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# 4.1 derivatives as rates linear motions

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### 4.1 derivatives as rates linear motions

1. 1. Derivatives as Rates and Linear Motions
2. 2. Derivatives as Rates and Linear MotionsLet’s investigate derivatives as “rates” of changes.Mathematics formulas appearing in the real worldhave specified input unit and output unit.
3. 3. Derivatives as Rates and Linear MotionsLet’s investigate derivatives as “rates” of changes.Mathematics formulas appearing in the real worldhave specified input unit and output unit.The input variable time, t, is often used in applicationswith units such as seconds, hours, years, etc..
4. 4. Derivatives as Rates and Linear MotionsLet’s investigate derivatives as “rates” of changes.Mathematics formulas appearing in the real worldhave specified input unit and output unit.The input variable time, t, is often used in applicationswith units such as seconds, hours, years, etc..The output unit could be in meters (m) for distancemeasurement as in D = D(t) = distance of a movingobject from a fixed location. D(t)
5. 5. Derivatives as Rates and Linear MotionsLet’s investigate derivatives as “rates” of changes.Mathematics formulas appearing in the real worldhave specified input unit and output unit.The input variable time, t, is often used in applicationswith units such as seconds, hours, years, etc..The output unit could be in meters (m) for distancemeasurement as in D = D(t) = distance of a movingobject from a fixed location.The output unit could be in centigrade (co) fortemperature measurement as inT = T(t) = temperature of an object at time t.
6. 6. Derivatives as Rates and Linear MotionsLet’s investigate derivatives as “rates” of changes.Mathematics formulas appearing in the real worldhave specified input unit and output unit.The input variable time, t, is often used in applicationswith units such as seconds, hours, years, etc..The output unit could be in meters (m) for distancemeasurement as in D = D(t) = distance of a movingobject from a fixed location.The output unit could be in centigrade (co) fortemperature measurement as inT = T(t) = temperature of an object at time t.The output might be W kilograms (kg) for weightmeasurement as inW = W(t) = weight of an object at time t.
7. 7. Derivatives as Rates and Linear MotionsLinear Motions
8. 8. Derivatives as Rates and Linear MotionsLinear MotionsWe begin with Linear Motions, which are problemsrelated to the movements of a point on a straight line.
9. 9. Derivatives as Rates and Linear MotionsLinear MotionsWe begin with Linear Motions, which are problemsrelated to the movements of a point on a straight line.We start with an example from physics.If a stone is thrown straight up on Earththen D = –16t2 + vt whereD = height in feet,t = time in secondsv = initial speed (ft /sec)
10. 10. Derivatives as Rates and Linear MotionsLinear MotionsWe begin with Linear Motions, which are problemsrelated to the movements of a point on a straight line.We start with an example from physics.If a stone is thrown straight up on Earththen D = –16t2 + vt where D = –16t2 + vtD = height in feet,t = time in secondsv = initial speed (ft /sec)
11. 11. Derivatives as Rates and Linear MotionsLinear MotionsWe begin with Linear Motions, which are problemsrelated to the movements of a point on a straight line.We start with an example from physics.If a stone is thrown straight up on Earththen D = –16t2 + vt where D = –16t2 + vtD = height in feet,t = time in secondsv = initial speed (ft /sec)A stone is thrown straight up at aspeed of 96 ft/sec, after t secondsit reaches the height D where D(t) =
12. 12. Derivatives as Rates and Linear MotionsLinear MotionsWe begin with Linear Motions, which are problemsrelated to the movements of a point on a straight line.We start with an example from physics.If a stone is thrown straight up on Earththen D = –16t2 + vt where D = –16t2 + vtD = height in feet,t = time in secondsv = initial speed (ft /sec)A stone is thrown straight up at aspeed of 96 ft/sec, after t secondsit reaches the height D where D(t) = –16t2 + 96t.
13. 13. Derivatives as Rates and Linear MotionsLinear MotionsWe begin with Linear Motions, which are problemsrelated to the movements of a point on a straight line.We start with an example from physics.If a stone is thrown straight up on Earththen D = –16t2 + vt where D = –16t2 + vtD = height in feet,t = time in secondsv = initial speed (ft /sec)A stone is thrown straight up at aspeed of 96 ft/sec, after t secondsit reaches the height D where D(t) = –16t2 + 96t.We make a few observations with some simple algebra.
14. 14. Derivatives as Rates and Linear MotionsThe amount of time the stone willstay in the air is the time between itleaveing the ground till it falls back to D = -16t2 + 96tthe ground, both with D = 0.
15. 15. Derivatives as Rates and Linear MotionsThe amount of time the stone willstay in the air is the time between itleaveing the ground till it falls back to D = -16t2 + 96tthe ground, both with D = 0.Solve D = –16t2 + 96t = 0
16. 16. Derivatives as Rates and Linear MotionsThe amount of time the stone willstay in the air is the time between itleaveing the ground till it falls back to D = -16t2 + 96tthe ground, both with D = 0.Solve D = –16t2 + 96t = 0–16t(t – 6) = 0or t = 0, 6.
