lOMoARcPSD|28760100 MCV4U UNIT 1 Rates of Change Assessment Opportunity Calculus And Vector (Best notes for high school - EN-CA) Studocu is not sponsored or endorsed by any college or university Downloaded by Inderjit Sidhu (indersidhu098@gmail.com) lOMoARcPSD|28760100 MCV4U UNIT 1: Rates of Change Assessment Opportunity Instructions: There are 10 questions in this Assessment Opportunity for a total of 60 marks. Show all work for each question. 3 2 1. Determine the average rate of change of y in the function π¦ = 2π₯ + 7π₯ + 2π₯ − 3 over the interval [3,5]. (5 marks) Given: 3 2 π¦ = 2π₯ + 7π₯ + 2π₯ − 3 x=3, x=5 3 2 ππ π₯ = 3, π¦ = 2(3) + 7(3) + 2(3) − 3 π¦ = 2(27) + 7(9) + 6 − 3 π¦ = 54 + 63 + 6 − 3 π¦ = 120 Point (3, 120) 3 2 ππ π₯ = 5, π¦ = 2(5) + 7(5) + 2(5) − 3 π¦ = 2(125) + 7(25) + 10 − 3 π¦ = 250 + 175 + 7 π¦ = 432 Point (5, 432) Average Rate of Change: 432−120 5−3 = 312 2 π(π)−π(π) π−π = 156 3 2 ∴ The average rate of change of y in the function π¦ = 2π₯ + 7π₯ + 2π₯ − 3 over the interval [3,5] is 156. Downloaded by Inderjit Sidhu (indersidhu098@gmail.com) lOMoARcPSD|28760100 2 2. Given the function π(π₯) = 2π₯ + 3π₯ + 1, a. Use the secant method to ο¬nd the instantaneous rate of change when x = 1. (4 marks) Given: 2 π(π₯) = 2π₯ + 3π₯ + 1 x= 1 Selected series of points close to x=1: 1.1, 1.01, 1.001 2 ππ π₯ = 1, π(1) = 2(1) + 3(1) + 1 π(1) = 2 + 3 + 1 π(1) = 6 Point: (1, 6) 2 ππ π₯ = 1. 1, π(1. 1) = 2(1. 1) + 3(1. 1) + 1 π(1. 1) = 2. 42 + 3. 3 + 1 π(1. 1) = 6. 72 Point: (1.1, 6.72) 2 ππ π₯ = 1. 01, π(1. 01) = 2(1. 01) + 3(1. 01) + 1 π(1. 01) = 2. 0402 + 3. 03 + 1 π(1. 01) = 6. 0702 Point: (1.01, 6.0702) 2 ππ π₯ = 1. 001, π(1. 001) = 2(1. 001) + 3(1. 001) + 1 π(1. 001) = 2. 004002 + 3. 003 + 1 π(1. 001) = 6. 007002 Point: (1.001, 6.007002) Downloaded by Inderjit Sidhu (indersidhu098@gmail.com) lOMoARcPSD|28760100 Find the average rates of change for each set of points Average Rate of Change: π(π)−π(π) π−π (π₯1, π¦1) = (1, 6) (π₯2, π¦2) = (1. 1, 6. 72) (π₯1, π¦1) = (1, 6) (π₯2, π¦2) = (1. 01, 6. 0702) 6.72−6 1.1−1 0.72 0.1 = 6.0702−6 1.01−1 = 7. 2 0.0702 0.01 = (π₯1, π¦1) = (1, 6) 6.007002−6 1.001−1 = = 7. 02 (π₯2, π¦2) = (1. 001, 6. 007002) 0.007002 0.001 = 7. 002 As the point gets closer to the tangent point, the value of the average rate of change is getting closer to 7. 2 ∴ The instantaneous rate of change when x=1 in the function π(π₯) = 2π₯ + 3π₯ + 1 is approximately 7. b. Use First Principles to ο¬nd the value of the derivative when x = 1. (4 marks) Derivative: π'(π₯) = lim β→0 π(π₯+β)−π(π₯) β Given: 2 π(π₯) = 2π₯ + 3π₯ + 1 x= 1 π'(1) = lim β→0 π'(1) = lim β→0 2 2 (2(1+β) +3(1)+1)−(2(1) +3(1)+1) β 2 (2(1+2β+β )+3+1)−(2+3+1) β Downloaded by Inderjit Sidhu (indersidhu098@gmail.com) lOMoARcPSD|28760100 π'(1) = lim β→0 