm Pratt Brooklyn Campus Office: ARC G-45 (or E-08 Lab) (*Please email me at least a day in advance if you plan on coming to office hours…) Send me an email … crubenst@pratt. edu Subject line: 300 Calc Copyright © 2016 C. P. Rubenstein 2
MSCI 300 – Fall 2015 - Class Schedule & Due Dates Monday (Week) NOTES 18 January (1) NO CLASS – Martin Luther King Day 25 January (2) Introduction; Review of Syllabus, Algebra & Trigonometry 1 February (3) Functions and Slopes (Quiz 1) 8 February (4) Approximating Slopes, Limits, The Derivative (Quiz 2) 15 February (5) Rules of Differentiation (Q 3) 22 February (6) Max/Min Prob, 2 nd + Derivatives (Q 4) 29 February (7) Trig, Trig Derivatives, Limit of sin(x)/x (Q 5) 7 March (8) Derivatives of Exponentials, Constant "e"; Take Home Exam (Q 6) 14 March (9) NO CLASSES – Pratt Spring Break – 14 -20 March 2016 21 March (10) Linear Approximation, Newton's Method; Midterm Due (Q 7) 28 March (11) l'Hôpital's Rule, Kinematics: Position, Velocity, Acceleration ; Midterm Review (Q 8) 4 April (12) CMFM Seminar Review Logs (Q 9) 11 April (13) NO CLASS – Instructor out of town 18 April (14) Inverse Function Derivatives, Implicit Differentiation (Q 10) 25 April (15) Areas, Intro to Integrals, Fundamental Theorem of Calculus 2 May (16) Using Integrals to find Volumes and Lengths, Review 9 May (17) In-class Final Examination (* Quizzes on Homework due; Reviewed in same session) Copyright © 2016 C. P. Rubenstein 3
Spring 2016 Math/Science TUTORING WHO: Professor Joe Guadagni BY APPOINTMENT ONLY: WHEN and WHERE: Mondays 4: 30 -6: 00 pm PMC 403 Email: jguadagni@pratt. edu Tuesdays 12 -1: 30 pm WTC North Hall Call: 718 -636 -3459 Copyright © 2016 C. P. Rubenstein 4
In Class #02 • • • DUE: Homework Set #01 Reading: Strang, Calculus Chapter 1 2 Do: Quiz #01 Review: Homework Set #01; Quiz #01 Lecture and Problem Review: Functions and Slopes Next class – Session 3: • • DUE: Homework Set #02 Reading: Strang - Chapter 2: Derivatives 2 Do: ; Quiz #02 2 Do: Review Programming your Calculator: SLOPE BRING YOUR CALCULATOR TO CLASS!!! • Lecture and Problem Review: Approximating Slopes, Limits, The Derivative Copyright © 2016 C. P. Rubenstein 5
Archives & Instructor Information Online Now at : www. Charles. Rubenstein. com/300 16 sp 01. pdf and /165 sp 01 h. pdf NOTES SETS: 01 Notes. pdf (Class Notes for Session #01) 02 Notes. pdf (Class Notes for Session #02) 03 Notes. pdf (Class Notes for Session #03) 04 Notes. pdf (Class Notes for Session #04) 05 Notes. pdf (Class Notes for Session #05) 06 Notes. pdf (Class Notes for Session #06) Available by Thursday evening: /16 sp 02. pdf and /16 sp 02 h. pdf (slide set as slides & handouts) Copyright © 2016 C. P. Rubenstein 6
Quick Review Some highlights of the Syllabus Copyright © 2016 C. P. Rubenstein 7
About Your Grades Assignments are designed to see what you have learned - NOT to see what you DIDN’T learn 1. Homework Quiz value = 30% of final grade – HOMEWORK IS DUE NEXT CLASS SESSION – There will be a five (5) minute quiz on one homework problem for each of ten (10) homework assignments. A set of ALL homework assignments will be provided the first class session. There is NO excuse for not doing homework, quiz. 2. Total Exams value = 70% of final grade – Take Home Midterm = 30% – Final Exam = 40% Copyright © 2016 C. P. Rubenstein 8
