MSCI 300 FALL 2015 Calculus 1 Charles Rubenstein
MSCI 300 – FALL 2015 Calculus 1 Charles Rubenstein, Ph. D. Professor of Engineering and Information Science Week 7: Session 5: Monday 10/05/15 Mondays 6: 30 pm-8: 50 pm PMC 705 A
Instructor Contact Information Dr. Charles Rubenstein <crubenst@pratt. edu> Professor of Engineering & Information Science Fall 2015 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 © 2015 C. P. Rubenstein 2
MSCI 300 – Fall 2015 - Class Schedule & Due Dates (* Quizzes on Homework due; Reviewed in same session) Copyright © 2015 C. P. Rubenstein 3
* Class Session Archives * http: //www. Charles. Rubenstein. com/300/ 15 fa 05. pdf (Class Power. Point slides) * 15 fa 05 h. pdf (slides in handout format) * *Archive materials normally online by Thursday evenings Copyright © 2015 C. P. Rubenstein 4
FALL 2015 TUTORING AVAILABLE Copyright © 2015 C. P. Rubenstein 5
In Class #05 CM/FM Seminar 6: 30 pm-7: 30 pm – PMC 213 • DUE: Homework Set #04 • Reading: Strang - Chapter 3: Applications of the Derivative • • In Class Quiz and Review Homework Set #04 Lecture and Problem Review: Max/Min Problems, Second and Higher Derivatives • Distribute: Take Home Midterm Exam NO Class – 12 October - Columbus Day/Fall Break For class Session #06: • • • DUE: MIDTERM! Homework Set #05 Reading: Strang - Chapter 3: Applications of the Derivative 2 Do: In Class Quiz and Review Homework Set #05 Lecture and Problem Review: Trigonometry Review Distribute: Remaining Class Note Sets Copyright © 2015 C. P. Rubenstein 6
Questions? Copyright © 2015 C. P. Rubenstein 7
Homework #04 Quiz Equations of Straight Lines 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 © 2015 C. P. Rubenstein 8
Homework QUIZ #04 #2 b 1. Find the value of the slope of the function f(x) = 2 x 3 + 3 x 2 + x +1 and calculate its value at x = 1 You have five minutes to solve! Copyright © 2015 C. P. Rubenstein 9
Homework QUIZ #04 From # 2. Find the value of the slope of the function f(x) = 2 x 3 + 3 x 2 + x +1 and calculate its value at x = 1 Ans: f(x) = 2 x 3 + 3 x 2 + x +1 f '(x) = 6 x 2 + 6 x + 1 + 0 and when x= 1 f '(1) = (6*1 + 1) = 13. Copyright © 2015 C. P. Rubenstein 10
Homework #04 REVIEW Equations of Straight Lines (9 problems; 14 parts) 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 © 2015 C. P. Rubenstein 11
Homework #04 Review #1 a. For the function graphed below 1 a. Estimate the values of x for which the derivative of y(x) is zero. Ans: By inspection of the graph: The derivative is zero (curve is flat) at approximately x = -0. 9 and x = 2. 8 Copyright © 2015 C. P. Rubenstein 12
Homework #04 Review #1 b. For the function graphed below 1 b. Identity the point for which the derivative has the highest (most positive) values and estimate the value of the derivative at this. Ans: By inspection of the graph: The derivative is most positive at the left side of the graph and its value there is approximately [-25 - (-70)] / [-3 -(-4)] = 45 Copyright © 2015 C. P. Rubenstein 13
Homework #04 Review #1 c. For the function graphed below 1 c. Find the point at which the derivative is a minimum (least positive or most negative) and estimate the value of the derivative at this point. Ans: By inspection of the graph: The derivative is most negative at approximately x =1 and its value there is approximately (0 -20)/(2 -0) = -10 Copyright © 2015 C. P. Rubenstein 14
