Applications of Differentiation Aishah Saim Rasyida A'alaf Engineering Mathematics
©Hak Cipta Politeknik Banting Selangor 2022 Cetakan Pertama / First Printing, 2022 All rights reserved. No part of this publication may be reproduced or transmitted in any form or by any means, electronic or mechanical including photocopy, recording, or any information storage and retrieval system, without permission from Politeknik Banting Selangor and Bahagian Instruksional dan Pembelajaran Digital, JPPKK. Author Aishah Binti Saim Rasyida Binti A'alaf Editor Ruhiyah Nazihah Binti Zahkai Published by: Politeknik Banting Selangor, Persiaran Ilmu, Jalan Sultan Abdul Samad, 42700 Banting Selangor 03-3120 1657 / 03-3120 1625 https://pbs.mypolycc.edu.my/ eMail: [email protected] Perpustakaan Negara Malaysia Data Pengkatalogan-dalam-Penerbitan / Cataloguing-inPublication Data i
PREFACE The purpose of this eBook is to provide an understanding to Differentiation. It will discuss the use of differentiation such as maxima and minima, rates of change and solving optimization problems. In addition, this eBook contains notes, examples and exercises to enhance the learning experience. The only way to learn mathematics is to do mathematics ~Paul Halmos~ Aishah Saim Rasyida A'alaf ii
Table of Contents 2.0 Maxima and Minima 3.0 Rates Of Change 2.1 Steps to find maximum an minimum values 04 18 4.0 Optimization Problems 30 References 52 1.0 Definition 01 2.1.1 Stationary point 2.1.2 The nature of stationary point 2.1.3 Sketching curves 2.1.4 Geometry formulas 3.1 Positive and negative rates of change 3.2 Related rates of change 4.1 Introduction 4.2 Steps to solve optimization problems iii
1.1 Maxima and minima 1.2 Rates of change 1.3 Optimization problems DEFINITION 2 1.0 1
1.0 DEFINITION Differentiation can be used for solving problems of rates of change of certain quantities with respect to time. The rate at which some quantity is changing by relating it to other quantities for which the rate of change is known. Solving the problems of rates of change requires to understand the given statement and translate it into a mathematical statement by using the chain rule formula. 1.2 RATES OF CHANGE 1.1 MAXIMA AND MINIMA Maximum and minimum values of functions, is a skill that allows to solve many kinds of problems in which we need to find the largest and/or smallest value in a real-world situation. Differentiation skills is applied to graphing and to approximating function values. 2
One of the most important applications of differentiation is to determine the optimization values of any given function. Optimization is the process of finding maximum and minimum values given constraints using calculus. Optimization means examining “best available” values of the specific objective function in a defined domain including multiple types of objective functions. DEFINITION 1.3 OPTIMIZATION PROBLEMS 3 “Math is the only place where truth and beauty mean the same thing.” Danica McKellar
MAXIMA AND MINIMA 2.0 2.1 Steps to find maximum and minimum 2.1.1 Stationary point 2.1.2 The Nature of Stationary Points 2.1.3 Sketching Curves Examples Exercises 2.1.4 Geometry formulas 4
2.0 MAXIMA AND MINIMA The Diagram 1 below shows part of a function y = f(x). The point A is a local maximum and the point B is a local minimum. At each of these points the tangent to the curve is parallel to the x-axis so the derivative of the function is zero. Both of these points are therefore stationary points of the function. Diagram 1 to the left of A the gradients are positive (+) between A and B the gradients are negative (−) to the right of B the gradients are positive (+) The gradients of tangents to the curve are: 5
Obtain the gradient function, Equate Solve for x. The values of x obtained will be the x coordinates of the stationary points. Substitute x values into the equation of the curve to determine the corresponding y coordinates of the stationary points. To obtain the coordinates of the stationary point, we need to find out where on the curve the gradient is zero. Therefore; 1. 2. 3. 4. 2.1 STEPS TO FIND MAXIMUM AND MINIMUM VALUES 2.1.1 Stationary Point Stationary points are points on a graph where the gradient is zero. A stationary point can be any one of a maximum, minimum or a point of inflection as shown in diagram 2. Diagram 2 6
To determine the nature of the stationary points, we can obtain the second derivative, and examine its sign at the turning point. It represents the derivative of the gradient function and it specifies the rate of change of the gradient function. 2.1.2 The Nature of Stationary Points Conclusion **Important reminder If there is still an x term exists after the second derivative is done, replace the x value that was found in the third step of finding stationary point. Example 1: Stationary point (1,1) Hence (1,1) is a minimum point 7
Substitute x = 0 in the curve's equation to find the y co-ordinate of the point where the curve meet the y axis. Substitute y = 0 in the curve's equation to find the x co-ordinate of the point where the curve meet the x axis. Find the point(s) where the curve meets the axes: 2.1.3 Sketching Curves Example 2: 8 Sketch Stationary point is (2, 4) So (2, 4 ) is a maximum point Solution:
Example 3: 9 Sketch Stationary points are (0, 2) & (2, 6) So (0, 2 ) is a minimum point (2, 6 ) is a maximum point Solution:
10 To sketch the graph: To sketch the curve using just the stationary points and it crosses the y axis at (0, 2) Ignore these values
Example 4: Find the stationary point and the nature of the curve Solution: Stationary points, Nature, Stationary point is (0, -4) Substitute x=0 Point (0, -4) is an inflection point 11
Example 5: Find the stationary point and the nature of the curve Solution: Stationary points, Nature, 12
Example 6: Find the stationary point and the nature of the curve Solution: Stationary points, Nature, 13
Example 7: Find the turning points and the nature of the curve Solution: Turning points, Nature, (3x-3) (x-3) = 0 3x-3 = 0 x = 1 x-3 = 0 x = 3 When x = 1 When x = 3 y = 1 - 6 + 9 + 5 = 9 y = 27 - 54 + 27 + 5 = 5 Turning points are (1, 9) and (3, 5) When x = 1 When x = 3 (1, 9) is a maximum point (3, 5) is a minimum point 14
Exercise 1: 1.Find the turning points for Answer: 2. Find the stationary point for the curve 3. Find the turning points of the function 4. Find the stationary of the curve and sketch the graph. 5. Find the stationary of the curve and determine the nature of the point. 6. Find the coordinate of the maximum or minimum point and determine the nature of and sketch the graph. 1. Turning point (1, -4) 2. Turning point (2, -17) and (-2, 15) 3. Turning point (1, 4) and (3, 0) 4. Stationary point (1, -12). (1,-12) is a minimum point 5. Stationary point (2, -32) and (-2, 32) (2, -32) is a minimum point. (-2, 32) is a maximum point 6. Stationary point (-1, -4) (-1, -4) is a minimum point 15
2.1.4 Geometry formulas Rectangle Square Triangle Trapezoid a b a b c h a a h b1 b2 h h 16
h r h Geometry formulas Cone Sphere Cylinder r r r Circle h r 17
RATES OF CHANGE 18 3.0 3.1 Positive and negative rates of change Examples Exercises 3.2 Related rates of change
Rates of change refers to the rate at which any variable changes say x with respect to time, t. The gradient function, is therefore a rate of change. The symbol, dx refers to the change in the variable and the symbol dt refers to the change in time For example: A length, l, of elastic is stretched at the rate of ... Stretched = increasing rate A fire is spreading at the rate of ... Spreading = increasing rate Water is leaking from a tank at the rate of... Leaking = decreasing rate 3.0 RATES OF CHANGE A positive and negative sign denotes an increasing and decreasing rate. Sometimes the word increasing or the word decreasing may not be used, but is replaced by a word that describes whether the variable is increasing or decreasing. 3.1 Positive and negative rates of change Courtesy of canva.com 19
Find the rate of change of volume with respect to time at the instant when t = 5. Given that where V is the volume of the oil leaking from a tanker in and t is the time in minutes. In such a situation, the ‘chain rule' is used to find the corresponding rate of change of one variable, given the change in the other variable. 3.2 Related rates of change In a situation involving changes in two quantities, we need to find the rate of change if one quantity changes and affects the change in the other quantity. Example 1: Solution: When t = Courtesy of canva.com 5 The rate of change after 5 min of volume with respect to time is 20
The volume of a cube is increasing at a rate of Example 2: Solution: How fast is the surface area increasing when the length of an edge is 10 cm. x x x Total surface area, A = 6 (x x x) Volume of cube, V = x x x x x The rate of change for volume The rate of change for surface area 21
An electric balloon pump blows spherical balloon whose surface area increases at the rate of Example 3: Solution: Courtesy of canva.com What is the rate of increase of the radius of a bubble when its radius is 3 cm. Science without religion is lame and religion without science is blind -Albert Einstein22
A spherical balloon is filled with air at the constant rate of Example 4: Solution: Calculate the rate at which the radius is increasing when the radius = 10 cm. The volume of spherical balloon, When r = 10 cm Courtesy of canva.com 10 23
An upturned cone with semivertical angle is being filled with water at a constant rate of Example 5: Solution: The depth of the water, h is 60 cm. Find the rate of: i. h if the depth of the water is increasing ii. r if radius of the surface of the water is increasing iii. S if the area of the water surface is increasing. h r The cross section of the cone is a right-angled. Hence r = h = 60 cm 45 h r h = depth r = radius S = surface area 24
i. The rate of h if the depth of the water is increasing ii. The rate of r if radius of the surface of the water is increasing. We have seen that r (t) = h(t). Thus, when h = 60 iii. The rate of S if the area of the water surface is increasing 25