17. 17. Derivatives as Rates and Linear MotionsThe amount of time the stone willstay in the air is the time between itleaveing the ground till it falls back to D = -16t2 + 96tthe ground, both with D = 0.Solve D = –16t2 + 96t = 0–16t(t – 6) = 0or t = 0, 6.Hence it will stay in air for 6 seconds.It reaches the maximum height in half of that time or3 seconds.
18. 18. Derivatives as Rates and Linear MotionsThe amount of time the stone willstay in the air is the time between itleaveing the ground till it falls back to D = -16t2 + 96tthe ground, both with D = 0.Solve D = –16t2 + 96t = 0–16t(t – 6) = 0or t = 0, 6.Hence it will stay in air for 6 seconds.It reaches the maximum height in half of that time or3 seconds. The corresponding maximum height itreaches is D(3) = –16(3)2 + 96(3) = 144 ft.
19. 19. Derivatives as Rates and Linear MotionsThe amount of time the stone willstay in the air is the time between itleaveing the ground till it falls back to D = -16t2 + 96tthe ground, both with D = 0.Solve D = –16t2 + 96t = 0–16t(t – 6) = 0or t = 0, 6.Hence it will stay in air for 6 seconds.It reaches the maximum height in half of that time or3 seconds. The corresponding maximum height itreaches is D(3) = –16(3)2 + 96(3) = 144 ft.Note that the speed of the stone slows from96 ft/sec to 0 ft/sec as it reaches the height of 144 ft.
20. 20. Derivatives as Rates and Linear MotionsThe speed or rate of the moving stone isdifferent at any given time t. D = -16t2 + 64t
21. 21. Derivatives as Rates and Linear MotionsThe speed or rate of the moving stone isdifferent at any given time t. We call this“the instantaneous rate (of change in D)” D = -16t2 + 64tor simply “the rate” at time t.
22. 22. Derivatives as Rates and Linear MotionsThe speed or rate of the moving stone isdifferent at any given time t. We call this“the instantaneous rate (of change in D)” D = -16t2 + 64tor simply “the rate” at time t.For example, the instantaneous rate att = 3 is 0 ft/sec when the stone reachesthe maximum height of 144 ft.
23. 23. Derivatives as Rates and Linear MotionsThe speed or rate of the moving stone isdifferent at any given time t. We call this“the instantaneous rate (of change in D)” D = -16t + 64t 2or simply “the rate” at time t.For example, the instantaneous rate att = 3 is 0 ft/sec when the stone reachesthe maximum height of 144 ft.This “instantaneous rate” is the derivative D(t).
24. 24. Derivatives as Rates and Linear MotionsThe speed or rate of the moving stone isdifferent at any given time t. We call this“the instantaneous rate (of change in D)” D = -16t + 64t 2or simply “the rate” at time t.For example, the instantaneous rate att = 3 is 0 ft/sec when the stone reachesthe maximum height of 144 ft.This “instantaneous rate” is the derivative D(t).
25. 25. Derivatives as Rates and Linear MotionsThe speed or rate of the moving stone isdifferent at any given time t. We call this“the instantaneous rate (of change in D)” D = -16t + 64t 2or simply “the rate” at time t.For example, the instantaneous rate att = 3 is 0 ft/sec when the stone reachesthe maximum height of 144 ft.This “instantaneous rate” is the derivative D(t).To see this, consider the “averagerate of change of a function f(t)from the time t = a to t = b”.
26. 26. Derivatives as Rates and Linear MotionsThe speed or rate of the moving stone isdifferent at any given time t. We call this“the instantaneous rate (of change in D)” D = -16t + 64t 2or simply “the rate” at time t.For example, the instantaneous rate att = 3 is 0 ft/sec when the stone reachesthe maximum height of 144 ft.This “instantaneous rate” is the derivative D(t).To see this, consider the “averagerate of change of a function y = f(t) (b, f(b))from the time t = a to t = b” which is (a, f(a)) f(b)–f(a) = Δy b-a=Δxf(b) – f(a) = Δy = the slope of the b–a Δtcord as shown. a b
27. 27. Derivatives as Rates and Linear MotionsThe average rate of D(t) = –16t 2 + 96tfrom t = 1 to t = 3 isD(3) – D(1) 3–1
28. 28. Derivatives as Rates and Linear MotionsThe average rate of D(t) = –16t 2 + 96tfrom t = 1 to t = 3 isD(3) – D(1) = 144 – 80 3–1 2= 32ft/sec
29. 29. Derivatives as Rates and Linear MotionsThe average rate of D(t) = –16t 2 + 96t t=3from t = 1 to t = 3 is 64 ft in 2 sec = 32 ft/secD(3) – D(1) = 144 – 80 3–1 2 t=1= 32ft/sec
30. 30. Derivatives as Rates and Linear MotionsThe average rate of D(t) = –16t 2 + 96t t=3from t = 1 to t = 3 is 64 ft in 2 sec = 32 ft/secD(3) – D(1) = 144 – 80 3–1 2 t=1= 32ft/secOn the graph of D = D(t) this isthe slope of the cord from(1, 80) to ( 3, 144).