Homework #04 Review #1 d. For the function graphed below 1 d. On the same graph, draw in a rough sketch of the derivative of y(x). Ans: By inspection of the graph: Note that the slope graph is zero when the original curve levels off. The slope graph is a positive number when the original graph slope is positive going and a negative number when it is negative going. Copyright © 2015 C. P. Rubenstein 15
Homework #04 Review #2 a. Estimate the value of the slope of the function f(x) = 2 x 3 + 3 x 2 + x +1 at the point 1, f(1) Use the rise over run formula, [ f(x+dx) - f(x) ] /dx. Let dx =. 001 Ans slope ≈ 2(x+dx)3 + 3(x+dx)2 + (x+dx) + 1 for x = 1, dx =. 001 therefore slope ≈ ([2(1+. 001)3 +3(1+. 001)2 + (1+. 001) +1] – [2(1)3 +3(1)2 + 1 +1]) /. 001 m = 13. 0090 Copyright © 2015 C. P. Rubenstein 16
Homework #04 Review #2 b 1. Find the derivative (derivative function) of this f(x) and calculate its value at x=1 (This was the QUIZ PROBLEM) Ans f '(x) = 6 x 2 +6 x +1 when x= 1 f '(x) = (6 + 1) = 13 #2 b 2. Calculate the percentage error of the estimate found in 2 a. Error is (m 1 – m 2) / m 2 2 a. : m 1= 13. 0090 and 2 b 1. : m 2= 13. 000 0. 0090/13 = 6. 9231 e-4 = 0. 00069 Percentage Error = (Error • 100) % 0. 069% Copyright © 2015 C. P. Rubenstein 17
Homework #04 Review #3 a. You have a 100 ft. roll of fencing with which to make a rectangular corral that includes a divider. Find the width of the corral that will produce the maximum total area. Hint: Let x be the height of the sides and the divider. The top and bottom will then each have width (100 -3 x)/2 where 3 x is the amount of material needed for the three vertical pieces. Ans: x = height of the sides and the divider; width of top and bottom = ½ (100 -3 x) and 3 x = material for 3 verticals The area will therefore be A = x (100 -3 x)/2 = 50 x -3 x 2/2 Taking the derivative, d. A/dx = (50) – ½ (2 • 3 x) = (50 – 3 x) The derivative will be zero for x = 50/3 feet The width is W = (100 -3 x)/2 = (100 -50)/2 = W = 25 ft The height is 50/3 = h = 16. 67 ft Copyright © 2015 C. P. Rubenstein 18
Homework #04 Review #3 a. You have a 100 ft. roll of fencing with which to make a rectangular corral that includes TWO dividers. Find the width of the corral that will produce the maximum total area. Again x = height of sides and dividers. Top and bottom width = (100 -4 x)/2 where 4 x is amount of material needed for each vertical piece. Ans: With two dividers, there are four vertical pieces; area is: A = x (100 - 4 x)/2 = ½ (x(100 – 4 x)) = ½ (100 x – 4 x 2) = 50 x - 2 x 2 The derivative is d. A/dx = 50 - 4 x (which is zero at x = 50/4 ) The width is then W = (100 -4 x)/2 = (100 -50)/2 = W = 25 ft (The same answer as when we had only one divider!) and the height is 50/4 = h = 12. 5 ft Copyright © 2015 C. P. Rubenstein 19
Homework #04 Review #4 a. Find the point x, y on the parabola y = x 2 that is closest to the point 0, 2 on the y-axis Ans: (Note that for a parabola x values are x) x y = x 2 0 0 0. 5 0. 25 1. 0 1 1. 5 2. 25 2. 0 4 2. 5 6. 25 3. 0 9 Copyright © 2015 C. P. Rubenstein (0, 2) 20
Homework #04 Review #4 b. (Parabola y = x 2 and point 0, 2 ) … Then use the distance formula (Pythagorean Theorem) to write a function that represents the square of length of the straight line segment connecting the two points. Ans: The distance from the point 0, 2 to a point on the parabola is given by the distance formula: D 2 = (x - 0)2 + (y - 2)2 D 2 = (x)2 + (y - 2)2 Since y = x 2 this becomes D 2 = x 2 + [ (x 2 -2) 2 ] D 2 = x 2 + [ (x 2 -2) ] D 2 = x 2 + [ (x 2) + (x 2)(-2) + (-2)(x 2) + (-2) ] or D 2 = x 2 + [ x 4 - 4 x 2 +4 ] which is D 2 = x 4 - 3 x 2 + 4. Copyright © 2015 C. P. Rubenstein 21