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Pak Ali placed a 9 foot ladder against his house so he could paint the wall. If the base of the ladder begins to slide away at a rate of 2 feet per second, how fast the ladder sliding down when the base of the ladder is 6 feet from the wall. Example 6: Solution: x y 9 (ladder slide away) Ladder sliding down: When x = 6 ft 27
Exercise 2: 1. The radius, r cm of a circle changes from 5cm to 5.1 cm. Find the approximate change in the area. State whether this change is an increase or decrease. 2. Given Calculate: a. When x = -2 b. The approximate change in y when x increases from -2 to -1.97 3. Find the rates of change of the area of the square when 4. Find the rates of change of the area of the circle when 5. The radius, r cm of the circular cone is increasing at the rate of 3 cm per minute. The cone's height, h is related to the radius by h = 4r. Find the rate of change of the volume when r = 8 cm. 6. The length x of a rectangle is decreasing at the rate of 3cm per minute. The width y is increasing at the rate of 2 cm per minute. When x = 10 cm and y = 6 cm, find the rates of change of perimeter. 28
Answers: 1. 2. a. b. 3. 4. 5. 6. Courtesy of canva.com 29
OPTIMIZATION PROBLEMS 2 4.0 4.1 Introduction Examples Exercises 4.2 Steps to solve optimization problems 30
4.0 OPTIMIZATION PROBLEMS 4.1 Introduction The optimization problems require us to understand the given statement and translate it into a mathematical statement. Draw or sketch a diagram to help in figuring out the problem and identify the variables and parameters that need to be determined. Write equations that are related to the variables. Find the first and second derivatives. Determine the maximum and minimum values. 4.2 Steps to solve optimization problem: 1. 2. 3. 4. Example 1: A farmer has 200m of fencing. He wants to enclose a rectangular paddock. Find the maximum possible area he can enclose. Solution: x y Figure : Rectangular Paddock Courtesy of canva.com 31
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A solid glass cylinder has a volume of . The cost of polishing the surface area of this glass cylinder is RM2 per for the curved surface area and RM3 per for the circular top and base area. b) Find the minimum cost of the polishing. Giving the answer to the nearest RM. Example 2: Given that the radius of the cylinder is r cm. a) Show that the cost of the polishing, RM C is given by Solution: r h Volume of cylinder Curve surface area of cylinder Area of the circular top and base 33
a) Cost for polishing for surface area Cost for polishing the top and base areas b) Minimum cost of the polishing 34
A rectangular box is to be constructed with a square base and an open top. The volume of the box is Example 3: a) Find the dimensions of the box to minimize the surface area of the box. b) What is the minimum surface area of the box. x x y 35
a) Surface area, A = (area of the base) + 4 (area of a side face) The dimensions of the box is x=7.55 cm and y=3.79cm Solution: Courtesy of canva.com 36
b) The minimum surface area of the box. Example 4: The rocket is launched vertically upwards at 800m per second. When it is 3000m up, what is the rate of change in the camera evaluation angle, with respect to time, t in order to keep the rocket in sight? Launch pad Camera 4000 m 3000 m Courtesy of canva.com 37
Solution: 38
A rectangular box has base dimensions and its volume is 1800 Example 5: h cm i. Prove that the total surface area, A : ii. Calculate the value of x of which A has a stationary value. Find the value of A and determine its nature. 3x cm 2x cm Solution: 39
When x=5 Courtesy of canva.com 40 Value of A
Aminah soup company makes a soup can with a volume of 250 What dimensions will allow for the minimum amount of metal to produce the can. Example 6: r Solution: 41 h Therefore
A 230g sardine can is contain 12 cubic inches. The cost of materials for the top and bottom of the can is 5 cents per square inch. The cost of the side is 3 cents per square inch. What are the dimensions of the can to minimize the cost. Example 7: h Solution: r h Total cost, C = cost of top + cost of bottom + cost of side Surface area of the top and Surface area of the side of the can bottom of the can 42
To find r, = 1.046 To verify the minimum value occurs, Substitute r = 1.046 = 1.887 > 0 minimum value To find h, substitute r = 1.046 = 3.49 As a conclusion, to minimize the cost, the can should have a radius of approximately 1.046 inch and a height of 3.49 inch. Courtesy of canva.com 43
A football field is formed in rectangular shape and two semi circles. If the perimeter of the field is 600 m, find: a. The area if the field, A in terms of r b. The value of r so that the area of the field is maximum. c. The maximum value of the area of the field. Example 8: x Solution: r 44 a. Area, A = Rectangular area + circle area Therefore
45 b. The value of r Therefore c. The maximum value of the area Courtesy of canva.com
Exercise 3: 1. A 45 meter square of material to build a box has a square base and no top. Determine the dimensions of the box that will maximize the enclosed volume. Answer: 46 l=w=3.873 h=1.937 Solution: