Republic of the Philippines NUEVA VIZCAYA STATE UNIVERSITY Bambang, Nueva Vizcaya INSTRUCTIONAL MODULE IM No.: MATH 1-1S-2020-2021 College: Engineering Campus: Bambang DEGREE PROGRAM SPECIALIZATION YEAR LEVEL I. BSME/BSEE/ BSECE/BSCPE Mechanical/Electrical Electronics/Computer 1st Year COURSE NO. MATH 1 COURSE TITLE Differential Calculus TIME FRAME 16 hrs. WK NO. 14 -17 IM NO. UNIT TITLE/CHAPTER TITLE Applications of Derivative II. III. LESSON TITLE I. Applications of Maxima and Minima II. Rate of Changes III. Rectilinear Motion IV. Rectilinear Rates V. Parametric Equations LESSON OVERVIEW This lesson provides the students an understanding the derivatives in its applications. IV. DESIRED LEARNING OUTCOMES At the end of the lesson, the students should be able to: 1. Apply the different properties of derivatives in solving related rates 2. Solving problems involving relationships between changing quantities. 3. Derive mathematical concepts related to rates. V. COURSE CONTENT 5 5 APPLICATIONS OF DERIVATIVES There are various applications of derivatives not only in math and real life but also in other fields like science, engineering, physics, etc. In previous classes, you must have learned to find the derivative of different functions, like, trigonometric functions, implicit functions, logarithm functions, etc. In this section, you will learn the use of derivatives with respect to mathematical concepts and in real-life scenarios. 5.1 MAXIMA AND MINIMA It was noted that at a point where its first derivative vanishes, a function assumes an extreme value, provided the derivative changes sign at that point. This result finds application in a great variety of problems, some of which will now be considered. When the derivative is equated to zero, it may happen, of course, that several critical values are obtained. in practice, the value that gives the desired maximum or minimum can often be selected at once by inspection NVSU-FR-ICD-05-00 (081220) “In accordance with Section 185, Fair Use of a Copyright Work of Republic Act 8293, the copy righted works included in this material may be reproduced for educational purposes only and not for commercial distribution” Page |1 Republic of the Philippines NUEVA VIZCAYA STATE UNIVERSITY Bambang, Nueva Vizcaya INSTRUCTIONAL MODULE IM No.: MATH 1-1S-2020-2021 EXAMPLE A box is to be made of piece of cardboard 9 𝑖𝑛2 by cutting equal squares out of the corners and turning up the sides. Find the volume of the largest box that can be made in this way. Solution: 𝑉 = (9 − 2𝑥)2 𝑥 𝑉 = 81𝑥 − 36𝑥 2 + 4𝑥 3 𝑑𝑉 = 81 − 72𝑥 + 12𝑥 2 = 0 𝑑𝑥 4𝑥 2 − 24𝑥 + 27 = 0 ÷3 Using quadratic formula −𝑏 ± √𝑏 2 − 4𝑎𝑐 𝑥= 2𝑎 −(−24) ± √(−24)2 − 4(4)(27) 𝑥= 2(4) 24 ± 12 𝑥= 8 𝑥 = 4.5 and 𝑥 = 1.5 Use 𝑥 = 1.5 𝑉 = (9 − 2𝑥)2 𝑥 2 𝑉 = (9 − 2(1.5)) (1.5) 𝟑 𝑽 = 𝟓𝟒 𝒊𝒏 EXAMPLE Find the rectangle of maximum perimeter inscribed in a given circle Solution: Diameter 𝐷 is constan, (circle is given) 𝑥 2 +𝑦 2 = 𝐷 2 2𝑥 + 2𝑦𝑦 ′ = 0 −𝑥 𝑦′ = 𝑦 Perimeter 𝑃 = 2𝑥 + 2𝑦 𝑑𝑃 = 2 + 2𝑦 ′ 𝑑𝑥 0 = 2 + 2𝑦 ′ −𝑥 2 +2( ) = 0 𝑦 𝑦=𝑥 The largest rectangle is a square. NVSU-FR-ICD-05-00 (081220) “In accordance with Section 185, Fair Use of a Copyright Work of Republic Act 8293, the copy righted works included in this material may be reproduced for educational purposes only and not for commercial distribution” Page |2 Republic of the Philippines NUEVA VIZCAYA STATE UNIVERSITY Bambang, Nueva Vizcaya INSTRUCTIONAL MODULE IM No.: MATH 1-1S-2020-2021 EXAMPLE If the hypotenuse of the right triangle is given, show that the area is maximum when the triangle is isosceles. Solution: 𝑥2 + 𝑦2 = 𝑐2 2𝑥 + 2𝑦𝑦 ′ = 0 −𝑥 𝑦′ = 𝑦 Area: 1 𝑥𝑦 2 𝑑𝐴 1 = [𝑥𝑦 ′ + 𝑦] 𝑑𝑏 2 1 0 = [𝑥𝑦 ′ + 𝑦] 2 0 = 𝑥𝑦 ′ + 𝑦 −𝑥 𝑥( )+𝑦 = 0 𝑦 𝑥2 𝑦= 𝑦 𝑦2 = 𝑥2 𝑦=𝑥 𝐴= The triangle is an isosceles right triangle. Do what is being asked for. 1. A box is to be made from a sheet of cardboard that measures 12" × 12". The construction will be achieved by cutting a square from each corner of the sheet and then folding up the sides. What is the box of greatest volume that can be constructed in this fashion? 