Homework #04 Review #4 c. Find the derivative of the length squared function and set it equal to zero to solve for x. Note: When the square of the length reaches a minimum, the length itself also must reach a minimum. It is easier here to work with the square of the length - no square root needed. Ans: Taking the derivative d. D 2/dx = 4 x 3 - 6 x d. D 2/dx = 2 x ( 2 x 2 - 3) This will be zero for x= 0 or x= ± (3/2) If you graph the function D 2(x) you will see that x = 0 is a point at which D 2 has a local maximum Copyright © 2015 C. P. Rubenstein 22
Homework #04 Review #5. You have a rectangular piece of sheet metal with dimensions 3” × 2” with which to make a box with an open top by cutting a square out of each corner and then folding up the sides. What size square will produce a box with the maximum volume? Hint: Begin by finding the formula that gives the volume of the box as a function of x. Then find the derivative function. Finally, find the value of x at which the derivative is zero. Ans: The volume will be given by multiplying the area by the height of the box’s side ‘x’ : V(x) = box area • x = (3 -2 x)(2 -2 x) • x V(x) = x • (6 -10 x + 4 x 2) = 6 x - 10 x 2 + 4 x 3 The derivative d. V(x)/dx = 6 - 20 x + 12 x 2 or d. V/dx = 12 x 2 - 20 x + 6 Using the quadratic formula, d. V(x)/dx is zero at x = 1. 274 or x = 0. 3924 (Only the second root makes physical sense) Copyright © 2015 C. P. Rubenstein 23
Homework #04 Review #6. Find the derivative of y(x) = (x+2)(x 3 + x -1) after first expanding the function by carrying out the indicated multiplication Ans: y(x) = (x+2)(x 3 + x -1) y(x) = (x) (x 3 + x -1) +2 (x 3 + x -1) y(x) = (x 4 + x 2 - x) + (2 x 3 + 2 x -2) Rearranging terms - y(x) = x 4 + 2 x 3 + x 2 + x – 2 therefore dy/dx = 4 x 3 + 6 x 2 + 2 x+ 1 Copyright © 2015 C. P. Rubenstein 24
Homework #04 Review #7. Calculate the (first) derivative and also the second, third, and fourth derivatives* of y(x) = x 5 - 3 x 4 + x 2 + 22 (NOTE: * The second derivative, y''(x) is the derivative of the derivative, y'(x), the third derivative, y'''(x), is the derivative of the second derivative, and the fourth derivative, y. IV(x), is the derivative of the third derivative. ) Ans: y(x) = dy/dx = y’ = d 2 y/dx = y'' = d 3 y/dx = y''' = d 4 y/dx = yiv = x 5 5 x 4 20 x 3 60 x 2 120 x - 3 x 4 + x 2 + 22 - 12 x 3 + 2 x - 36 x 2 + 2 - 72 x - 72 Copyright © 2015 C. P. Rubenstein 25
Homework #04 Review #8. Find the first, second, and third derivatives of the function f(x) = 1 + x 2/2 + x 3/6 + x 4/24 Ans: We can rewrite this formula in decreasing exponential format: f(x) = x 4/24 + x 3/6 + x 2/2 + x + 1 f '(x) = x 3/6 + x 2/2 + x f ''(x) = x 2/2 + x + 1 f'''(x) = x + 1 and thus… fiv(x) = 1 +1 Copyright © 2015 C. P. Rubenstein 26
Homework #04 Review #9. Find a function y(x) whose derivative is dy/dx = 3 x 2 + x + 1 BACKGROUND: This is the reverse from differentiation: Recall that if y = xa then dy/dx = a xa-1 Starting with dy/dx = C xa we can ‘reverse’ the process to find y = C [ 1/(a+1)] xa +1 We know that x 0 = 1 thus If dy/dx = Cx 0 then, using the above formula we find : y = C [ 1/(0+1)] x 0+1 and thus y = C [ 1 ] x 1 or y = Cx If dy/dx = Cx 1 then, using the above formula we find: y = C [ 1/(1+1)] x 1+1 and thus y = C [ ½ ] x 2 If dy/dx = Cx 2 then, y = C [ 1/(2+1)] x 2+1 or y = C [ 1/3 ] x 3 Ans: Therefore when dy/dx = 3 x 2 + x + 1 or dy/dx = 3 x 2 + 1 x 1 + 1 x 0 y = 3[1/(2+1)] x(2+1) + 1[1/(1+1)] x(1+1) + 1 [ 1/(0+1) ] x(0+1) y= x 3 + ½ x 2 + Copyright © 2015 C. P. Rubenstein x . 27
Questions? Copyright © 2015 C. P. Rubenstein 28
FALL 2015 CM/FM Seminar Series DISCUSSION Uninvited Guest, Putting Pest Control on Your Radar. Ensuring that they Check In but Don’t Check Out • Hope Bowman, Regional Entomologist • Maria lgnez De Oliveira, Sr. Account Executive, Western Pest Services The issues of mitigating pest controls systems in your facility or construction site will be discussed. The question of “why” it is really important to develop a plan for integrated pest management control in the built environment is presented. Copyright © 2015 C. P. Rubenstein 29
Class #05: Rules for Differentiation (continued) Rules (and Proofs) for Differentiation 1. Derivative of a Constant (and Proof – last week) 2. Derivative of a constant times a function (and Proof – last week) 3. The Sum Rule: Proof and Application Graphing a Function and its Derivative First Application of Differential Calculus: Max/Min problems 4. The Product Rule: Proof and Application 5. The Reciprocal Rule: Proof and Application 6. The Quotient Rule: Proof The Chain Rule Slopes to Derivatives to rules The Derivative of x Copyright © 2015 C. P. Rubenstein 30
REVIEW: Rules for Differentiation Rule 1. The Derivative of a Constant is zero d/dx [C] = 0 (The graph of a constant is a horizontal line whose slope = 0) Rule 2. Constants can be removed from equation d/dx [C f(x) ] = C df/dx Where C is any constant Rule 3. The “Sum Rule” d/dx [ f(x) + g(x) ] = df/dx + dg/dx Rule 3 a. Combining Rules 2 and 3 above we have d/dx [C 1 f(x) + C 2 g(x)] = C 1 df/dx + C 2 dg/dx Copyright © 2015 C. P. Rubenstein 31
More Rules for Differentiation Rule 4. The “Product Rule” d/dx [f(x) g(x)] = g(x)df/dx + f(x)dg/dx Today, we will prove: Rule 5. The “Reciprocal Rule” d/dx [1/g(x)] = -dg/dx / [g(x)]2 Rule 6. The “Quotient Rule” Combining rules 4 and 5 produces: d/dx [f(x) / g(x)] = [g(x) df/dx - f(x) dg/dx] / [g(x)]2 Rule 7. The “Chain Rule” Copyright © 2015 C. P. Rubenstein 32
The Reciprocal Rule Let us now show that the derivative of one over a function equals The negative reciprocal squared times the derivative of the function: Using the derivative definition is: Cross multiplying: d/dx [1/f(x+dx) – 1/f(x)] = d/dx [-f(x) + f(x+dx)] / [f(x+dx) f(x)] = [-f(x) + f(x+dx)] / dx [f(x+dx) f(x)] Thus: d/dx [1/f(x)] = f(x+dx) - f(x) / f(x+dx) f(x) dx or : Copyright © 2015 C. P. Rubenstein 33
The Reciprocal Rule… continued Only dx is involved with resolving the zero-over-zero limit, so we can pull the two terms f(x+dx)f(x) outside the parentheses and then take the product of the limits of the two factors: The limit of f(x+dx)f(x), as dx goes to zero is just f(x)2 so we have which proves the reciprocal rule . Copyright © 2015 C. P. Rubenstein 34
Example of the Reciprocal Rule Example: Let y(x) = 1/x 3 (i. e. , y = 1/f(x) where f(x) = x 3) The derivative of f(x) is 3 x 2 so the reciprocal rule gives us If we write the equation as d/dx (x -3) we see that this follows the familiar pattern d/dx (xn) = n • x n-1 with n= -3: d/dx (x-3) = -3 • x -3 -1 d/dx (x-3) = -3 x -4. Copyright © 2015 C. P. Rubenstein 35
The Quotient Rule The quotient rule, d/dx [ f(x)/g(x) ] = f ' /g - g ' f / g 2 can be seen by combining the product rule and the reciprocal rule: NOTE: It is probably easier to apply the product rule to f(x) • 1/g(x) than to remember the quotient rule formula. Copyright © 2015 C. P. Rubenstein 36
The Chain Rule The chain rule lets us find the derivative of a function such as f(x) = [ sin(x 3) ] which is the sine function with an argument of x 3 instead of just x If we let g(x) denote the “cube function” and h(x) denote the sine function, then f(x) = sin( [x 3] ) = h(g[x]) Therefore, for any arbitrary functions h and g: Copyright © 2015 C. P. Rubenstein 37