31. 31. Derivatives as Rates and Linear MotionsThe average rate of D(t) = –16t 2 + 96t t=3from t = 1 to t = 3 is 64 ft in 2 sec = 32 ft/secD(3) – D(1) = 144 – 80 3–1 2 t=1= 32ft/secOn the graph of D = D(t) this isthe slope of the cord from(1, 80) to ( 3, 144). D D(t) = -16t2 + 96t t 1 3 6
32. 32. Derivatives as Rates and Linear MotionsThe average rate of D(t) = –16t 2 + 96t t=3from t = 1 to t = 3 is 64 ft in 2 sec = 32 ft/secD(3) – D(1) = 144 – 80 3–1 2 t=1= 32ft/secOn the graph of D = D(t) this isthe slope of the cord from slope =32 (3, 144) = avg. rate(1, 80) to ( 3, 144). D D(t) = -16t2 + 96t (1, 80) t 1 3 6
33. 33. Derivatives as Rates and Linear MotionsThe average rate of D(t) = –16t 2 + 96t t=3from t = 1 to t = 3 is 64 ft in 2 sec = 32 ft/secD(3) – D(1) = 144 – 80 3–1 2 t=1= 32ft/secOn the graph of D = D(t) this isthe slope of the cord from slope =32 (3, 144) = avg. rate(1, 80) to ( 3, 144). DThe average rate from t = 1 to t = 2 D(t) = -16t2 + 96tis D(2) – D(1)= 128 – 80 = 48 ft/sec (1, 80) 2–1 1 t 1 3 6
34. 34. Derivatives as Rates and Linear MotionsThe average rate of D(t) = –16t 2 + 96t t=3from t = 1 to t = 3 is 64 ft in 2 sec = 32 ft/secD(3) – D(1) = 144 – 80 3–1 2 t=1= 32ft/secOn the graph of D = D(t) this isthe slope of the cord from slope =32 (3, 144) = avg. rate(1, 80) to ( 3, 144). DThe average rate from t = 1 to t = 2 D(t) = -16t2 + 96tis D(2) – D(1)= 128 – 80 = 48 ft/sec (1, 80) 2–1 1which is as expected because it’s ttraveling faster in the beginning. 1 3 6
35. 35. Derivatives as Rates and Linear MotionsGiven y = f(x), we define the instantaneous rate attime t = a as the derivative f (a) = lim f(a + h) – f(a) 0 hif it exists.
36. 36. Derivatives as Rates and Linear MotionsGiven y = f(x), we define the instantaneous rate attime t = a as the derivative f (a) = lim f(a + h) – f(a) 0 hif it exists.With the above example D(t) = –16t2 + 96t, we haveD(t) = –16*2t + 96 = –32t + 96
37. 37. Derivatives as Rates and Linear MotionsGiven y = f(x), we define the instantaneous rate attime t = a as the derivative f (a) = lim f(a + h) – f(a) 0 hif it exists.With the above example D(t) = –16t2 + 96t, we haveD(t) = –16*2t + 96 = –32t + 96Hence D(1) = 64 ft/sec or that the upward speed ofthe stone is 64 ft/sec at t = 1.
38. 38. Derivatives as Rates and Linear MotionsGiven y = f(x), we define the instantaneous rate attime t = a as the derivative f (a) = lim f(a + h) – f(a) 0 hif it exists.With the above example D(t) = –16t2 + 96t, we haveD(t) = –16*2t + 96 = –32t + 96Hence D(1) = 64 ft/sec or that the upward speed ofthe stone is 64 ft/sec at t = 1.The rate of 64 ft/sec means that if the gravitydisappeared at the instant t = 1, then the stone wouldgo upward at a constant speed of 64 ft/sec.
39. 39. Derivatives as Rates and Linear MotionsGiven y = f(x), we define the instantaneous rate attime t = a as the derivative f (a) = lim f(a + h) – f(a) 0 hif it exists.With the above example D(t) = –16t2 + 96t, we haveD(t) = –16*2t + 96 = –32t + 96Hence D(1) = 64 ft/sec or that the upward speed ofthe stone is 64 ft/sec at t = 1.The rate of 64 ft/sec means that if the gravitydisappeared at the instant t = 1, then the stone wouldgo upward at a constant speed of 64 ft/sec.It also means that at the instant t = 1,per unit change in the input would give 64 units ofchange in output, i.e. 1 sec. in time t to 64 ft. in height.