2. A rectangular garden is to be constructed against the side of a garage. The gardener has 100𝑓𝑡 of fencing and will construct a three-sided fence; the side of the garage will form the fourth side. What dimensions will give the garden of greatest area? 3. A farmer has enough money to build only 100𝑚 of fence. What are the dimensions of the field he can enclose the maximum area? 4. Find the minimum amount of tin sheet that can be made into a closed cylinder having a volume of 108 cu. inches in square inches. 5. A box is to be constructed from a piece of zinc 20𝑖𝑛2 by cutting equal squares from each corner and turning up the zinc to form the side. What is the volume of the largest box that can be so constructed? 6. A poster is to contain 300 (cm square) of printed matter with margins of 10𝑐𝑚 at the top and bottom and 5𝑐𝑚 at each side. Find the overall dimensions if the total area of the poster is minimum. 7. A normal window is in the shape of a rectangle surmounted by a semi-circle. What is the ratio of the width of the rectangle to the total height so that it will yield a window admitting the most light for a given perimeter? NVSU-FR-ICD-05-00 (081220) “In accordance with Section 185, Fair Use of a Copyright Work of Republic Act 8293, the copy righted works included in this material may be reproduced for educational purposes only and not for commercial distribution” Page |3 Republic of the Philippines NUEVA VIZCAYA STATE UNIVERSITY Bambang, Nueva Vizcaya INSTRUCTIONAL MODULE IM No.: MATH 1-1S-2020-2021 8. Determine the diameter of a closed cylindrical tank having a volume of 11.3 𝑐𝑢. 𝑚 to obtain minimum surface area. 9. An open top rectangular tank with square bases is to have a volume of 10𝑚 3 . The materials for its bottom are to cost 𝑃15/𝑚2 and that for the sides, 𝑃6/𝑚2 . Find the most economical dimensions for the tank. 10. A fencing is limited to 20𝑓𝑡 length. What is the maximum rectangular area that can be fenced in using two perpendicular corner sides of an existing wall? 5.2 RATE OF CHANGES The derivative of a function is identical to its rate of changes. Thus, the rate of change of the volume (V) of a sphere with respect to its radius (r) is dV/dt. EXAMPLE The radius of a sphere increases at the rate of 3 cm per second. From zero initially. How fast is the volume increasing after 2 seconds? How fast is the surface area increasing after 3 sec? Given: 𝑑𝑟 𝑐𝑚 =3 𝑑𝑡 𝑠𝑒𝑐 Required: 𝑑𝑉 a. @ 𝑡 = 2 𝑠𝑒𝑐 𝑑𝑡 𝑑𝐴 @ 𝑡 = 3 𝑠𝑒𝑐 b. 𝑑𝑡 Solution: a. Volume of sphere: 4 𝑉 = 𝜋𝑟 3 3 𝑑𝑉 4 𝑑𝑟 = 𝜋 (3𝑟 2 ) 𝑑𝑡 3 𝑑𝑡 𝑑𝑉 𝑑𝑟 = 4𝜋𝑟 2 𝑑𝑡 𝑑𝑡 when t = 0, r = 0, dr/dt = 3 cm/sec when t = 2, r = (2)(3) = 6cm, dr/dt = 3 cm/sec sec 𝑑𝑉 = 4𝜋(62 )(3) 𝑑𝑡 𝒅𝑽 𝒄𝒎 = 𝟒𝟑𝟐𝝅 𝒅𝒕 𝒔𝒆𝒄 b. Area of sphere 𝐴 = 4𝜋𝑟 2 𝑑𝐴 𝑑𝑟 = 4𝜋 (2𝑟 ) 𝑑𝑡 𝑑𝑡 𝑑𝐴 𝑑𝑟 = 8𝜋𝑟 𝑑𝑡 𝑑𝑡 when t = 0, r = 0 when t = 3, r = (3)(3) = 9cm 𝑑𝐴 = 8𝜋 × 9 × 3 𝑑𝑡 𝒅𝑨 𝒔𝒒. 