π'(1) = lim β→0 2 2+4β+2β +3+1−2−3−1 β 2 4β+2β β π'(1) = lim 4 + 2β β→0 π'(1) = lim 4 + 2(0) β→0 π'(1) = 7 ∴The value of the derivative when x = 1 using the First Principles is 7. c. What did you notice? Ans. I noticed that the value of the derivative and the instantaneous rate of change are similar in value (7) when x = 1, which means that the derivative of the function of the curve and the slope of the tangent line at a point on the curve are the same. 3. Explain the di erence between a secant line and a tangent line. How do they relate to the rate of change of a function? Include a sketch of each type of line in your solution. (6 marks) Ans. A secant line on a curve is a straight line that intersects the curve at two points and is also known as the line that joins the two points; while a tangent line on a curve is a straight line that touches the curve at one point, the point it touches is called the tangent point. In relations to the rate of change of a function, the slope of the secant line is equivalent to the average rate of change between two points of the curve; the tangent line is equivalent to the instantaneous rate of change at one point of the curve. Downloaded by Inderjit Sidhu (indersidhu098@gmail.com) lOMoARcPSD|28760100 RED LINE= Tangent Line BLUE LINE= Secant Line 4. The path of a baseball relative to the ground can be modelled by the function π(π‘) =− π‘ + 8π‘ + 1, where d(t) represents the height of the ball in metres, and t represents 2 time in seconds. a. Find the average rate of change of the ball between 1 and 3 seconds. (4 marks) Given: 2 π(π‘) =− π‘ + 8π‘ + 1 t= 1, t= 3 2 ππ π‘ = 1, π(1) =− 1 + 8(1) + 1 = 8 Point: (1, 8) 2 ππ π‘ = 3, π(3) =− 3 + 8(3) + 1 = 16 Point: (3, 16) Downloaded by Inderjit Sidhu (indersidhu098@gmail.com) lOMoARcPSD|28760100 Average Rate of Change: = 16−8 3−1 8 2 = π(π)−π(π) π−π = 4 ∴ The average rate of change of the ball between 1 and 3 seconds is 4m/s. b. Using the secant method, ο¬nd the instantaneous rate of change at 2 seconds. (8 marks) Given: 2 π(π‘) =− π‘ + 8π‘ + 1 t= 2 Selected series of points close to t=2: 2.1, 2.01, 2.001 2 ππ π‘ = 2, π(2) =− 2 + 8(2) + 1 = 13 Point: (2, 13) 2 ππ π‘ = 2. 1, π(2. 1) =− 2. 1 + 8(2. 1) + 1 = 13. 39 Point: (2.1, 13.39) 2 ππ π‘ = 2. 01, π(2. 01) =− 2. 01 + 8(2. 01) + 1 = 13. 0399 Point: (2.01, 13.0399) 2 ππ π‘ = 2. 001, π(2. 001) =− 2. 001 + 8(2. 001) + 1 = 13. 003999 Point: (2.001, 13.003999) Find the average rates of change for each set of points Average Rate of Change: π(π)−π(π) π−π Downloaded by Inderjit Sidhu (indersidhu098@gmail.com) lOMoARcPSD|28760100 (π₯1, π¦1) = (2, 13) (π₯2, π¦2) = (2. 1, 13. 39) (π₯1, π¦1) = (2, 13) (π₯2, π¦2) = (2. 01, 13. 0399) 13.39−13 2.1−2 0.39 0.1 = 13.0399−13 2.01−2 = = 3. 9 0.0399 0.01 = 3. 