Attendance Requirements You are responsible for attending and participating in all class sessions. TWO (2) unexcused absences (no Dr. note, etc. ) will result in ONE grade lower final grade (e. g. , if you earned a B+, get a C+) THREE (3) unexcused absences (no Dr. note, etc. ) will result in the final grade of “F” Copyright © 2016 C. P. Rubenstein 9
Recommended Textbook* Pub Date: 1991 Publisher: Wellesely-Cambridge Press *FREE Online at this Web. Link: http: //ocw. mit. edu/ans 7870/resources/Strang/Edited/Calculus. pdf Calculus Strang, Gilbert Copyright © 2016 C. P. Rubenstein 10
Math 300 – Chapter Topics Calculus (1991). Strang, Gilbert. Wellesely-Cambridge Press. http: //ocw. mit. edu/ans 7870/resources/Strang/strangtext. htm First six Chapters*: CHAPTER 1 : Introduction to Calculus CHAPTER 2 : Derivatives CHAPTER 3 : Applications of the Derivative CHAPTER 4 : The Chain Rule CHAPTER 5 : Integrals CHAPTER 6 : Exponentials and Logarithms *You are expected to be reading at least the chapter we will be working on in class, with new chapters assigned every other week and should be doing chapter problems as needed. Copyright © 2016 C. P. Rubenstein 11
Questions? Copyright © 2016 C. P. Rubenstein 12
Homework #01 Quiz Algebra/Trigonometry Review Problems When you DO NOT show work, I have to guess. When you DO show work, I can try to see what you are doing and give an “O. K. ” Copyright © 2016 C. P. Rubenstein 13
Homework QUIZ #01 #12. “Factor the following expression” 6 x 2 + 5 x + 1 You have five minutes to solve Copyright © 2016 C. P. Rubenstein 14
Homework QUIZ #01 Review #12. “Factor the following expression” 6 x 2 + 5 x + 1 Answer: Try possible “x” factor coefficients: 6, 1 and 3, 2 (3 x +1)(2 x +1) NOTE: this is a partial equation as there is only one side. Only if the question was “Solve 6 x 2 + 5 x + 1 = 0” would we be able to state the roots were -1/3 and -1/2 Copyright © 2016 C. P. Rubenstein 15
Homework #01 REVIEW Algebra/Trigonometry Review Problems (18 problems; 18 b = 2 bonus points) When you DO NOT show work, I have to guess. When you DO show work, I can try to see what you are doing and give an “O. K. ” Copyright © 2016 C. P. Rubenstein 16
Homework #01 Review #1. “Find the value (or values) of x that satisfy the following equation” 2 x = 9 Ans: Divide both sides by 2: 2 x/2 = 9/2 x = 9/2 Or x = 4. 5 Copyright © 2016 C. P. Rubenstein 17
Homework #01 Review #2. “Find the value (or values) of x that satisfy the following equation” x/3 = 22 Ans: Multiply both sides by 3: 3 • x/3 = 3 • 22 which yields x = 66 Copyright © 2016 C. P. Rubenstein 18
Homework #01 Review #3. “Find the value (or values) of x that satisfy the following equation” 2 x + 3 = x + 8 Ans: Subtract ‘x’ and Subtract ‘ 3’ from both sides: 2 x (–x) + 3 (– 3) = x (- x) + 8 (- 3) x +0=0+5 Yields: x=5 Copyright © 2016 C. P. Rubenstein 19
Homework #01 Review #4. “Find the value (or values) of x that satisfy the following equation” 3/(x+2) = 5 Ans: Multiply both sides by (x+2) [3/(x+2)] (x+2) = 5 (x+2) 3 = 5 x+10 5 x = -7/5 Copyright © 2016 C. P. Rubenstein 20
Homework #01 Review #5. “Find the value (or values) of x that satisfy the following equation” x (x -5) = 0 Ans: By inspection is: (x 0)(x -5) = 0 Therefore: x = 0 and x = 5; You could multiply through to get x 2 - 5 x = 0 from which x 2 = 5 x; then Divide by x, to get x = 5 BUT you’ll lose the solution x = 0. The problem is the division… If x = 0, both sides of the equation are multiplied by 0/0 which is undefined. Copyright © 2016 C. P. Rubenstein 21
Homework #01 Review #6. “Find the value (or values) of x that satisfy the following equation” x 2 = x Ans: factoring yields: Therefore: x 2 = x x (x-1) = 0 (x – 0) (x-1) = 0 Thus, by inspection: x = 0 and x = 1 (see note in solution to Prob. 5 about 0/0 … ) Copyright © 2016 C. P. Rubenstein 22
Homework #01 Review #7. “Find the value (or values) of x that satisfy the following equation” x 2 + 4 x + 4 =0 Ans: Which can be factored: (x+2) = 0 and therefore, x = -2 which is a double root. Copyright © 2016 C. P. Rubenstein 23
Homework #01 Review #8. “Find the value (or values) of x that satisfy the following equation” x/3 = 3/x Ans: Multiply both sides by 3 x: x/3 • 3 x = 3/x • 3 x x 2 = 9 Therefore x = 3 and x = -3 Note: another way to write this is x = 3 since squaring a negative number yields a positive result… Copyright © 2016 C. P. Rubenstein 24
Homework #01 Review #9. “Find the value (or values) of x that satisfy the following equation” x 2 + 2 x = 5 Ans: You can complete the square by adding ‘ 1’ to both sides: x 2 +2 x + 1 = 5 + 1 (x+1)2 = 6 Taking square root of both sides: x+1 = ± 6 Therefore, x = -1 ± 6 x = -1 ± 2. 45 Thus: x = +1. 45 , x = -3. 45 You could have also used the quadratic formula. Copyright © 2016 C. P. Rubenstein 25
Homework #01 Review #10. “Find the value (or values) of x that satisfy the following equation” 2 x 2 + πx - 3 = 0 Ans: Use the quadratic formula : x= x= (where π ≈ 3. 14159; π/4 = 0. 7854 ≈ 0. 79; π2 ≈ 9. 87; 9. 87+ 24 = 33. 87; 33. 87/16 ≈ 2. 12) x ≈ - 0. 79 2. 12; or x ≈ - 0. 79 1. 46 x ≈ - 2. 25; x ≈ + 0. 67 Copyright © 2016 C. P. Rubenstein 26
Homework #01 Review #11. “Factor the following expressions” x 2 - 2 x +1 Ans: By inspection: (x-1) = (x-1)2 Copyright © 2016 C. P. Rubenstein 27
Homework #01 Review #12. “Factor the following expressions” 6 x 2 + 5 x + 1 Ans: (3 x +1)(2 x +1) Copyright © 2016 C. P. Rubenstein 28
Homework #01 Review #13. “If a/b = c/d , complete the following” a=? Ans: Multiply both sides by b to get b(a/b) = b(c/d) Thus: a = cb/d Copyright © 2016 C. P. Rubenstein 29
Homework #01 Review #14. “If a/b = c/d , complete the following” b=? Ans: Invert the equation a/b = c/d : b/a = d/c Multiply by a: a • [b/a] = a • [d/c] thus ab/a = ad/c or b = ad/c Copyright © 2016 C. P. Rubenstein 30
Homework #01 Review #15 “If a/b = c/d , complete the following” a/c = ? Ans: Divide the ans. to Prob. 13 a = cb/d by c: a/c = cb/dc Thus: a/c = b/d Copyright © 2016 C. P. Rubenstein 31
Homework #01 Review #16. “If a/b = c/d , complete the following” ad / bc = ? Ans: to get Cross multiply a/b = c/d ad = bc Then (ad) / (bc) = (ad) / (ad) = 1 Copyright © 2016 C. P. Rubenstein 32
Homework #01 Review #17. “Expand the Expression” (a +b +c)3 = (a+b+c)(a 2 +b 2 +c 2 + 2 ab +2 bc +2 ac) = a 3 + ab 2 + ac 2 + 2 a 2 b +2 abc +2 a 2 c +ba 2 +b 3 +bc 2 +2 bab + 2 b 2 c +2 bac +ca 2 +cb 2 +c 3 +2 cab +2 cbc +2 cac (a +b +c)3 = a 3 +b 3 +c 3 + 6 abc +3 a 2 b + 3 a 2 c + 3 bc 2 + 3 ac 2 +3 b 2 a +3 b 2 c Copyright © 2016 C. P. Rubenstein 33