The Chain Rule… continued The chain rule for arbitrary functions h and g is: If we regard the right-hand side as the product of two fractions, we cancel the dg terms, leaving just dh/dx If dx is small but not infinitesimally small, this formula is only a good approximation In the limit that dx goes to zero, this formula is the derivative of h(g[x]) and in the same way [g(x+dx) -g(x)]/dx is only an approximation until we take the limiting case where dx becomes infinitesimally small. Copyright © 2015 C. P. Rubenstein 38
Example of the Chain Rule Example: Find the derivative of f(x) = sin(x 3) = h(g[x]) Use the chain rule and that the derivative of the sine is the cosine. Applying the chain rule we have Let us test this result by numerically approximating the slope of sin(x 3) at x = 2: Approx. slope = (sin[(2 + 0. 0001)3] - sin[23] ) / 0. 0001 = -1. 75 The value of the our derivative function at x = 2 is cos(23) • 3 • 22 = -1. 746 which is good agreement so we have confidence that we applied the chain rule correctly. (Check this approximation using the TI calculator program ‘SLOPE’ we set up in Session 4. Be sure you have the calculator set to radians and not degrees!) Copyright © 2015 C. P. Rubenstein 39
Slopes to Derivatives to Rules We have moved through three levels of abstraction while studying the slopes of curves. We began by finding the numerical value of the slope at a particular point on a particular curve, f(x). Next, we learned to find a formula, the derivative function, whose value at any point x is equal to the slope of the curve at the point x, f(x). Finally, we derived rules such as the sum, product, and chain rules, which let us find the derivatives of functions constructed from arbitrary functions Copyright © 2015 C. P. Rubenstein 40
The Derivative of the Square Root of x We have learned so far to find the derivative of xn and (with the aid of the reciprocal rule) x-n. But what do we do about fractional powers? Let us look at the derivative of y(x) = x½ which is known as the square root function, y(x) = x By squaring both sides we can rewrite the equation y = x½ equals y 2 = x We can now regard x as a function of y: x(y) = y 2 . Copyright © 2015 C. P. Rubenstein 41
Derivative of the Square Root of x … continued We know how to find the derivative of the square function, so we can write What we are after is dy/dx, which is the ratio of a change in y to a change in x (when the changes are infinitesimally small). But this ratio, dy/dx, is just the reciprocal of dx/dy. That is, Another way to write this is: y'(x½) = ½x(½ - 1) y’ = ½x(-½) Note that this again fits the familiar pattern d/dx (xa) = a xa-1 which is also written: Copyright © 2015 C. P. Rubenstein 42
Questions? Copyright © 2015 C. P. Rubenstein 43
For Class Session #06: NO Class – 12 October - Columbus Day/Fall Break For class Session #06: • • • DUE: MIDTERM! Homework Set #05 Reading: Strang - Chapter 3: Applications of the Derivative 2 Do: In Class Quiz and Review Homework Set #05 Lecture and Problem Review Trigonometry Review, Derivatives of Exponentials, “e” Distribute: Remaining Class Note Sets For class Session #07: • DUE: Homework Set #06 • Reading: Strang - Chapter 5 : Integrals • 2 Do: In Class Quiz and Review Homework Set #06 • Lecture and Problem Review Linear Approximations, Newton’s Method, Midterm Review Copyright © 2015 C. P. Rubenstein 44
Any Questions? Send me an email … crubenst@pratt. edu or c. rubenstein@ieee. org Copyright © 2015 C. P. Rubenstein 45
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