40. 40. Derivatives as Rates and Linear MotionsExample A. A point moves up and down and itsposition at time t is y(t) = –t2–t+12.a. Find the position of thepoint at t = 0. For whatvalue(s) of t is y = 0?
41. 41. Derivatives as Rates and Linear MotionsExample A. A point moves up and down and itsposition at time t is y(t) = –t2–t+12.a. Find the position of thepoint at t = 0. For whatvalue(s) of t is y = 0?When t = 0 the position isy(0) = 12.
42. 42. Derivatives as Rates and Linear MotionsExample A. A point moves up and down and itsposition at time t is y(t) = –t2–t+12.a. Find the position of thepoint at t = 0. For what yvalue(s) of t is y = 0? y= 12 atWhen t = 0 the position is t=0y(0) = 12. 0 y = –t2–t+12
43. 43. Derivatives as Rates and Linear MotionsExample A. A point moves up and down and itsposition at time t is y(t) = –t2–t+12.a. Find the position of thepoint at t = 0. For what yvalue(s) of t is y = 0? y= 12 atWhen t = 0 the position is t=0y(0) = 12. 0If the position y = 0, then–t2 – t + 12 = 0 or y = –t2–t+12
44. 44. Derivatives as Rates and Linear MotionsExample A. A point moves up and down and itsposition at time t is y(t) = –t2–t+12.a. Find the position of thepoint at t = 0. For what yvalue(s) of t is y = 0? y= 12 atWhen t = 0 the position is t=0y(0) = 12. 0 If the position y = 0, then –t2 – t + 12 = 0 or t2 + t –12 = 0 y = –t2–t+12(t – 3) (t + 4) = 0or when t = 3 or –4 that y = 0.
45. 45. Derivatives as Rates and Linear MotionsExample A. A point moves up and down and itsposition at time t is y(t) = –t2–t+12.a. Find the position of thepoint at t = 0. For what y yvalue(s) of t is y = 0? y= 12 atWhen t = 0 the position is t=0y(0) = 12. 0 y= 0 at 0 If the position y = 0, then t= –4 –t2 – t + 12 = 0 or t2 + t –12 = 0 y = –t2–t+12(t – 3) (t + 4) = 0or when t = 3 or –4 that y = 0.
46. 46. Derivatives as Rates and Linear MotionsExample A. A point moves up and down and itsposition at time t is y(t) = –t2–t+12.a. Find the position of thepoint at t = 0. For what y y yvalue(s) of t is y = 0? y= 12 atWhen t = 0 the position is t=0y(0) = 12. 0 y= 0 at 0 0 y= 0 at If the position y = 0, then t= –4 t= 3 –t2 – t + 12 = 0 or t2 + t –12 = 0 y = –t2–t+12(t – 3) (t + 4) = 0or when t = 3 or –4 that y = 0.
47. 47. Derivatives as Rates and Linear Motionsb. Find the rate of change in y at t = –4, 0, 3.Is the point going up or down at those points?
48. 48. Derivatives as Rates and Linear Motionsb. Find the rate of change in y at t = –4, 0, 3.Is the point going up or down at those points?The rate at t is y(t) =–2t – 1 the derivative at t.
49. 49. Derivatives as Rates and Linear Motionsb. Find the rate of change in y at t = –4, 0, 3.Is the point going up or down at those points?The rate at t is y(t) =–2t – 1 the derivative at t.The rates at t = –4, 0, 3 arey(–4) = 7y(0) = –1y(3) = –7
50. 50. Derivatives as Rates and Linear Motionsb. Find the rate of change in y at t = –4, 0, 3.Is the point going up or down at those points?The rate at t is y(t) =–2t – 1 the derivative at t.The rates at t = –4, 0, 3 arey(–4) = 7 yy(0) = –1y(3) = –7Respectively, the point is 0 y= 0 atgoing up when t = –4 as it t= –4passed y = 0. y = –t2–t+12
51. 51. Derivatives as Rates and Linear Motionsb. Find the rate of change in y at t = –4, 0, 3.Is the point going up or down at those points?The rate at t is y(t) =–2t – 1 the derivative at t.The rates at t = –4, 0, 3 arey(–4) = 7 y yy(0) = –1 y= 12 aty(3) = –7 t=0Respectively, the point is 0 y= 0 at 0going up when t = –4 as it t= –4passed y = 0. It’s on theway down when t = 0 as itpasses y =12. y = –t2–t+12
52. 52. Derivatives as Rates and Linear Motionsb. Find the rate of change in y at t = –4, 0, 3.Is the point going up or down at those points?The rate at t is y(t) =–2t – 1 the derivative at t.The rates at t = –4, 0, 3 arey(–4) = 7 y y yy(0) = –1 y= 12 aty(3) = –7 t=0Respectively, the point is 0 y= 0 at 0 0 y= 0 atgoing up when t = –4 as it t= –4 t= 3passed y = 0. It’s on theway down when t = 0 as itpasses y =12. When t = 3, y = –t2–t+12it passes y = 0 going down.
53. 53. Derivatives as Rates and Linear Motionsc. When does y reach the maximum value andwhat is the maximum y value?
54. 54. Derivatives as Rates and Linear Motionsc. When does y reach the maximum value andwhat is the maximum y value?When y reaches the maximum value, its rate y(t)must be 0 because the rate can’t be + or –(it’s not going up nor down at that instance).
55. 55. Derivatives as Rates and Linear Motionsc. When does y reach the maximum value andwhat is the maximum y value?When y reaches the maximum value, its rate y(t)must be 0 because the rate can’t be + or –(it’s not going up nor down at that instance). So itreaches the max. when y(t) = -2t –1 = 0 or t = –½,
56. 56. Derivatives as Rates and Linear Motionsc. When does y reach the maximum value andwhat is the maximum y value?When y reaches the maximum value, its rate y(t)must be 0 because the rate can’t be + or –(it’s not going up nor down at that instance). So itreaches the max. when y(t) = -2t –1 = 0 or t = –½,at y(–½ ) = –(–½)2 – (–½) + 12 = 12¼. y y y (–1/2,12.25) y y(–1/2)=0 y= 12 at (0,12) t=0 y(0)=–30 y= 0 at 0 0 y= 0 at t (–4, 0) (3,0) t= –4 t= 3 y(–4)=7 y(3)= –7 y = –t2–t+12 y = –t2–t+12
57. 57. Derivatives as Rates and Linear MotionsWe may interpret the movement as the shadow castonto the y–axis as the point moves along theparabola.
58. 58. Derivatives as Rates and Linear MotionsWe may interpret the movement as the shadow castonto the y–axis as the point moves along theparabola. y (–1/2,12.25) y Max at y(–1/2)=0 y=12.25 (0,12) t =–1/2 y(0)=–3 cast0 y= 0 at t t= –4 shadow (–4, 0) (3,0) y(–4)=7 y(3)= –7 y = –t2–t+12 y = –t2–t+12
59. 59. Derivatives as Rates and Linear MotionsWe may interpret the movement as the shadow castonto the y–axis as the point moves along theparabola. y y (–1/2,12.25) y Max at y(–1/2)=0 y=12.25 y= 12 at (0,12) t =–1/2 t=0 y(0)=–3 cast0 y= 0 at 0 t t= –4 shadow (–4, 0) (3,0) y(–4)=7 y(3)= –7 y = –t2–t+12 y = –t2–t+12
60. 60. Derivatives as Rates and Linear MotionsWe may interpret the movement as the shadow castonto the y–axis as the point moves along theparabola. y y y (–1/2,12.25) y Max at y(–1/2)=0 y=12.25 y= 12 at (0,12) t =–1/2 t=0 y(0)=–3 cast0 y= 0 at 0 0 y=0 t t= –4 shadow (–4, 0) (3,0) at t=3 y(–4)=7 y(3)= –7 y = –t2–t+12
61. 61. Derivatives as Rates and Linear MotionsWe may interpret the movement as the shadow castonto the y–axis as the point moves along theparabola. y y y (–1/2,12.25) y Max at y(–1/2)=0 y=12.25 y= 12 at (0,12) t =–1/2 t=0 y(0)=–3 cast0 y= 0 at 0 0 y=0 t t= –4 shadow (–4, 0) (3,0) at t=3 y(–4)=7 y(3)= –7 y = –t2–t+12We define the velocity of the linear motion as the1st derivative y of the position function y.
62. 62. Derivatives as Rates and Linear MotionsWe may interpret the movement as the shadow castonto the y–axis as the point moves along theparabola. y y y (–1/2,12.25) y Max at y(–1/2)=0 y=12.25 y= 12 at (0,12) t =–1/2 t=0 y(0)=–3 cast0 y= 0 at 0 0 y=0 t t= –4 shadow (–4, 0) (3,0) at t=3 y(–4)=7 y(3)= –7 y = –t2–t+12We define the velocity of the linear motion as the1st derivative y of the position function y.The velocity y gives the rates of changes in thepositions y as in the phrase “the velocity is 60 mph”.
63. 63. Derivatives as Rates and Linear MotionsThe 2nd derivative y of y is the acceleration(or deceleration depending on the signs of y).
64. 64. Derivatives as Rates and Linear MotionsThe 2nd derivative y of y is the acceleration(or deceleration depending on the signs of y). The acceleration gives the rates of changes in the velocities.
65. 65. Derivatives as Rates and Linear MotionsThe 2nd derivative y of y is the acceleration(or deceleration depending on the signs of y). The acceleration gives the rates of changes in the velocities. Hence if the deceleration is 11 ft /sec2, it would take 8 seconds for a car traveling at 60 mph to come to a complete stop. (1 m = 5280 ft)
66. 66. Derivatives as Rates and Linear MotionsThe 2nd derivative y of y is the acceleration(or deceleration depending on the signs of y). The acceleration gives the rates of changes in the velocities. Hence if the deceleration is 11 ft /sec2, it would take 8 seconds for a car traveling at 60 mph to come to a complete stop. (1 m = 5280 ft) Example B. A point moves along the x–axis and its position is given as x(t) = t4/4 – t3/3..