Homework #01 Review #18 a. Show that (p 2 - q 2)2 + (2 pq)2 = (p 2 + q 2)2 Carrying out the multiplications: p 2 2 - 2 p 2 q 2 + q 2 2 + 4 p 2 q 2 = p 2 2 + 2 p 2 q 2 + q 2 2 We see that the left and right sides are identical. Note that (p 2 + q 2) is, therefore, the hypotenuse of a right triangle whose other sides are p 2 - q 2 and 2 pq. If p and q are integers with p > q > 0, the resulting right triangle has sides whose lengths are all integers which equals the 3 – 4 – 5 “perfect” right triangle. Copyright © 2016 C. P. Rubenstein 34
Homework #01 Review #18 b (bonus) “Use 18 a to find three more perfect right triangles (“Pythagorean triples”)” Let p=2, q=1 : p 2 -q 2 = 4 -1 = 3 ; 2 pq = 4 ; & p 2 + q 2 = 4+1= 5 This is the familiar 3 – 4 - 5, right triangle. Let p=3, q=2 : p 2 -q 2 = 9 -4 =5 ; 2 pq = 12 ; & p 2 + q 2 = 9+4 = 13 Let p=4, q=1 : p 2 -q 2 = 16 -1 =15 ; 2 pq = 8 ; & p 2 + q 2 = 16+1 = 17 Let p=4, q=3 : p 2 -q 2 = 16 -9 = 7 ; 2 pq = 24 ; & p 2 + q 2 = 16+9 = 25 Let p=3, q=1 : p 2 -q 2 = 9 -1 = 8 ; 2 pq = 6 ; & p 2 + q 2 = 9+1 = 10 These are just 3, 4, 5 triangles scaled by 2 so they are not a new shape. To get new shapes, p and q must be relatively prime with one odd and one even value. Copyright © 2016 C. P. Rubenstein 35
Questions? Copyright © 2016 C. P. Rubenstein 36
Class #02: Functions and Slopes Graphing in the x-y plane Graphs of Equations Finding the Slope Equations of a Straight Line The Point Slope Formula The Two-Point Formula The Distance Formula (Pythagorean Theorem) Perpendicular lines (negative reciprocal slopes) The Midpoint of a Line Segment (average x, y) Finding the Intersection of two lines Equation of a Vertical Line Formulas and Functions Combinations of Functions of More than One Variable Copyright © 2016 C. P. Rubenstein 37
Graphing in the x-y plane Rectangular x-y graph is the most common way to represent related pairs of numbers: x, y Origin: point of intersection of the x-axis and the y-axis Four points shown on the graph: 0, 0 – origin -2, 1 -3, 4 and 2, -3 Copyright © 2016 C. P. Rubenstein 38
Graphs of Equations y = 2 x+3 is an equation with two variables, x and y. There an infinite number of points x, y , called a locus of points, which satisfy this equation. They are plotted on the graph and form a continuous line on the graph. Five points satisfying this equation are x values: 1, 0, -1, -2, and -3 y = 2 x+3 x y 1 5 0 3 -1 1 -2 -1 -3 -3 Copyright © 2016 C. P. Rubenstein 39
Graphs of Equations y = mx + b is the equation for a straight line When x =0, y = b, so b is called the y-intercept of the line. The constant m is called the slope of the line The larger the value of m, the faster the line climbs as we move in the positive x direction. For any segment of the line, the rise (change in y) divided by the run (change in x) is equal to m Copyright © 2016 C. P. Rubenstein 40
Finding the Slope of a Line The slope is rise over run. If we have two points, x 1, y 1 and x 2, y 2 the slope of the straight line connecting them is (y 2 -y 1)/(x 2 -x 1) = m In our example, x 1, y 1 is 1, 5 ; x 2, y 2 is -3, -3 and the slope, (y 2 -y 1)/(x 2 -x 1) is (-3 – 5) / (-3 – 1) = -8/-4 = m = 2 If m = 2, when x = 0, y = 3 Thus the equation is: y = 2 x+3 Slope Example: Find the slope between the points 2, 1 and 5, 3 Ans: m = (3 – 1) / (5 – 2) = 2/3 Copyright © 2016 C. P. Rubenstein 41
Equations of a Straight Line If we need the equation of a straight line and do not have m and b, we need to calculate them from whatever we do have. The Point Slope Formula Given the slope m, that passes through a point x 1, y 1 We can write: y = mx + b Substituting x 1, y 1 for x and y we have: y 1 = mx 1 + b Subtract the 2 nd equation from the 1 st: y - y 1 = mx - mx 1 Factor the right hand side: y - y 1 = m(x - x 1) This result is called the Point Slope Formula Rearrange to get the y = mx +b form: y = mx + (y 1 - mx 1) From which we can determine the intercept: b = y 1 -mx 1 Copyright © 2016 C. P. Rubenstein 42
Point Slope Formula Examples 1. Find the equation of the line having slope 3 that passes through the point 2, 5 Using the Point Slope Formula y - y 1 = m(x - x 1) y - 5 = 3(x - 2) Rearranging, y = 3 x - 1 2. Find the equation of the line having slope 5 that passes through the point 1, 2 Using the Point Slope Formula y - y 1 = m(x - x 1) y - 2 = 5(x - 1) Rearranging, y = 5 x - 3 Copyright © 2016 C. P. Rubenstein 43
Two-Point Formula To find the equation of a straight line given two points, One can use the Two-Point Formula: But this is harder to remember than first calculating the slope, m = (y 2 - y 1)/(x 2 - x 1) and then using the Point Slope Formula y - y 1 = m(x - x 1) and finally calculate the intercept b = (y 1 x 2 - x 1 y 2) / (x 2 -x 1) Example: Find the equation of the line that passes through the points 2, 5 and 3, -1. Ans: The slope is (-1 - 5)/ (3 - 2) = -6. The Point Slope Formula gives y - 5 = -6(x - 2) = -6 x + 12 Rearranging, this becomes y = -6 x +17 Copyright © 2016 C. P. Rubenstein 44
The Distance Formula The (straight-line) distance between two points, x 1, y 1, and x 2, y 2 is given by D 2 = (x 1 - x 2)2 + (y 1 - y 2)2 The figure here shows the calculation of the distance between the points 3, 1 and 2, -3 The distance formula is nothing more than the Pythagorean Theorem applied to a right triangle whose hypotenuse is the distance to be found. Copyright © 2016 C. P. Rubenstein 45
Proving the Pythagorean Theorem A common proof begins by drawing the right triangle and then constructing a square, c × c, on its hypotenuse. The next step is to add three more identical triangles to form a larger square, (a+b) × (a+b). The area of the larger square must equal the area of the smaller square plus the area of the four triangles: Outer square area = inner square area + 4 x area of triangle (a+b)2 = c 2 + 4 (ab/2) Using FOIL to expand the left hand side produces a 2 + 2 ab + b 2 = c 2 + 2 ab Subtracting 2 ab from both sides leaves a 2 + b 2 = c 2 Copyright © 2016 C. P. Rubenstein 46
Perpendicular Line Slopes DEFINITION: A straight line has slope m, therefore, any straight line perpendicular to it must have slope -1/m Perpendicular Line slopes are negative reciprocals of each other. β γ Slope of Line 1 = a/b Slope of Line 2 = - a/c But a/c = b/a (similar triangles) therefore Slope of Line 2 = - b/a = -1 / (Slope of Line 1) Copyright © 2016 C. P. Rubenstein 47
About Perpendicular Lines In the figure below, Line 1 and Line 2 are perpendicular. The slope of Line 1, its rise over run, can be seen to be a/b. The slope of Line 2 can be seen to be -a/c, where the negative sign is included to show that an increase in the x direction produces a decrease in the y direction. Line 1 a Line 2 b Copyright © 2016 C. P. Rubenstein γ c 48
The Perpendicular Line Proof The right triangles that define the slopes of Line 1 and Line 2 are similar triangles*, so a/c = b/a. Therefore, the slope of Line 2 , -a/c , is equal to -b/a, the negative reciprocal of the slope of Line 1. *To see that these triangles are similar, study the figure and: Note that as the angle γ is shared in ADB and ADC: angle β must be the same as angle Line 1 PROOF for Right Triangles: For adc: 180 – ( γ + 90 ) = β For adb: 180 – ( γ + 90 ) = From which: ≡ β Copyright © 2016 C. P. Rubenstein γ a Line 2 b c 49
Perpendicular Line Example: Find the equation of the line that passes through the origin (0, 0) and is perpendicular to the line y = 5 x +2. You have 5 Minutes for this problem: a. If the equation for Line 1 is y = 5 x +2 then m = 5 and b = 2 Slope of Line 2 = -1 / (Slope of Line 1) = -1 / m The slope of the perpendicular line, Line 2 is -1/5 b. Using the Point Slope Formula y - y 1 = m(x - x 1) where the origin point is 0, 0; y - 0 = -1/5 ( x - 0) Thus the Equation for Line 2 is: y = -x/5 or y = -1/5 x Final Ans Intercept = 0 indicates that the line goes through the origin Copyright © 2016 C. P. Rubenstein 50
Midpoints of Line Segments The figure below illustrates that: the midpoint's x-coordinate is the average of the x-coordinates the midpoint‘s y-coordinate is the average of the y-coordinates Copyright © 2016 C. P. Rubenstein 51
Intersection of Two Lines At the intersection point of two lines, lines have equal y values (and the same x value). Equate the “ y = “ expressions, solve for x Example: Find the intersection of y = x + 1 and y = 3 x - 3 Ans: Equate the y values: x +1 = 3 x - 3 ; this produces x = 2 for the x value of the intersection. The y value follows from using x = 2 in either equation: y = x + 1 = 2 + 1 = 3 or y = 3 x - 3 = 3 • 2 - 3 = 3 Copyright © 2016 C. P. Rubenstein 52
Equations of Horizontal and Vertical Lines The horizontal line y = b is a special case of y = mx +b where m = 0 Likewise, x = a represents a vertical line How can we prove this with the general form, y = mx +b when m is infinite? Copyright © 2016 C. P. Rubenstein 53
Equations of Horizontal and Vertical Lines PROOF: Write the equation for a slanted line passing though a, b where b is an arbitrary point: y-b = m(x - a) Now divide both sides by m to get (y-b)/m = x - a In the limiting case, where m goes to infinity (increasingly vertical) the left-hand side of this equation goes to zero This leaves 0 = x-a which is the same as: x = a Copyright © 2016 C. P. Rubenstein 54
Formulas To graph the y vs. x relation contained in an equation such as 2 x + y = 3 x 2 First rearrange it to read y = 3 x 2 – 2 x a formula that, given a value for x, produces a corresponding y. We call such a formula a function of x, where x is the independent variable, We use parentheses to indicate the functional relationship y(x) = 3 x 2 - 2 x where y is known as the dependent variable. Copyright © 2016 C. P. Rubenstein 55
Expressions and Arguments The expression y(x) is read as “y of x” where x is called the argument of the function The parentheses do not indicate multiplication; they simply indicate that the value of y depends on the value of x. We have been using example that are polynomial functions but there are many other basic kinds of functions such as sin(x), a trigonometric function, and 2 x , an exponential function Copyright © 2016 C. P. Rubenstein 56
