Velocity and Acceleration

Velocity and Acceleration Part 1: Limits, Derivatives, and Antiderivatives In R 3 ; vector-valued functions are of the form r (t) = hf (t) ; g (t) ; h (t)i ; t in [a; b] If f (t) ; g (t) ; and h (t) are continuous over [a; b], then r (t) ; t in [a; b] ; parameterizes a curve C. If r (t 1 ) 6= r (t 2 ) for all t 1 6= t 2 in (a; b) ; then C is an oriented curve with the orientation going from r (a) to r (b). Conversely, r (t) ; t in [a; b] ; is called a parametrization of C: Sometimes we will refer to r (t) as a vector-valued dependent variable. The variable t is called the parameter of r (t) : EXAMPLE 1 for t in [; 2] : Sketch the curve parametrized by r (t) = hcos (t) ; sin (t) ; ti Solution: The computer algebra system Maple yields the following curve (Note: Click and drag on the image above to see it from di erent views). 1

Because we have little experience with 3-dimensional curves, eliminating t from x = f (t) ; y = g (t) ; and z = h (t) tells us little about the curve. Instead, we must develop the calculus of vector-valued functions in R 3 as a means of studying curves and their parameterizations. To begin with, we de ne the limit of a vector-valued function r (t) = hf (t) ; g (t) ; h (t)i to be lim r (t) = lim f (t) ; lim g (t) ; lim h (t) (1) t!p t!p t!p t!p when each of the limits exist. As a result, we de ne the derivative of r (t) = hf (t) ; g (t) ; h (t)i to be dr dt = lim t! r (t + t) t which via the de nition of the limit (1) yields dr dt = f (t + t) f (t) g (t + t) lim ; lim t! t t! t r (t) g (t) ; lim t! h (t + t) t h (t) We often denote dr=dt by v and call it the velocity of r (t) : Likewise, the derivative of v (t) is often called the acceleration of r (t) : The computation above shows us that we have the following: Theorem 6.1: The velocity v (t) of the function r (t) = hf (t) ; g (t) ; h (t)i is v (t) = dr dt = hf (t) ; g (t) ; h (t)i Likewise, the acceleration of r (t) is given by a (t) = dv dt = hf (t) ; g (t) ; h (t)i If v (t) exists for each t in (a; b) ; then we say that r (t) is di erentiable over (a; b). EXAMPLE 2 Compute the velocity and acceleration of r (t) = t 2 ; e 2t ; t 3 Solution: To begin with, the velocity of r (t) is d v (t) = dt t2 ; d dt e2t ; d dt t3 = 2t; 2e 2t ; 3t 2 2

As a result, the acceleration of r (t) is a (t) = dv dt = d dt 2t; d dt 2e2t ; d dt 3t2 = 2; 4e 2t ; 6t Check your Reading: What is v (t) if r (t) = t 2 ; t 2 ; t 2? Velocity as a Tangent Vector If r (t) is the position vector of a moving object at time t; then the vector r (t + t) r (t) is a secant vector whose initial and terminal points are on the curve traced by the terminal end of r (t) : Thus, as t approaches, the resulting secant vectors more and more resemble the curve itself in the vicinity of r (t) ; so that as t approaches, the di erence quotient r (t + t) t r (t) approaches the tangent vector shown in red above. Since the di erence quotient also approaches v (t) as t approaches, we obtain the following theorem: 3

Theorem 6.2: The velocity vector v = dr=dt is tangent to the parametrized curve at the point with position r (t). A curve which has a non-zero tangent vector at each point is said to be smooth, and a parameterization r (t) ; t in [a; b], is said to be smooth if it parameterizes a smooth curve. EXAMPLE 3 Sketch v (1) at the point of tangency to r (t) = t 3 ; t 2 : Solution: To do so, we rst di erentiate to obtain v (t) = dr dt = 3t 2 ; 2t We then let t = 1 to obtain v (1) = h3; 2i which when graphed versus r (t) yields the following: 4