67. 67. Derivatives as Rates and Linear MotionsThe 2nd derivative y of y is the acceleration(or deceleration depending on the signs of y). The acceleration gives the rates of changes in the velocities. Hence if the deceleration is 11 ft /sec2, it would take 8 seconds for a car traveling at 60 mph to come to a complete stop. (1 m = 5280 ft) Example B. A point moves along the x–axis and its position is given as x(t) = t4/4 – t3/3. a. Find its velocity x. When is it heading right and . when is it heading left?
68. 68. Derivatives as Rates and Linear MotionsThe 2nd derivative y of y is the acceleration(or deceleration depending on the signs of y). The acceleration gives the rates of changes in the velocities. Hence if the deceleration is 11 ft /sec2, it would take 8 seconds for a car traveling at 60 mph to come to a complete stop. (1 m = 5280 ft) Example B. A point moves along the x–axis and its position is given as x(t) = t4/4 – t3/3. a. Find its velocity x. When is it heading right and . when is it heading left? When does it change direction?
69. 69. Derivatives as Rates and Linear MotionsThe 2nd derivative y of y is the acceleration(or deceleration depending on the signs of y). The acceleration gives the rates of changes in the velocities. Hence if the deceleration is 11 ft /sec2, it would take 8 seconds for a car traveling at 60 mph to come to a complete stop. (1 m = 5280 ft) Example B. A point moves along the x–axis and its position is given as x(t) = t4/4 – t3/3. a. Find its velocity x. When is it heading right and . when is it heading left? When does it change direction? What are the right most and left most positions of the point on the x–axis?
70. 70. Derivatives as Rates and Linear MotionsThe 2nd derivative y of y is the acceleration(or deceleration depending on the signs of y). The acceleration gives the rates of changes in the velocities. Hence if the deceleration is 11 ft /sec2, it would take 8 seconds for a car traveling at 60 mph to come to a complete stop. (1 m = 5280 ft) Example B. A point moves along the x–axis and its position is given as x(t) = t4/4 – t3/3. a. Find its velocity x. When is it heading right and . when is it heading left? When does it change direction? What are the right most and left most positions of the point on the x–axis? b. Find the acceleration x and when does x = 0? What’s the meaning of x = 0?
71. 71. Derivatives as Rates and Linear Motionsa. Find its velocity x. When is it heading right andwhen is it heading left? When does it changedirection? What are the right most and left mostpositions of the point on the x–axis?The velocity is x (t) = t3 – t2.
72. 72. Derivatives as Rates and Linear Motionsa. Find its velocity x. When is it heading right andwhen is it heading left? When does it changedirection? What are the right most and left mostpositions of the point on the x–axis?The velocity is x (t) = t3 – t2. It’s heading right whenthe velocity is positive and heading left when thevelocity is negative.
73. 73. Derivatives as Rates and Linear Motionsa. Find its velocity x. When is it heading right andwhen is it heading left? When does it changedirection? What are the right most and left mostpositions of the point on the x–axis?The velocity is x (t) = t3 – t2. It’s heading right whenthe velocity is positive and heading left when thevelocity is negative. Solve x (t) = 0 and draw thesign–chart:
74. 74. Derivatives as Rates and Linear Motionsa. Find its velocity x. When is it heading right andwhen is it heading left? When does it changedirection? What are the right most and left mostpositions of the point on the x–axis?The velocity is x (t) = t3 – t2. It’s heading right whenthe velocity is positive and heading left when thevelocity is negative. Solve x (t) = 0 and draw thesign–chart: t3 – t2 = 0  t2(t – 1) = 0  t = 0, 1
75. 75. Derivatives as Rates and Linear Motionsa. Find its velocity x. When is it heading right andwhen is it heading left? When does it changedirection? What are the right most and left mostpositions of the point on the x–axis?The velocity is x (t) = t3 – t2. It’s heading right whenthe velocity is positive and heading left when thevelocity is negative. Solve x (t) = 0 and draw thesign–chart: t3 – t2 = 0  t2(t – 1) = 0  t = 0, 1Sign–chart – – + + + x(t) = t3–t2 0 1
76. 76. Derivatives as Rates and Linear Motionsa. Find its velocity x. When is it heading right andwhen is it heading left? When does it changedirection? What are the right most and left mostpositions of the point on the x–axis?The velocity is x (t) = t3 – t2. It’s heading right whenthe velocity is positive and heading left when thevelocity is negative. Solve x (t) = 0 and draw thesign–chart: t3 – t2 = 0  t2(t – 1) = 0  t = 0, 1Sign–chart – – + + + x(t) = t3–t2 0 1Hence the point is heading left when t < 1, i.e. from“t = –∞” up to t = 1,
77. 77. Derivatives as Rates and Linear Motionsa. Find its velocity x. When is it heading right andwhen is it heading left? When does it changedirection? What are the right most and left mostpositions of the point on the x–axis?The velocity is x (t) = t3 – t2. It’s heading right whenthe velocity is positive and heading left when thevelocity is negative. Solve x (t) = 0 and draw thesign–chart: t3 – t2 = 0  t2(t – 1) = 0  t = 0, 1Sign–chart – – + + + x(t) = t3–t2 0 1Hence the point is heading left when t < 1, i.e. from“t = –∞” up to t = 1, except at t = 0, it paused (why?).
78. 78. Derivatives as Rates and Linear Motionsa. Find its velocity x. When is it heading right andwhen is it heading left? When does it changedirection? What are the right most and left mostpositions of the point on the x–axis?The velocity is x (t) = t3 – t2. It’s heading right whenthe velocity is positive and heading left when thevelocity is negative. Solve x (t) = 0 and draw thesign–chart: t3 – t2 = 0  t2(t – 1) = 0  t = 0, 1Sign–chart – – + + + x(t) = t3–t2 0 1Hence the point is heading left when t < 1, i.e. from“t = –∞” up to t = 1, except at t = 0, it paused (why?).After t = 1, it changes direction and heads to theright. Here is the corresponding graphic projection.
79. 79. Derivatives as Rates and Linear MotionsThe graph of x(t) = t4/4 – t3/3 xx(t) = t4/4 – t3/3is shown to the right. t
80. 80. Derivatives as Rates and Linear MotionsThe graph of x(t) = t /4 – t /3 4 3 xx(t) = t4/4 – t3/3is shown to the right.Below is the projection of thisgraph (a compressed version) tonto the x–axis. x
81. 81. Derivatives as Rates and Linear MotionsThe graph of x(t) = t /4 – t /3 4 3 xx(t) = t4/4 – t3/3is shown to the right.Below is the projection of thisgraph (a compressed version) tonto the x–axis. The point is coming from the right,pauses (why?) at x = 0 when t = 0, x=0 x at t =0
82. 82. Derivatives as Rates and Linear Motions The graph of x(t) = t /4 – t /3 4 3 x x(t) = t4/4 – t3/3 is shown to the right. Below is the projection of this graph (a compressed version) t onto the x–axis. The point is coming from the right, pauses (why?) at x = 0 when t = 0, continues on to the left. At t = 1 it reaches the left most position x = –1/12,x = –1/12 x=0 xat t =1 at t =0
83. 83. Derivatives as Rates and Linear Motions The graph of x(t) = t /4 – t /3 4 3 x x(t) = t4/4 – t3/3 is shown to the right. Below is the projection of this graph (a compressed version) t onto the x–axis. The point is coming from the right, pauses (why?) at x = 0 when t = 0, continues on to the left. At t = 1 it reaches the left most position x = –1/12, then it changes direction and heads right after t = 1.x = –1/12 x=0 xat t =1 at t =0
84. 84. Derivatives as Rates and Linear Motionsb. Find the acceleration x and when does x = 0?What’s the meaning of x = 0?
85. 85. Derivatives as Rates and Linear Motionsb. Find the acceleration x and when does x = 0?What’s the meaning of x = 0?The acceleration is x(t) = 3t2 – 2t.
86. 86. Derivatives as Rates and Linear Motionsb. Find the acceleration x and when does x = 0?What’s the meaning of x = 0?The acceleration is x(t) = 3t2 – 2t.Set 3t2 – 2t = 0  t = 0, 2/3.
87. 87. Derivatives as Rates and Linear Motionsb. Find the acceleration x and when does x = 0?What’s the meaning of x = 0?The acceleration is x(t) = 3t2 – 2t.Set 3t2 – 2t = 0  t = 0, 2/3.Sign–chart + – + x (t) = 3t –2t 2 0 2/3
88. 88. Derivatives as Rates and Linear Motionsb. Find the acceleration x and when does x = 0?What’s the meaning of x = 0?The acceleration is x(t) = 3t2 – 2t.Set 3t2 – 2t = 0  t = 0, 2/3.Sign–chart + – + x (t) = 3t –2t 2 0 2/3Hence both of them are inflection points of x(t).