Functions In many problems we deal with more than just one function. We can distinguish them with different letters such as f(x), g(x), h(x), etc. When we are going to graph a function vs. x, we often denote the function as y(x), since the points on the graph are then the set x, y(x). But when we are discussing the properties of a function rather than merely graphing it, we often denote it as f(x) rather than y(x). Copyright © 2016 C. P. Rubenstein 57
Combination of Functions We can combine functions to make new functions such as h(x) = f(x) + g(x) or h(x) = f(x) • g(x) or h(x) = f [g(x) ], which is a “function of a function” Examples of arbitrary functions constructed from the functions f(x) = 3 x 2 - 2 x and g(x) = 10 x S(x) = f(x) + g(x) = (3 x 2 - 2 x) + 10 x P(x) = f(x) • g(x) = (3 x 2 - 2 x) • 10 x Q(x) = f [ (g(x) ] = (3 • (10 x)2 - 2 • (10 x)) R(x) = g [ (f(x) ] = 10 Copyright © 2016 C. P. Rubenstein 58
Functions of more than one variable In this course we concentrate on functions of a single variable, usually denoted as x. But functions can depend on several variables, for example, z(x, y) = x 2 - y 2 = a function of two variables. For any point x, y in the x y plane, the value of z can be plotted in the z direction. The set of all such points forms a surface. The surface formed by this example is a hyperbolic paraboloid … a saddle-shape surface (similar to the shape that forms the roof of Pratt's ARC building in Brooklyn) Copyright © 2016 C. P. Rubenstein 59
Questions? Copyright © 2016 C. P. Rubenstein 60
From Class #03: Slopes and Rates of Increase For straight line functions y = mx +b The slope, m, is just a single number which, for any segment of the line, is the ratio of the rise to run. Rise is the increase in the y value produced by a run in the positive x direction. Note that the constant b just shifts the line up or down in the y direction but doesn't affect the slope or its value. Copyright © 2016 C. P. Rubenstein 61
Class #03: Finding the Slope of a function The figure below shows a graph of the straight line passing through the points 1, 3 and 3, 4. Looking at this segment of the line, we see that the rise over run is (4 -3)/(3 -1) = 1/2. Using this value for the slope, the point slope formula leads us to function that describes the line, y(x) = (1/2)x +5/2 Note that we could have selected any two points to find the slope. Copyright © 2016 C. P. Rubenstein 62
From Class #03: Slope Calculation From the figure below, we can express the slope symbolically using the functional notation, y(x) where x 2 and x 1 are arbitrarily chosen points: Please note: Using the Point Slope Formula we do not need to remember additional formulas… Copyright © 2016 C. P. Rubenstein 63
For Class Session #03: Next class – Session 3: • DUE: Homework Set #02 • Reading: Strang - Chapter 2: Derivatives • 2 Do: ; Quiz #02 Review: Quiz and Homework #02 • 2 Do: Review Programming your Calculator: SLOPE BRING YOUR CALCULATOR TO CLASS!!! • Lecture and Problem Review: Approximating Slopes, Limits, The Derivative For class Session #04: • DUE: Homework Set #03 • 2 Do: Rules of Differentiation • In Class Quiz and Review: Homework Set #03 • Lecture and Problem Review Copyright © 2016 C. P. Rubenstein 64
Any Questions? Send me an email … crubenst@pratt. edu or c. rubenstein@ieee. org Copyright © 2016 C. P. Rubenstein 65
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