Module 6 - Time Rate Problems (PDF)

MODULE 6: TIME RATE PROBLEMS INSTRUCTOR: ENGR. GRANT LLOYD L. LAWAGUEY Topic Learning Outcome: 1. Solve related rates problems involving the rates of change of di erent variables with respect to time. (CLO3) RATES OF CHANGE this gives the rate of change in 𝒚 with respect...

MODULE 6: TIME RATE PROBLEMS INSTRUCTOR: ENGR. GRANT LLOYD L. LAWAGUEY Topic Learning Outcome: 1. Solve related rates problems involving the rates of change of di erent variables with respect to time. (CLO3) RATES OF CHANGE this gives the rate of change in 𝒚 with respect to the variable 𝑥 The derivative of a function can be used to find di erent rates of change of the function with respect to di erent variables such as time. KINEMATICS Definition of terms: Displacement – the distance moved by a particle or body in a specific direction Velocity – the rate of change in the displacement of a particle or body in a given direction Acceleration – the rate of change in the speed of a particle or body over time Given a function, 𝑠(𝑡), that describes the position, 𝑠, of a body or particle over time, the gradient of the graph of 𝑠(𝑡) at any specific time, 𝑡, represents the instantaneous rate of change of displacement at that moment. This rate of change is denoted as , which is actually the velocity of the particle at time t. Similarly, if 𝑣(𝑡) is a function that represent the velocity of an object over time, the rate of change of 𝑣(𝑡) with respect to time can be denoted by , which corresponds to the acceleration of the particle at time, 𝑡. SIGN INTERPRETATION If we have an object moving in a straight line horizontally starting at the origin. If 𝑠(𝑡) = 0 The object is at the origin or starting point. 𝑠(𝑡) > 0 The object is located at the right of the origin. 𝑠(𝑡) < 0 The object is located at the left of the origin. 𝑣(𝑡) = 0 The object is not moving or is instantaneously at rest. 𝑣(𝑡) > 0 The object is moving to the right. 𝑣(𝑡) < 0 The object is moving to the left.  𝑎(𝑡) = 0 The velocity of the object is constant. 𝑎(𝑡) > 0 The velocity is increasing. 𝑎(𝑡) < 0 The velocity is decreasing. EXAMPLE 1. A ball is thrown, its height above the ground is given by 𝑠(𝑡) = 1.2 + 28.1𝑡 − 4.9𝑡 meters where 𝑡 is the time in seconds. a) From what distance above the ground was the ball released? b) Find 𝑡 when = 0. What is the significance of this result? c) What is the maximum height reached by the ball. d) What is the velocity of the ball when 𝑡 = 5𝑠 e) What is the acceleration of the ball when 𝑡 = 2𝑠. Sol’n: a. The distance of the ball was released when 𝑡 = 0. 𝑠(0) = 1.2 + 28.1(0) − 4.9(0) = 1.2 𝑚𝑒𝑡𝑒𝑟𝑠 b. = 28.1 − 2(4.9)𝑡 = 28.1 − 9.8𝑡 0 = 28.1 − 9.8𝑡 9.8𝑡 = 28.1. 𝑡= = 2.87 𝑠. This point is a critical point. c. We can investigate the critical point from letter b if it is the maximum of the function 𝑠(𝑡). From first derivative test 0 < 𝑡 < 2.87 𝑠 𝑡 = 1 𝑠 𝑠 (1) = 28.1 − 9.8(1) = + 𝑡 > 2.87 𝑠 𝑡 =3𝑠 𝑠 (3) = 28.1 − 9.8(3) = − Therefore when 𝑡 = 2.87 𝑠, the ball is at maximum height. 𝑠(2.87 𝑠) = 1.2 + 28.1(2.87) − 4.9(2.87) = 41.49 𝑚𝑒𝑡𝑒𝑟𝑠 d. 𝑣(𝑡) = = 28.1 − 9.8𝑡 𝑣(5) = 28.1 − 9.8(5) = −20.9 e. 𝑎(𝑡) = = −9.8 𝑎(2) = −9.8 TIME RATE PROBLEMS Suggested steps in solving time rate problems: 1. Read and analyze the problem. Draw a large clear diagram to visualize the problem. Sometimes two or more diagrams are necessary. 2. Identify the relevant information, variables and constants. Label the diagram with your variables and constant. 3. Formulate the equation. Use the variables identified in step 2. You will often need the concepts from your trigonometry and solid mensuration. 4. Di erentiate the equation with respect to time. Substitute any known variables and solve for the unknowns. 5. Verify the solution. Check your answer to ensure that it is reasonable and logically consistent with the problem context. EXAMPLE 1. A 5m long ladder rests against a vertical wall with its feet on the ground. The feet on the ground slip, and at the instant when they are 3m from the wall, they are moving at 10 m/s. At what speed is the other end of the ladder moving at an instant? Figure 1 Figure 2 Figure 1 shows the position of the ladder in terms of the variables 𝑥 𝑎𝑛𝑑 𝑦. Figure 2 shows the position of the ladder when the feet is 3m away from the wall. From Figure 1, by applying the Pythagorean theorem, we have an equation 𝑥 +𝑦 =5 Di erentiating the equation with respect to time, 𝑡 (𝑥 ) + (𝑦 ) = (5 ) 2𝑥 + 2𝑦 =0 the rate of change in position of the feet of the ladder the rate of change in position of the other end of the ladder In figure 2, the distance, 𝑥 = 3𝑚 is given while the distance, 𝑦 = 4𝑚 is determined using the Pythagorean theorem. At this instance, the rate of change of 𝑥 with respect to time, , is 10 m/s. Therefore, substituting these values to the equation gives: 2(3)(10) + 2(4) = 0  Solving for 60 + 8 =0 = − = −7.5 𝑚/𝑠 The negative value of the answer means that the distance 𝑦 is decreasing in value. Therefore, the other end of the ladder is moving down 7.5 𝑚/𝑠 at that instant. 2. Air is being pumped into a spherical weather balloon at a constant rate of 6𝜋 𝑚 per minute. Find the rate of change of its surface area at the instant when the radius of the balloon is 2 𝑚. Given: = 6𝜋 𝑚 rate of change of Volume of the balloon 𝑟 =2𝑚 Req’d: rate of change of radius of the balloon rate of change of Surface Area of the balloon Formulas needed: 𝑉 = 𝜋𝑟 𝑆𝐴 = 4𝜋𝑟 Equation: Solving for from the volume of the sphere. = 𝜋 3𝑟 = 4𝜋𝑟 taking the derivative with respect to t 6𝜋 = 4𝜋(2 𝑚) substitute = 6𝜋 𝑎𝑛𝑑 𝑟 = 2 𝑚 = = Solving for from the surface area of the sphere = 4𝜋 2𝑟 taking the derivative with respect to t = 4𝜋(2)(2𝑚) = 6𝜋 substitute = 𝑎𝑛𝑑 𝑟 = 2 𝑚  3. A trough of length 6m has a uniform cross-section which is an equilateral triangle with sides of length 1m. Water leaks from the bottom of the trough at a constant rate of 0.1. Find the rate at which the water level is falling at the instant when the water is 20 cm deep. Given: 𝑙 (𝑙𝑒𝑛𝑔𝑡ℎ) = 6𝑚, 𝑙 𝑖𝑠 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝐷(𝑑𝑒𝑝𝑡ℎ) = 20 𝑐𝑚 = 0.2 𝑚 (𝑟𝑎𝑡𝑒 𝑜𝑓 𝑐ℎ𝑎𝑛𝑔𝑒 𝑜𝑓 𝑉𝑜𝑙. 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟) = 0.1 Req’d: (𝑟𝑎𝑡𝑒 𝑜𝑓 𝑐ℎ𝑎𝑛𝑔𝑒 𝑜𝑓 𝑑𝑒𝑝𝑡ℎ 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟) Formulas By similar triangles √3 = √ solve for b √ 𝑏= 𝑜𝑟 √ Equation √ 𝑉 = (𝑏)(𝐷)(𝑙) = (𝐷)(6) = 2𝐷 √3 volume of water Solving for = 4𝐷 √3. = = (. ) = 𝑜𝑟 0.0722 √ √ √ REFERENCES Dawkins, Paul. Pauls Online Math Notes. https://tutorial.math.lamar.edu. Accessed September 30, 2023 Love, C.E. and Rainville, E. D. (1962). Differential and Integral Calculus. 6th ed. Macmillan Martin, D. et all. (2012). Mathematics for the International Student Mathematics HL (Core) 3rd Edition. Haese Mathematics