𝒄𝒎 = 𝟐𝟏𝟔𝝅 𝒅𝒕 𝒔𝒆𝒄 NVSU-FR-ICD-05-00 (081220) “In accordance with Section 185, Fair Use of a Copyright Work of Republic Act 8293, the copy righted works included in this material may be reproduced for educational purposes only and not for commercial distribution” Page |4 Republic of the Philippines NUEVA VIZCAYA STATE UNIVERSITY Bambang, Nueva Vizcaya INSTRUCTIONAL MODULE IM No.: MATH 1-1S-2020-2021 EXAMPLE The dimension of a box is x, x + 1, x + 2. Find how fast is the total volume increasing as x increases. Given: 𝑑𝑖𝑚𝑒𝑛𝑠𝑖𝑜𝑛 = 𝑥, 𝑥 + 1, 𝑥 + 2 Required: 𝑑𝑉 𝑑𝑥 Solution: 𝑉 = (𝑥)(𝑥 + 1)(𝑥 + 2) 𝑉 = 𝑥 3 + 3𝑥 2 + 2𝑥 𝒅𝑽 = 𝟑𝒙𝟐 + 𝟔𝒙 + 𝟐 𝒄𝒖𝒃𝒊𝒄 𝒖𝒏𝒊𝒕𝒔 𝒅𝒙 EXAMPLE A right circular cylinder has a height of 4 cm. Find the rate of change of its volume with respect to a radius r of its base. Given: ℎ𝑒𝑖𝑔ℎ𝑡 = 4 𝑐𝑚 Required: 𝑑𝑉 𝑑𝑟 Solution: Volume of a cylinder 𝑉 = 𝜋𝑟 2 ℎ 𝑑𝑉 = 2𝜋𝑟ℎ 𝑑𝑟 𝑑𝑉 = 2𝜋𝑟(4) 𝑑𝑟 𝒅𝑽 = 𝟖𝝅𝒓 𝒄𝒖𝒃𝒊𝒄 𝒖𝒏𝒊𝒕𝒔. 𝒅𝒓 EXAMPLE In a right circular cone expands in such a way that the altitude remains equal to the diameter of the base, find the rate of change of the volume with respect to the radius of the base, if the radius is equal to 5 cm. Given: 𝑟 = 5 𝑐𝑚 Required: 𝑑𝑉 𝑑𝑟 Solution: Volume of a right circular cone 𝜋𝑟 2 ℎ 𝑉= , ℎ = 2𝑟 3 2 𝜋𝑟 (2𝑟) 𝑉= 3 2𝜋𝑟 3 𝑉= 3 𝑑𝑉 = 2𝜋𝑟 2 𝑑𝑟 At r = 5 𝑑𝑉 = 2𝜋(52 ) 𝑑𝑟 NVSU-FR-ICD-05-00 (081220) “In accordance with Section 185, Fair Use of a Copyright Work of Republic Act 8293, the copy righted works included in this material may be reproduced for educational purposes only and not for commercial distribution” Page |5 Republic of the Philippines NUEVA VIZCAYA STATE UNIVERSITY Bambang, Nueva Vizcaya INSTRUCTIONAL MODULE IM No.: MATH 1-1S-2020-2021 𝒅𝑽 = 𝟓𝟎𝝅 𝒄𝒖𝒃𝒊𝒄 𝒖𝒏𝒊𝒕𝒔. 𝒅𝒓 EXAMPLE For what values of x does the function 3x + x-1 have a rate of change of -1 with respect to x? Solution: Let 𝑦 = 3𝑥 + 𝑥 −1 1 𝑦 = 3𝑥 + 𝑥 𝑑𝑦 1 =3− 2 𝑑𝑥 𝑥 When dy/dx = -1 𝑑𝑦 1 = −1 = 3 − 2 𝑑𝑥 𝑥 −𝑥 2 = 3𝑥 2 − 1 4𝑥 2 = 1 𝟏 𝒙=± 𝟐 EXAMPLE If y =7x2 – 9x, find the value of x for which dy/dx = 33. Solution: 𝑦 = 7𝑥 2 − 9𝑥 𝑑𝑦 = 14𝑥 − 9 𝑑𝑥 At dy/dx = 33 𝑑𝑦 = 33 = 14𝑥 − 9 𝑑𝑥 14𝑥 = 42 𝒙=𝟑 EXAMPLE If the base of a triangle is constantly 4 cm, find the rate at which the area of the triangle is changing with respect to the altitude. Solution: 𝑏ℎ 𝐴= 2 At b = 4 4ℎ 𝐴= 2 𝐴 = 2ℎ 𝒅𝑨 =𝟐 𝒅𝒉 EXAMPLE Find the average rate of change of the function y = x 2 between x = 2 and x = 5. Find the instantaneous rate of change at x = 2. Solution: 𝑦 = 𝑥2 𝑑𝑦 = 2𝑥 𝑑𝑥 At x = 2 at x = 4 NVSU-FR-ICD-05-00 (081220) “In accordance with Section 185, Fair Use of a Copyright Work of Republic Act 8293, the copy righted works included in this material may be reproduced for educational purposes only and not for commercial distribution” Page |6 Republic of the Philippines NUEVA VIZCAYA STATE UNIVERSITY Bambang, Nueva Vizcaya INSTRUCTIONAL MODULE IM No.: MATH 1-1S-2020-2021 𝑑𝑦 =4 𝑑𝑥 𝑑𝑦 = 10 𝑑𝑥 10 + 4 𝒂𝒗𝒆𝒓𝒂𝒈𝒆 𝒓𝒂𝒕𝒆 𝒐𝒇 𝒄𝒉𝒂𝒏𝒈𝒆 = =𝟕 2 When x = 2 𝑑𝑦 = 4, 𝑖𝑛𝑠𝑡𝑎𝑛𝑡𝑒𝑛𝑒𝑜𝑢𝑠 𝑟𝑎𝑡𝑒 𝑜𝑓 𝑐ℎ𝑎𝑛𝑔𝑒 𝑑𝑥 Do what is being asked for. 1. A spherical toy balloon is inflated so that its volume is a function of time given by the formula V = πt3/6. Find the rate of change of V at the time t = 4 sec. 2. The surface area of the balloon which is inflated at s = πt2. How fast is S changing when t = 4 sec? 3. A thin sheet of ice is in the form of a circle. If the ice is melting in such a way that the area of the sheet is decreasing at a rate of 0.5 m2/sec at what rate is the radius decreasing when the area of the sheet is 12 m2? 4. A person is standing 350 feet away from a model rocket that is fired straight up into the air at a rate of 15 ft/sec. At what rate is the distance between the person and the rocket increasing (a) 20 seconds after liftoff? (b) 1 minute after liftoff? 5. A plane is 750 meters in the air flying parallel to the ground at a speed of 100 m/s and is initially 2.5 kilometers away from a radar station. At what rate is the distance between the plane and the radar station changing (a) initially and (b) 30 seconds after it passes over the radar station? 6. Two cars start out 500 miles apart. Car A is to the west of Car B and starts driving to the east (i.e. towards Car B) at 35 mph and at the same time Car B starts driving south at 50 mph. After 3 hours of driving at what rate is the distance between the two cars changing? Is it increasing or decreasing? 7. The sides of a square are increasing at a rate of 10 cm/sec. How fast is the area enclosed by the square increasing when the area is 150 cm2. 8. The sides of an equilateral triangle are decreasing at a rate of 3 in/hr. How fast is the area enclosed by the triangle decreasing when the sides are 2 feet long? 9. A spherical balloon is being filled in such a way that the surface area is increasing at a rate of 20 cm2/sec when the radius is 2 meters. At what rate is air being pumped in the balloon when the radius is 2 meters? 10. A cylindrical tank of radius 2.5 feet is being drained of water at a rate of 0.25 ft3/sec. How fast is the height of the water decreasing? 11. A hot air balloon is attached to a spool of rope that is 125 feet away from the balloon when it is on the ground. The hot air balloon rises straight up in such a way that the length of rope increases at a rate of 15 ft/sec. How fast is the hot air balloon rising 20 seconds after it lifts off? NVSU-FR-ICD-05-00 (081220) “In accordance with Section 185, Fair Use of a Copyright Work of Republic Act 8293, the copy righted works included in this material may be reproduced for educational purposes only and not for commercial distribution” Page |7 Republic of the Philippines NUEVA VIZCAYA STATE UNIVERSITY Bambang, Nueva Vizcaya INSTRUCTIONAL MODULE IM No.: MATH 1-1S-2020-2021 12. A rock is dropped straight off a bridge that is 50 meters above the ground and falls at a speed of 10 m/sec. Another person is 7 meters away on the same bridge. At what rate is the distance between the rock and the second person increasing just as the rock hits the ground? 13. A person is 550 meters away from a road and there is a car that is initially 800 meters away approaching the person at a speed of 45 m/sec. At what rate is the distance between the person and the car changing (a) 5 seconds after the start, (b) when the car is directly in front of the person and (c) 10 seconds after the car has passed the person. 14. Two cars are initially 1200 miles apart. At the same time Car A starts driving at 35 mph to the east while Car B starts driving at 55 mph to the north (see sketch below for this initial setup). At what rate is the distance between the two cars changing after (a) 5 hours of travel, (b) 20 hours of travel and (c) 40 hours of travel? NVSU-FR-ICD-05-00 (081220) “In accordance with Section 185, Fair Use of a Copyright Work of Republic Act 8293, the copy righted works included in this material may be reproduced for educational purposes only and not for commercial distribution” Page |8 Republic of the Philippines NUEVA VIZCAYA STATE UNIVERSITY Bambang, Nueva Vizcaya INSTRUCTIONAL MODULE IM No.: MATH 1-1S-2020-2021 15. Repeat problem 14 above except for this problem assume that Car A starts traveling 4 hours after Car B starts traveling. For parts (a), (b) and (c) assume that these are travel times for Car B. 16. Two people are on a city block. See the sketch below for placement and distances. Person A is on the northeast corner and Person B is on the southwest corner. Person A starts walking towards the southeast corner at a rate of 3 ft/sec. Four seconds later Person B starts walking towards the southeast corner at a rate of 2 ft/sec. At what rate is the distance between them changing (a) 10 seconds after Person A starts walking and (b) after Person A has covered half the distance? 17. A person is standing 75 meters away from a kite and has a spool of string attached to the kite. The kite starts to rise straight up in the air at a rate of 2 m/sec and at the same time the person starts to move towards the kites launch point at a rate of 0.75 m/sec. Is the length string increasing or decreasing after (a) 4 seconds and (b) 20 seconds. 18. A person lights the fuse on a model rocket and starts to move away from the rocket at a rate of 3 ft/sec. Five seconds after lighting the fuse the rocket launches straight up into the air at a rate of 10 ft/sec. Is the distance between the person and the rocket increasing or decreasing (a) 6 seconds after launch and (b) 12 seconds after launch? 19. A light is on a pole and is being lowered towards the ground at a rate of 9 in/sec. A 6 foot tall person is on the ground and 8 feet away from the pole. At what rate is the persons shadow increasing then the light is 15 feet above the ground? NVSU-FR-ICD-05-00 (081220) “In accordance with Section 185, Fair Use of a Copyright Work of Republic Act 8293, the copy righted works included in this material may be reproduced for educational purposes only and not for commercial distribution” Page |9 Republic of the Philippines NUEVA VIZCAYA STATE UNIVERSITY Bambang, Nueva Vizcaya INSTRUCTIONAL MODULE IM No.: MATH 1-1S-2020-2021 20. A light is fixed on a wall 10 meters above the floor. Twelve meters away from the wall a pole is being raised straight up at a rate of 45 cm/sec. When the pole is 6 meters tall at what rate is the tip of the shadow moving (a) away from the pole and (b) away from the wall? Note the sketch below is not to scale… 21. A light is on the top of a 15 foot tall pole. A 5 foot tall person starts at the pole and moves away from the pole at a rate of 2.5 ft/sec. After moving for 8 seconds at what rate is the tip of the shadow moving (a) away from the person and (b) away from the pole? 22. A tank of water is in the shape of a cone (assume the “point” of the cone is pointing downwards) and is leaking water at a rate of 35 cm3/sec. The base radius of the tank is 1 meter and the height of the tank is 2.5 meters. When the depth of the water is 1.25 meters at what rate is the (a) depth changing and (b) the radius of the top of the water changing? NVSU-FR-ICD-05-00 (081220) “In accordance with Section 185, Fair Use of a Copyright Work of Republic Act 8293, the copy righted works included in this material may be reproduced for educational purposes only and not for commercial distribution” P a g e | 10 Republic of the Philippines NUEVA VIZCAYA STATE UNIVERSITY Bambang, Nueva Vizcaya INSTRUCTIONAL MODULE IM No.: MATH 1-1S-2020-2021 5.3 RECTILINEAR MOTION Rectilinear motion is a motion of a particle or object along a straight line. Position is the location of object and is given as a function of time s (t) or x (t). Velocity is the derivative of position: 𝑣= 𝑑𝑥 𝑑𝑡 Acceleration is the derivative of velocity: 𝑎= 𝑑𝑣 𝑑𝑡 EXAMPLE A particle’s position (in inches) along 𝑥 axis after 𝑡 𝑠𝑒𝑐 of travel is given by the equation 𝑥 = 24𝑡 2 − 𝑡 3 + 10. a. What is the particle’s average velocity during the first 3 𝑠𝑒𝑐 of travel? b. Where is the particle and how fast is it moving after 3 𝑠𝑒𝑐 of travel? c. Where is the particle and how fast is it moving after 20 𝑠𝑒𝑐 of travel? d. When is the velocity of the particle 0? What is the particle’s position at that time? Solution: a. 𝑡 = 0 𝑠𝑒𝑐 𝑥 = 24𝑡 2 − 𝑡 3 + 10 𝑥 = 24(0)2 − (0)3 + 10 𝑥 = 10 𝑖𝑛 𝑡 = 3 𝑠𝑒𝑐 𝑥 = 24𝑡 2 − 𝑡 3 + 10 𝑥 = 24(3)2 − (3)3 + 10 𝒙 = 𝟏𝟗𝟗 𝒊𝒏 ∆𝑠 ∆𝑡 199 − 10 𝑉𝑎𝑣 = 3−0 189 𝑉𝑎𝑣 = 3 𝑽𝒂𝒗 = 𝟔𝟑 𝒊𝒏/𝒔𝒆𝒄 𝑉𝑎𝑣 = b. 𝑡 = 3 𝑠𝑒𝑐 𝒙 = 𝟏𝟗𝟗 𝒊𝒏 𝑥 = 24𝑡 2 − 𝑡 3 + 10 𝑑𝑥 = 48𝑡 − 3𝑡 2 𝑑𝑡 𝑑𝑥 = 48(3) − 3(3)2 𝑑𝑡 𝒅𝒙 = 𝟏𝟕𝟕 𝒊𝒏/𝒔𝒆𝒄 𝒅𝒕 c. 𝑡 = 20 𝑠𝑒𝑐 𝑥 = 24𝑡 2 − 𝑡 3 + 10 𝑥 = 24(20)2 − (20)3 + 10 𝒙 = 𝟏 𝟔𝟏𝟎 𝒊𝒏 NVSU-FR-ICD-05-00 (081220) “In accordance with Section 185, Fair Use of a Copyright Work of Republic Act 8293, the copy righted works included in this material may be reproduced for educational purposes only and not for commercial distribution” P a g e | 11 Republic of the Philippines NUEVA VIZCAYA STATE UNIVERSITY Bambang, Nueva Vizcaya INSTRUCTIONAL MODULE IM No.: MATH 1-1S-2020-2021 𝑥 = 24𝑡 2 − 𝑡 3 + 10 𝑑𝑥 = 48𝑡 − 3𝑡 2 𝑑𝑡 𝑑𝑥 = 48(20) − 3(20)2 𝑑𝑡 𝒅𝒙 = −𝟐𝟒𝟎 𝒊𝒏/𝒔𝒆𝒄 𝒅𝒕 d. 𝑥 = 24𝑡 2 − 𝑡 3 + 10 𝑑𝑥 = 48𝑡 − 3𝑡 2 𝑑𝑡 0 = 48𝑡 − 3𝑡 2 0 = 3𝑡(16 − 𝑡) 0 3𝑡 = 3 3 𝑡=0 When 𝒕 = 𝟎 𝑥 = 24𝑡 2 − 𝑡 3 + 10 𝑥 = 24(0)2 − (0)3 + 10 𝒙 = 𝟏𝟎 𝒊𝒏 When 𝒕 = 𝟏𝟔 𝑥 = 24𝑡 2 − 𝑡 3 + 10 𝑥 = 24(16)2 − (16)3 + 10 𝒙 = 𝟐 𝟎𝟓𝟖 𝒊𝒏 EXAMPLE Find the point of inflection of 𝑦 = 36 + 12𝑥 − 𝑥 3 Solution: 𝑦 = 36 + 12𝑥 − 𝑥 3 𝑦 ′ = 12 − 3𝑥 2 𝑦 ′′ = −6𝑥 0 = −6𝑥 0 −6𝑥 = −6 −6 𝒙=𝟎 If 𝒙 = 𝟏 𝑦 = 36 + 12𝑥 − 𝑥 3 𝑦 = 36 + 12(0) − (0)3 𝒚 = 𝟑𝟔 EXAMPLE Compute the acceleration of the particle in No.1 at 𝑡 = 3, 𝑡 = 5, 𝑡 = 10, and 𝑡 = 15. Solution: 𝑑𝑥 = 48𝑡 − 3𝑡 2 𝑑𝑡 ′′ 𝑎 = 48 − 6𝑡 𝑉= 𝒕=𝟑 𝑎′′ = 48 − 6𝑡 𝑎′′ = 48 − 6(3) 𝒂′′ = 𝟑𝟎 𝒊𝒏/𝒔𝒆𝒄 𝒕=𝟓 𝑎′′ = 48 − 6𝑡 𝑎′′ = 48 − 6(5) 𝒂′′ = 𝟏𝟖 𝒊𝒏/𝒔𝒆𝒄 𝒕 = 𝟏𝟎 𝑎′′ = 48 − 6𝑡 NVSU-FR-ICD-05-00 (081220) “In accordance with Section 185, Fair Use of a Copyright Work of Republic Act 8293, the copy righted works included in this material may be reproduced for educational purposes only and not for commercial distribution” P a g e | 12 Republic of the Philippines NUEVA VIZCAYA STATE UNIVERSITY Bambang, Nueva Vizcaya INSTRUCTIONAL MODULE IM No.: MATH 1-1S-2020-2021 𝑎′′ = 48 − 6(10) 𝒂′′ = −𝟏𝟐 𝒊𝒏/𝒔𝒆𝒄 𝒕 = 𝟏𝟓 𝑎′′ = 48 − 6𝑡 𝑎′′ = 48 − 6(15) 𝒂′′ = −𝟒𝟐 𝒊𝒏/𝒔𝒆𝒄 Find the point of inflection of the following functions: 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. If the displacement (in meters) at time 𝑡 (in seconds) of an object is given by 𝑠 = 4𝑡 3 + 7𝑡 2 − 2𝑡. Find the acceleration at time 𝑡 = 10. If 𝑠 = 5𝑡 3 + 3𝑡 + 8, find the velocity and acceleration if 𝑡 = 4 𝑠𝑒𝑐. A man was running along the road with a given law of 𝑠 = 2𝑡 3 − 4𝑡 2 − 5𝑡, determine the velocity and acceleration in kilometers at a given time of 𝑡 = 4 ℎ𝑟. A rock fell down from a cliff 3 600 𝑓𝑡 high. Determine the velocity the rock will hit the ground (𝑉𝑜 = 0). Electricity travelled through a straight cable line according to the given law 𝑠 = 4𝑡 3 + 2𝑡 2 − 3𝑡 km, determine the velocity and acceleration at the time 𝑡 = 0.5 𝑠𝑒𝑐. If 𝑠 = 4𝑡 − 𝑡 2 + 2𝑡 3 m. Find the velocity when acceleration is 1 𝑚/𝑠𝑒𝑐 If 𝑠 = 2𝑡 2 + 4𝑡 − 10 m, when will be the velocity be 2 𝑚/𝑠𝑒𝑐? If 𝑠 = 𝑡 3 − 3𝑡 2 + 6𝑡 − 5, when will be the acceleration be 6 𝑚/𝑠𝑒𝑐? A speeding car with an acceleration of 2 𝑚/𝑠𝑒𝑐 with a given law 𝑠 = 5𝑡 2 − 3𝑡 3 + 4𝑡. Find its velocity. If 𝑠 = 𝑡 4 − 12𝑡 2 − 3𝑡, find the velocity when the acceleration is 24 𝑚/𝑠𝑒𝑐. 5.4 RECTILINEAR RATES If a quantity 𝑥 is a function of time 𝑡, the time of change is given by 𝑑𝑥/𝑑𝑡. 𝑑 𝑑𝑥 𝑓(𝑥) = 𝑓′(𝑥) 𝑑𝑡 𝑑𝑡 EXAMPLE A spherical balloon is inflated with gas at rate of 900 𝑐𝑚3 /𝑚𝑖𝑛. How fast is the of the balloon with a radius of 12 𝑐𝑚? Solution: 4 𝑉 = 𝜋𝑟 3 3 𝑑𝑅 𝑑𝑡 𝑑𝑉 𝑑𝑅 = 4𝜋𝑟 2 𝑑𝑡 𝑑𝑡 𝑉 ′ = 4𝜋𝑟 2 900 𝑐𝑚3 /𝑚𝑖𝑛 = 4𝜋𝑟 2 𝑑𝑅 900 𝑐𝑚3 /𝑚𝑖𝑛 = 𝑑𝑡 4𝜋(12𝑐𝑚)2 𝑑𝑅 900 𝑐𝑚3 /𝑚𝑖𝑛 = 𝑑𝑡 4𝜋(144𝑐𝑚2 ) 𝒅𝑹 = 𝟎. 𝟓𝟎 𝒄𝒎/𝒎𝒊𝒏 𝒅𝒕 𝑑𝑅 𝑑𝑡 NVSU-FR-ICD-05-00 (081220) “In accordance with Section 185, Fair Use of a Copyright Work of Republic Act 8293, the copy righted works included in this material may be reproduced for educational purposes only and not for commercial distribution” P a g e | 13 Republic of the Philippines NUEVA VIZCAYA STATE UNIVERSITY Bambang, Nueva Vizcaya INSTRUCTIONAL MODULE IM No.: MATH 1-1S-2020-2021 EXAMPLE The radius of a circle is decreasing at 4 𝑐𝑚/𝑚𝑖𝑛. How fast is the area and circumference of the circle changing when the radius is 8 𝑐𝑚? Solution: 𝐴 = 𝜋𝑟 2 𝑑𝐴 𝑑𝑟 = 𝜋2𝑟 𝑑𝑡 𝑑𝑡 𝑑𝐴 = 𝜋2(8 𝑐𝑚)(−4 𝑐𝑚/𝑚𝑖𝑛) 𝑑𝑡 𝒅𝑨 = −𝟔𝟒 𝒄𝒎𝟐 /𝒎𝒊𝒏 𝒅𝒕 𝑐 = 2𝜋𝑅 𝑑𝑐 𝑑𝑟 = 2𝜋(1) 𝑑𝑡 𝑑𝑡 𝑑𝑐 = 2𝜋 (−4 𝑐𝑚/𝑚𝑖𝑛) 𝑑𝑡 𝒅𝒄 = −𝟖 𝒄𝒎/𝒎𝒊𝒏 𝒅𝒕 EXAMPLE The surface of a snowball decrease at a rate of 6 𝑓𝑡 2 /ℎ𝑟. How fast is the diameter changing when the radius is 2 𝑓𝑡. Solution: 𝑆𝐴 = 4𝜋𝑟 2 𝐷 2 𝑆𝐴 = 4𝜋 ( ) 2 4𝜋𝐷 2 𝑆𝐴 = 4 𝑆𝐴 = 𝜋𝐷 2 𝑑𝑆𝐴 𝑑𝐷 = 2𝐷𝜋 𝑑𝑡 𝑑𝑡 −6𝑓𝑡 2 /ℎ𝑟 = 𝜋2(4) 𝑑𝐷 𝑑𝑡 𝑑𝐷 −6𝑓𝑡 2 /ℎ𝑟 = 𝑑𝑡 2𝜋(4 𝑓𝑡) 𝒅𝑫 = −𝟎. 𝟐𝟒 𝒇𝒕/𝒉𝒓 𝒅𝒕 Do what is being asked for. 1. A ladder 20𝑓𝑡 long is placed against a wall. The foot of the ladder begins to slide away from the wall at the rate of 1𝑓𝑡/𝑠𝑒𝑐. How fast is the top of the ladder sliding down the wall when the foot of the ladder is 12𝑓𝑡 from the wall? 2. At a certain instant, car A is 60 miles north of a car B. A is travelling south at a rate of 20𝑚𝑖𝑙𝑒𝑠/ℎ𝑜𝑢𝑟 while B is travelling east at 30𝑚𝑖𝑙𝑒𝑠/ℎ𝑜𝑢𝑟. How fast is the distance between them changing 1ℎ𝑜𝑢𝑟 later? 3. A plane, P, flies horizontally at an altitude of 2𝑚𝑖𝑙𝑒𝑠 with a speed of 480𝑚𝑖𝑙𝑒𝑠/ℎ𝑜𝑢𝑟. At a certain moment it passes directly over a radar station, R. How fast is the distance between the plane and the radar station increasing 1𝑚𝑖𝑛𝑢𝑡𝑒 later? NVSU-FR-ICD-05-00 (081220) “In accordance with Section 185, Fair Use of a Copyright Work of Republic Act 8293, the copy righted works included in this material may be reproduced for educational purposes only and not for commercial distribution” P a g e | 14 Republic of the Philippines NUEVA VIZCAYA STATE UNIVERSITY Bambang, Nueva Vizcaya INSTRUCTIONAL MODULE 4. IM No.: MATH 1-1S-2020-2021 A point is moving along the circle 𝑥 2 + 𝑦 2 = 25 in the first quadrant in such a way that its 𝑥 coordinate changes at the rate of 2𝑐𝑚/𝑠𝑒𝑐. How fast its 𝑦 coordinate changing as the point passes through (3,4)? 5. The dimensions of a rectangle are continuously changing. The width increases at the rate of 3𝑖𝑛/𝑠𝑒𝑐 while the length decreases at the rate of 2𝑖𝑛/𝑠𝑒𝑐. At one instant the rectangle is a 20𝑖𝑛2 . How fast is it are changing 3𝑠𝑒𝑐 later? Is the area increasing or decreasing? 6. Two cars begin a trip from the same point P. If car A travels north at the rate of 30𝑚𝑖𝑙𝑒𝑠/ℎ𝑜𝑢𝑟 and car B travels west at the rate of 40𝑚𝑖𝑙𝑒𝑠/ℎ𝑜𝑢𝑟, how fast is the distance between them changing 2ℎ𝑜𝑢𝑟𝑠 later? 7. A man on a wharf 20𝑓𝑡 above the water pulls in a rope, to which a boat is attached, at the rate of 4𝑓𝑡/𝑠𝑒𝑐. At what rate is the boat approaching the wharf when there is 25𝑓𝑡 of rope out? 8. An LRT train 6𝑚 above the ground crosses a street at 9𝑚/𝑠 at the instant that a car approaching at a speed of 4𝑚/𝑠 is 12𝑚 up the street. Find the rate of the LRT train and the car separating 1𝑠𝑒𝑐 later? 9. Water is flowing into a conical cistern at the rate of 8𝑚3 /𝑚𝑖𝑛. If the height of the inverted cone is 12𝑚 and the radius of its circular opening is 6𝑚. How fast is the water level rising when the water is 4𝑚 deep? 10. Water is pouring into a conical vessel 15𝑐𝑚 deep and having a radius of 3.75𝑐𝑚 across the top. If the rate at which the water rises is 2𝑐𝑚/𝑠𝑒𝑐, how fast is the water flowing into the conical vessel when the water is 4𝑐𝑚 deep? 5.5 PARAMETRIC EQUATIONS A curve in the plane is said to be parameterized if the set of coordinates on the curve, (𝑥, 𝑦), are represented as functions of a variable 𝑡. Namely, 𝑥 = 𝑓(𝑡) 𝑦 = 𝑔(𝑡) 𝑡∈ 𝐷. where 𝐷 is a set of real numbers. The variable 𝑡 is called a parameter and the relations between 𝑥, 𝑦 and 𝑡 are called parametric equations. The set 𝐷 is called the domain of 𝑓 and 𝑔 and it is the set of values 𝑡 takes. Derivation in Parametric Form 𝑑𝑥 - is the rate of change of 𝑥 with respect to 𝑡. 𝑑𝑡 𝑑𝑦 - is the rate of change of 𝑦 with respect to 𝑡. 𝑑𝑡 First Derivative 𝑑𝑦 𝑑𝑦 𝑑𝑡 = 𝑑𝑥 𝑑𝑥 𝑑𝑡 Second Derivative 𝑑2𝑦 𝑑 𝑑𝑦 𝑑 𝑑𝑦 𝑑𝑡 = 𝑦 ′′ = ( )= ( )( ) 2 𝑑𝑥 𝑑𝑥 𝑑𝑥 𝑑𝑥 𝑑𝑥 𝑑𝑥 NVSU-FR-ICD-05-00 (081220) “In accordance with Section 185, Fair Use of a Copyright Work of Republic Act 8293, the copy righted works included in this material may be reproduced for educational purposes only and not for commercial distribution” P a g e | 15 Republic of the Philippines NUEVA VIZCAYA STATE UNIVERSITY Bambang, Nueva Vizcaya INSTRUCTIONAL MODULE IM No.: MATH 1-1S-2020-2021 EXAMPLE 𝑥 = 𝑥 3 − 3𝑥 − 17 𝑎𝑛𝑑 𝑦 = 7𝑥 4 − 10𝑥 2 − 0 Solution: 𝑑𝑥 = 𝑥2 − 3 𝑑𝑡 𝑑𝑦 = 28𝑥 3 − 20𝑥 𝑑𝑡 𝑑𝑦 𝑑𝑦 𝑑𝑡 = 𝑑𝑥 𝑑𝑥 𝑑𝑡 𝒅𝒚 𝟐𝟖𝒙𝟑 − 𝟐𝟎𝒙 = 𝒅𝒙 𝒙𝟐 − 𝟑 (𝑥 2 − 3)(84𝑥 2 − 20) − (28𝑥 3 − 20𝑥)(2𝑥) (𝑥 2 − 3)2 4 2 2 84𝑥 − 20𝑥 − 252𝑥 + 60 − 56𝑥 4 − 40𝑥 2 𝑦 ′′ = (𝑥 2 − 3)2 4 − 312𝑥 2 + 60 28𝑥 1 𝑦 ′′ = × 2 (𝑥 2 − 3)2 𝑥 −3 𝟐𝟖𝒙𝟒 − 𝟑𝟏𝟐𝒙𝟐 + 𝟔𝟎 𝒚′′ = (𝒙𝟐 − 𝟑)𝟑 𝑦 ′′ = EXAMPLE 𝑥 = 𝑐𝑜𝑠5𝑥 𝑎𝑛𝑑 𝑦 = 𝑐𝑜𝑠 5𝑥 Solution: 𝑑𝑥 = −5𝑠𝑖𝑛5𝑥 𝑑𝑡 𝑑𝑦 = −5𝑠𝑖𝑛5𝑥 𝑑𝑡 𝑑𝑦 𝑑𝑦 𝑑𝑡 = 𝑑𝑥 𝑑𝑥 𝑑𝑡 𝑑𝑦 −5𝑠𝑖𝑛5𝑥 = 𝑑𝑥 −5𝑠𝑖𝑛5𝑥 𝒅𝒚 =𝟎 𝒅𝒙 ′′ 𝒚 =𝟎 Solve the following parametric equations 1. = 𝑡 2 + 𝑡 𝑦 = 2𝑡 − 1 2. 𝑥 = 4 − 2𝑡 𝑦 = 3 + 6𝑡 − 4𝑡 2 3. 𝑥 = 𝑒 2𝑡 𝑦 = 𝑒 3𝑡 4. 𝑥 = 3 𝑠𝑖𝑛𝜃 𝑦 = 4 𝑐𝑜𝑠𝜃 NVSU-FR-ICD-05-00 (081220) “In accordance with Section 185, Fair Use of a Copyright Work of Republic Act 8293, the copy righted works included in this material may be reproduced for educational purposes only and not for commercial distribution” P a g e | 16 Republic of the Philippines NUEVA VIZCAYA STATE UNIVERSITY Bambang, Nueva Vizcaya INSTRUCTIONAL MODULE IM No.: MATH 1-1S-2020-2021 5. 𝑥 = √𝑡 + 1 1 𝑦= 𝑡+1 6. 𝑥 = 2 + 𝑡 𝑦 = 𝑡 2 − 4𝑡 + 7 7. 𝑥 = 2𝑡 + 1 𝑦 = 4𝑡 − 3 8. 𝑥 = 𝑡 2 𝑦 = 𝑡3 9. 𝑥 = 1 − 𝑡 2 𝑦 = 3 + 2𝑡 10. 𝑥 = 3 𝑡𝑎𝑛𝑥 𝑦 = 3 𝑠𝑖𝑛2 A) Book/Printed Resources AL., L. G. (1989). Calculus for Business Economics and the Social Life Science, 3rd Edition. McGraw_hill Book Corporation. Anton, H. (1992). Multivariable Calculus, 4th Edition. New York. Anton, H. (n.d.). Multivariable Calculus, 4th Edition. 1992. New York: John Wiley and Sons Incorporated. Berkey, D. D. (1990). Calculus for Management and Social Sciences, Second Edition. Saunders College Publishing. Farlow, S. J. (1993). Calculus and its Application. Prentice-Hall Incorporated. Leithold, L. (2001). The Calculus, 7th Edition. Addison-Wesley. Margaret B. Cozzens. (1987). Mathematics with Calculus. D.C. Health. Robert Ellis et. AL. (1990). Calculus with AnalyticGeometry. Hardcourt Brace Jovanovich Incorported. B) e-Resources http://www.soton.ac.uk/~cjg/eng1/modules/modules.html http://www.mit.opencourseware.com http://www.mathalino.com/reviewer/advance-engineering-mathematics/advance-engineeringmathematics http://ecereviewcourse.blogspot.com/p/math.html NVSU-FR-ICD-05-00 (081220) “In accordance with Section 185, Fair Use of a Copyright Work of Republic Act 8293, the copy righted works included in this material may be reproduced for educational purposes only and not for commercial distribution” P a g e | 17