99 (π₯1, π¦1) = (2, 13) 13.003999−13 2.001−2 = (π₯2, π¦2) = (2. 001, 13. 003999) 0.003999 0.001 = 3. 999 As the point gets closer to the tangent point, the value of the average rate of change is getting closer to 4. 2 ∴ The instantaneous rate of change when x=1 in the function π(π‘) =− π‘ + 8π‘ + 1 is approximately 4 m/s. 5. 3 a. Evaluate lim (3π₯ + 7π₯ − 16). π₯→2 3 ππ π₯ = 2, lim (3(2) + 7(2) − 16) π₯→2 = lim (24 + 14 − 16) = 22 π₯→2 3 ∴ lim (3π₯ + 7π₯ − 16) = 22 π₯→2 b. What is the meaning of this value? Ans. This value means that as x approaches 2 from the right or left, y approaches 22. Downloaded by Inderjit Sidhu (indersidhu098@gmail.com) lOMoARcPSD|28760100 6. Evaluate the following limits. a. lim π₯→4 2 π₯ −16 π₯−4 Substitute ππ π₯ = 4, lim π₯→4 = lim π₯→4 16−16 4−4 2 4 −16 4−4 = 0 ← Inconclusive Factoring lim π₯→4 2 π₯ −16 π₯−4 = lim π₯→4 (π₯ −4)(π₯+4) π₯−4 = lim (π₯ + 4) π₯→4 Substitute ππ π₯ = 4, lim (4 + 4) = 8 π₯→4 b. ∴ lim π₯→4 lim π₯→∞ 2 π₯ −16 π₯−4 3 = 8 8π₯ −5π₯+17 3 2 6π₯ +2π₯ −4π₯ Substitute lim π₯→∞ 3 8(∞) −5(∞)+17 3 2 6(∞) +2(∞) −4(∞) = No Conclusion Downloaded by Inderjit Sidhu (indersidhu098@gmail.com) lOMoARcPSD|28760100 Dividing Everything lim π₯→∞ 8π₯ π₯ 3 − 3 6π₯ π₯ 3 + 3 5π₯ π₯ 2 3 2π₯ π₯ 2 3 + − 5 17 π₯ 4π₯ π₯ 8− π₯ + = lim π₯→∞ 6+ 2 π₯ 3 3 17 − 3 π₯ 4 π₯ 2 Substitute ππ π₯ = ∞, lim π₯→∞ 5 17 8− ∞ + 6+ 2 π₯∞ 3 ∞ 4 − 2 ∞ A constant value divided by a really large number results in a number really small. 8−0+0 6+0−0 = lim π₯→∞ c. 3 8π₯ −5π₯+17 ∴ lim π₯→∞ lim β→0 8 6 = 3 = 2 6π₯ +2π₯ −4π₯ = 4 3 4 3 49+β−7 β Substitute ππ β = 0, lim β→0 = lim β→0 7−7 0 = 49+0−7 0 0 0 ← Inconclusive Rationalizing (multiply both numerator and denominator by the conjugate– ( 49 + β + 7) ) lim β→0 ( 49+β−7)( 49+β+7) = lim β→0 = lim β→0 = lim β→0 β( 49+β+7) (49+β)+(7( 49+β))+(−7( 49+β))−49 (49+β)−49 β( 49+β+7) β( 49+β+7) β β 49+β+7 Downloaded by Inderjit Sidhu (indersidhu098@gmail.com) lOMoARcPSD|28760100 1 = lim β→0 49+β+7 Substitute ππ β = 0, lim β→0 1 7+7 = lim β→0 d. 49+β−7 β ∴ lim β→0 lim π¦ → −2 = 1 49+0+7 1 14 1 14 = 3 π¦ −8 2 2π¦ −7π¦+12 Substitute ππ π¦ = 2, lim π¦ → −2 0 6 = lim π¦ → −2 ∴ lim π¦ → −2 e. lim β→0 = 0 2(2) −7(2)+12 3 π¦ −8 2 = 0 2π¦ −7π¦+12 4 2+β 3 2 −8 2 −2 β Substitute ππ β = 0, lim β→0 4 2+0 −2 0 = 0← Inconclusive Simplifying lim β→0 4 2+β −2 β Downloaded by Inderjit Sidhu (indersidhu098@gmail.com) lOMoARcPSD|28760100 4−2(2+β) 2+β = lim β→0 β 4−4−2β 2+β = lim β→0 β −2β 2+β = lim β→0 = lim − β→0 β 2β β(2+β) 2 2+β = lim − β→0 Substitute πΌπ β = 0, lim − β→0 ∴ lim β→0 4 2+β −2 β 2 2+0 =− 1 =− 1 2 7. Find the slope of the tangent line at point (-2,2) on the curve π(π₯) = 2π₯ + 3π₯ using First Principles. (4 marks) Given: 2 π(π₯) = 2π₯ + 3π₯ Point: (-2,2) First Principles: π = lim β→0 ππ π₯ =− 2, π = lim β→0 = lim β→0 = lim β→0 2 π(π₯+β)−π(π₯) β 2 2 (2(−2+β) +3(−2+β))−(2(−2) +3(−2)) β (2(4−4β+β )−6+3β)−(8−6) β 2 8−8β+2β −6+3β−8+6 β Downloaded by Inderjit Sidhu (indersidhu098@gmail.com) lOMoARcPSD|28760100 = lim β→0 = lim β→0 2 −8β+2β +3β β 2 2β −5β β = lim 2β − 5 β→0 Substitute ππ β = 0, π = lim 2(0) − 5 =− 5 β→0 2 ∴The slope of the tangent line at point (-2,2) on the curve π(π₯) = 2π₯ + 3π₯ using First Principles is -5. 