89. 89. Derivatives as Rates and Linear Motionsb. Find the acceleration x and when does x = 0?What’s the meaning of x = 0?The acceleration is x(t) = 3t2 – 2t.Set 3t2 – 2t = 0  t = 0, 2/3.Sign–chart + – + x (t) = 3t –2t 2 0 2/3Hence both of them are inflection points of x(t).The sign–chart tells us that the point “accelerates”from t = –∞ to t = 0, x=0 x at t =0
90. 90. Derivatives as Rates and Linear Motionsb. Find the acceleration x and when does x = 0?What’s the meaning of x = 0?The acceleration is x(t) = 3t2 – 2t.Set 3t2 – 2t = 0  t = 0, 2/3.Sign–chart + – + x (t) = 3t –2t 2 0 2/3Hence both of them are inflection points of x(t).The sign–chart tells us that the point “accelerates”from t = –∞ to t = 0, i.e. as it goes left it slows down. The point is “accelerating” because the 2nd derivative x(t) > 0, however the speed of the point is decreasing, or “decelerating” in daily usage of the word. It’s heading left and slowing down is a more apt answer. x=0 x at t =0
91. 91. Derivatives as Rates and Linear Motionsb. Find the acceleration x and when does x = 0?What’s the meaning of x = 0?The acceleration is x(t) = 3t2 – 2t.Set 3t2 – 2t = 0  t = 0, 2/3.Sign–chart + – + x (t) = 3t –2t 2 0 2/3Hence both of them are inflection points of x(t).The sign–chart tells us that the point “accelerates”from t = –∞ to t = 0, i.e. as it goes left it slows down.At t = 0, it stops at x = 0, and the acceleration is 0. x=0 x at t =0
92. 92. Derivatives as Rates and Linear MotionsThen it “decelerates” from t = 0 to t = 2/3, + – + x (t) = 3t2–2t 0 2/3 t=2/3 x=0 x at t =0
93. 93. Derivatives as Rates and Linear MotionsThen it “decelerates” from t = 0 to t = 2/3, Again, the point is “decelerating” because the 2nd derivative x(t) < 0, however the speed of the point is increasing, or “accelerating” in daily usage of the word. It’s heading left and speeding up is a more apt answer. + – + x (t) = 3t2–2t 0 2/3 t=2/3 x=0 x at t =0
94. 94. Derivatives as Rates and Linear MotionsThen it “decelerates” from t = 0 to t = 2/3, that is, itspeeds up the leftward motion from t = 0 until t = 2/3. + – + x (t) = 3t2–2t 0 2/3 t=2/3 x=0 x at t =0
95. 95. Derivatives as Rates and Linear MotionsThen it “decelerates” from t = 0 to t = 2/3, that is, itspeeds up the leftward motion from t = 0 until t = 2/3.After t = 2/3 it “accelerates”, i.e. slowing down theleftward motion. + – + x (t) = 3t2–2t 0 2/3 t=2/3 x=0 x at t =0
96. 96. Derivatives as Rates and Linear MotionsThen it “decelerates” from t = 0 to t = 2/3, that is, itspeeds up the leftward motion from t = 0 until t = 2/3.After t = 2/3 it “accelerates”, i.e. slowing down theleftward motion. At t = 1 the “acceleration” slows theleftward motion or the velocity to 0 with x-min. at -1/12. + – + x (t) = 3t2–2t 0 2/3x = –1/12 t=2/3 x=0 xat t =1 at t =0
97. 97. Derivatives as Rates and Linear MotionsThen it “decelerates” from t = 0 to t = 2/3, that is, itspeeds up the leftward motion from t = 0 until t = 2/3.After t = 2/3 it “accelerates”, i.e. slowing down theleftward motion. At t = 1 the “acceleration” slows theleftward motion or the velocity to 0 with x-min. at -1/12.Then the “acceleration” causes the point to changedirection and head right after t = 0. + – + x (t) = 3t2–2t 0 2/3x = –1/12 t=2/3 x=0 xat t =1 at t =0
98. 98. Derivatives as Rates and Linear MotionsThen it “decelerates” from t = 0 to t = 2/3, that is, itspeeds up the leftward motion from t = 0 until t = 2/3.After t = 2/3 it “accelerates”, i.e. slowing down theleftward motion. At t = 1 the “acceleration” slows theleftward motion or the velocity to 0 with x-min. at -1/12.Then the “acceleration” causes the point to changesdirection and head right, speeding up after t = 1.At t = 0, 2/3, + – + x (t) = 3t2–2tthe acceleration 0 2/3becomes decelerationand vice versa.x = –1/12 t=2/3 x=0 xat t =1 at t =0
99. 99. Derivatives as Rates and Linear MotionsHere is a summary about the acceleration x(t).1. Given that x(t) > 0 and a. its velocity is negative, i.e. x (t) < 0, it means its leftward motion is slowing down. b. its velocity is positive, i.e. x (t) > 0, it means its rightward motion is speeding up.2. Given that x(t) < 0 and a. its velocity is negative, i.e. x (t) < 0, it means its leftward motion is speeding up. b. its velocity is positive, i.e. x (t) > 0, it means its rightward motion is slowing down.3. Given that x(t) = 0. If it’s an inflection point, thenthe acceleration ceased and it begins to decelerate,or vice versa. If it’s not an inflection point, it’s a pause.