2 8. Find the derivative of the function π(π₯) = π₯ − 10π₯ + 3 using First Principles. (4 marks) Given: 2 π(π₯) = π₯ − 10π₯ + 3 First Principle: π¦' = lim β→0 2 π(π₯+β)−π(π₯) β 2 lim β→0 ((π₯+β) −10(π₯+β)+3)−(π₯ −10π₯+3) β lim β→0 2π₯β+β −10β β lim β→0 2 2 2 π₯ +2π₯β+β −10π₯−10β+3−π₯ +10π₯−3 β 2 lim 2π₯ + β − 10 β→0 Substitute ππ β = 0, lim 2π₯ + (0) − 10 = 2π₯ − 10 β→0 Downloaded by Inderjit Sidhu (indersidhu098@gmail.com) lOMoARcPSD|28760100 2 ∴ The derivative of the function π(π₯) = π₯ − 10π₯ + 3 using First Principles is 2x-10. 9. Using ο¬rst principles, determine the equation of the tangent line at point (2,2) on the curve. 2 π(π₯) = π₯ − 7π₯ + 12 Given: 2 π(π₯) = π₯ − 7π₯ + 12 Point (2,2) First Principles: π = lim β→0 π(π₯+β)−π(π₯) β equation: π¦ = π(π₯ − π₯1) + π¦1 Calculuate the slope ππ π₯ = 2, π = lim β→0 = lim β→0 = lim β→0 2 2 2 ((2+β) −7(2+β)+12)−(2 −7(2)+12) β 4+4β+β −14−7β+12−4+14−12 β 2 4β+β −7β β = lim 4 + β − 7 β→0 = lim β − 3 β→0 Substitute ππ β = 0, π = lim (0) − 3 β→0 π =− 3 The slope of the function is -3, with the point (2, 2) we can substitute the values in the formula. Downloaded by Inderjit Sidhu (indersidhu098@gmail.com) lOMoARcPSD|28760100 π¦ = π(π₯ − π₯1) + π¦1 π¦ =− 3(π₯ − 2) + 2 π¦ =− 3π₯ + 6 + 2 π¦ =− 3π₯ + 8 2 ∴ The equation of the tangent line at point (2,2) on the curve π(π₯) = π₯ − 7π₯ + 12 is y=-3x+8. 2 10. At what point on the parabola π¦ = 3π₯ + 2π₯ is the tangent line parallel to the line π¦ = 10π₯ − 2? Use ο¬rst principles in your solution. Given: 2 Parabola: π¦ = 3π₯ + 2π₯ Line: π¦ = 10π₯ − 2 First Principle: π¦' = lim β→0 π(π₯+β)−π(π₯) β Find the derivative of the parabola π¦' = lim β→0 = lim β→0 = lim β→0 2 2 2 (3(π₯+β) +2(π₯+β))−(3π₯ +2π₯) β 2 2 3π₯ +6π₯β+3β +2π₯+2β−3π₯ −2π₯ β 2 6π₯β+3β +2β β = lim 6π₯ + 3β + 2 β→0 ππ β = 0, π¦' = lim 6π₯ + 3(0) + 2 β→0 = 6π₯ + 2 The slope of the line π¦ = 10π₯ − 2 is its derivative, meaning that the parallel lines will have the same slopes. Downloaded by Inderjit Sidhu (indersidhu098@gmail.com) lOMoARcPSD|28760100 Finding x coordinate: π¦' = 6π₯ + 2 π = 6π₯ + 2 10 = 6π₯ + 2 8 = 6π₯ π₯ = 4 3 Use this value to ο¬nd the y coordinate. ππ π₯ = 4 4 3 = 6( 3 ) + 2 π¦ = 8 4 , π¦ = 6( 3 ) + 2 2 ∴ At point ( 3 ,8) on the parabola π¦ = 3π₯ + 2π₯ is the tangent line parallel to the line 4 π¦ = 10π₯ − 2. Downloaded by Inderjit Sidhu (indersidhu098@gmail.com)