Notice in example 3 that v () = h; i, so r (t) = t 3 ; t 2 is not smooth at r () : As the gure shows, the curve has a cusp at r () and there is no tangent line (or a tangent vector) at a cusp. EXAMPLE 4 Sketch v (5=2) at the point of tangency to the helix r (t) = h4 cos (t) ; 4 sin (t) ; ti. Solution: To do so, we notice that v (t) = h 4 sin (t) ; 4 cos (t) ; 1i so that the velocity vector at t = 5=2 is 5 5 v = 4 sin ; 4 cos 2 2 5 2 which is graphed along with the curve below: ; 1 = h 4; ; 1i Check your Reading: Which plane is v () parallel to if r (t) = hcos (t) ; sin (t) ; ti? 5

Antiderivatives of Vector-Valued Functions In addition, we de ne antiderivatives of q (t) = hf (t) ; g (t) ; h (t)i by Z Z Z Z q (t) dt = f (t) dt; g (t) dt; h (t) dt + hc 1 ; C 2 ; C 3 i where C 1 ; C 2 ; C 3 are arbitrary constants. The fundamental theorem of calculus then implies that Z * b Z b Z b Z + b q (t) dt = f (t) dt; g (t) dt; h (t) dt a or equivalently, if Q (t) = q (t) ; then Z b a a a a q (t) dt = Q (b) Q (a) (2) If a (t) is the acceleration of an object with an initial position r = r () and an initial velocity v = v () ; then (2) impies that a () d = v (t) v () where is the lowercase Greek letter tau and is used here because t is being used as the parameter of v (t) : Similarly for r (t) ; so that v (t) = v + a () d and r (t) = r + are the velocity and position of the objects at time t: v () d EXAMPLE 5 Find the velocity and position of an object subject to an acceleration of a (t) = h 8 sin (2t) ; 8 cos (2t) ; 32i given that v = h; ; i and r = h; ; 1i : Solution: The velocity is given by v (t) = h; 4; i + h 8 sin (2) ; 8 cos (2) ; 32i d = 8 sin (2) d; 8 cos (2) d; 32d D E = 4 cos (2)j t ; 4 sin (2)jt ; 32jt 6

Thus, v (t) = h4 cos (2t) 4; 4 sin (2t) ; 32ti, so that r (t) = h; ; 1i + v () d = (4 cos (2) 4) d; 4 sin (2) d; 1 D = 2 sin (2) 4j t ; 2 cos (2)jt ; 1 16t2E = 2 sin (2t) 4t; 2 cos (2t) 2; 1 16t 2 32d A projectile is an object whose motion is due solely to the force of gravity. The scientist Galileo showed that if a projectile moves only a short distance, then its acceleration is practically constant. That is, its acceleration is given by a (t) = g (3) where g is a constant vector called the acceleration due to gravity. The velocity v (t) of the projectile follows from integration of (3) and a given initial velocity v : The position r (t) at time t follows from integration of v (t) and a given initial position r : EXAMPLE 6 Near the surface of Mars, g = ; ; 12:2 ft= sec 2 : Find the position of a projectile with an initial position of r = h; ; i and an initial velocity of v = h1; 2; 64i. Solution: To begin with, (3) says that a (t) = h; ; 12:2i : The velocity is Z Z v (t) = a (t) dt = h; ; 12:2i dt = h; ; 12:2ti + hc 1 ; C 2 ; C 3 i 7

and since v = h1; 2; 64i ; we must have h1; 2; 64i = v () = h; ; 12:2 i + hc 1 ; C 2 ; C 3 i = hc 1 ; C 2 ; C 3 i Thus, the velocity of the object is v (t) = h; ; 12:2ti + h1; 2; 64i = h1; 2; 12:2t + 64i To nd r (t) ; we antidi erentiate again to obtain Z Z r (t) = v (t) dt = h1; 2; 12:2t + 64i dt = 1t; 2t; 6:1t 2 + 64t +hc 1 ; C 2 ; C 3 i Since r = h; ; i ; we must have hc 1 ; C 2 ; C 3 i = h; ; i ; so that the position of the object at time t is r (t) = 1t; 2t; 64t 6:1t 2 Check your Reading: Would you weigh more or less if you were on Mars? Plane of Motion and Maximum Altitude Let s suppose that r (t) is the position of a projectile at time t. Since v and a span the same plane at each time t, the motion of the projectile is in the plane through the point r with normal n = v a The maximum altitude occurs when v is horizontal, which is when the 3rd component of v is identically zero. The time to maximum altitude is denoted t max and is obtained by setting the third component of v equal to and solving for t: 8

Finally, the point at which the projectile strikes the ground (i.e., the xy-plane) is the point at which the third component of r (t) is equal to. Setting the third component of r (t) equal to and solving for t thus yields the time of impact, denoted t imp : The point of impact is r (t imp ) : EXAMPLE 7 Suppose a projectile has a position at time t of r (t) = 1t; 2t; 64t 16t 2 Find the plane of motion, the time to maximum height, the maximum height, the time of impact, and the point of impact of the projectile. Solution: To begin with, the velocity of the object is v (t) = h1; 2; 64 32ti and the acceleration is as expected, a (t) = h; ; 32i : Since n = v a = h1; 2; 64i h; ; 32i = h 64; 32; i the plane of motion is the plane through r = (; ; ) with normal n = h 64; 32; i : Thus, the equation of the plane of motion is 64 (x ) + 32 (y ) + (z ) = which is the vertical plane passing through the line y = 2x: To nd t max ; we set the third component of v (t) equal to and solve for t max : 64 32t = 32t = 64 t max = 2 Since t max = 2; the maximum altitude itself is simply the z-coordinate of r (t max ) : Since r (t max ) = r (2) = h2; 4; 128 64i = h2; 4; 64i the maximum altitude of the projectile is 64 feet. Finally, the time of impact occurs when the third component of r (t) is equal to : 64t 16t 2 = 16t (t 4) = t = or t = 4 9

Initially, the projectile is on the ground (i.e., r = h; ; i ). Thus, the time of impact must be t imp = 4; so that the point of impact is r (t imp ) = r (4) = (4; 8; ) Exercises Find the velocity and acceleration of the following vector-valued functions: 1. r (t) = t 2 ; t 3 ; t 4 2. r (t) = t 2 ; t 5 ; t 4 3. r (t) = t 1 ; t 2 ; t 3=2 D 4. r (t) = t 2 1 1=2 ; t 2 + 1 E 1 ; 1 5. r (t) = h3 cos (t) ; 5 sin (t) ; 4 cos (t)i 6. r (t) = h3 cos (t) ; 5 sin (t) ; 4 cos (t)i 7. r (t) = htan (t) ; cot (t) ; csc (t)i 8. r (t) = e t ; e t ; ln t 2 + 1 9. r (t) = h1; 1; ln [sec (t)]i 1. r (t) = h2; 3; 1i e t h1; 4; 7i e t 11. r (t) = t 2 i + tan 1 (t) j 12. r (t) = i + t j + t 2 k 13. r (t) = e t hsin (t) ; cos (t) ; ti 14. r (t) = tan (t) hsin (t) ; cos (t) ; sec (t)i Find the velocity at the given time and sketch it along with the curve r (t) : 15. r (t) = t 2 ; t 4 ; t = 1 16. r (t) = 2t; 64 16t 2 ; t = 1 17. r (t) = hcos (t) ; sin (t)i ; t = 6 18. r (t) = hcos (t) ; sin (t)i ; t = 6 19. r (t) = he t ; e t i ; t = 2. r (t) = htan (t) ; sec (t)i ; t = 4 21. r (t) = h3t + 1; 2t + 3i ; t = 2 22. r (t) = bt 2 ; 2bt ; t = 1 1

Find the velocity v (t) and the position r (t) for the given acceleration and initial conditions, and determine the plane of motion (if it exists). In exercises 23-26, also nd the time to maximum height, the maximum height, the time of impact, and the point of impact of the projectile (note: 27 and 28 are projectile motions with a small amount of atmospheric drag). 23. a (t) = h; ; 32i 24. a (t) = h; ; 32i r = h; ; i ; v = h1; 2; 64i r = h; ; i ; v = h1; 1; 96i 25. a (t) = h; ; 9:8i 26. a (t) = h; ; 9:8i r = h1; 3; i ; v = h72; 38; 65i 27. a (t) = ; ; 32e t=1 28. r = h2; 1; 3i ; v = h1; 1; i a (t) = ; ; 32e t=2 r = h; ; 64i ; v = h32; 32; i r = h; ; 4i ; v = h; ; i 29. a (t) = h 3 sin (t) ; 5 cos (t) ; 4 sin (t)i 3. a (t) = h 3 sin (t) ; 3 cos (t) ; 12:2i r = h; ; i ; v = h; ; i r = h; ; i ; v = h; ; i 31. In example 6 a projectile is moving near the surface of Mars, and we found that its position at time t is given by r (t) = 1t; 2t; 64t 6:1t 2 Find the plane of motion, the time to maximum height, the maximum height, the time of impact, and the point of impact of the projectile, and then compare the result to example 7, in which a projectile with the same initial conditions is traveling near the earth. 32. When r () is in the xy-plane, then the range of the projectile i.e., the distance the projectile travels between leaving the earth and striking the earth is the length of the vector r (t imp ) r () : That is, range = kr (t imp ) r ()k Find the range of the projectile in example 7, and then nd the range of the projectile in problem 31. How much farther would the projectile travel on Mars 11

than it would on the earth? Exercises 33-38 explore Major League Baseball s "GameDay" pitch tracking system, in which the trajectory of a pitch is calculated using an imaging system, the coordinate system below, and projectile motion. Speci cally, the Game Day pitch tracking system estimates the trajectory of a baseball in terms of a constant acceleration a (shown in black in the image above), a position r when tracking begins (the beginning of the blue curve), and an initial velocity v when tracking begins (4, 5, or 55 feet from the back of home plate). 33. On September 9, 29 in a game between Anaheim and Seattle, Mark Jipsen through Ken Gri ey, Jr., a sweeping curveball with the following gameday vectors: r = h 2:924; 5; 5:895i ; v = h2:264; 121:647; 2:486i a = h1:529; 3:971; 41:439i Find the parameterization of the ball s trajectory r (t) by integrating a twice and using the given initial data. What is the plane of motion of the curveball? What is its position as it crosses the front of the plate ( i.e., when y = 1:417 feet)? How much did it move from left to right 34. Repeat Exercise 33 for Mark Jipsen s last pitch of that game, which is r = h 2:91; 5; 5:773i ; v = h8:42; 141:714; 4:76i a = h 1:669; 36:44; 13:16i Is this also a curveball (in your opinion)? 35. On July 3, 29 in a game between Minnesota and Detroit, Joel Zumaya threw a pitch clocked at well over 1 mph (and possibly the fastest pitch ever recorded a possible 15 mph when it left his hand at approximately y = 55 feet). The data for that pitch is 12

r = h 1:91; 5; 5:928i ; v = h9:588; 147:939; 5:642i a = h 15:819; 39:136; 6:834i How fast was the pitch traveling initially (i.e., when y = 5 feet)? How long did it take to travel from y = 55 (approximate point of release) to y = 1:417 feet (the front of the plate)? How fast was it traveling when it crossed the front of the plate? (hint: 1 mile = 528 feet). 36. Try it out! The data and information above can be obtained by beginning at Alan M. Nathan s website http://webusers.npl.illinois.edu/~a-nathan/pob/tracking.htm. Try nding the information above for an outing of your favorite pitcher and then repeating exercise 33. 37. Magnus Force: The pitches in exercise 33 to 36 do not fall straight down because the spin of a baseball creates a force called the Magnus force which is perpendicular both to the direction of motion and to the axis about which the ball spins. For example, an angular velocity of about 15 revolutions per minute induces a Magnus force which is about 1/3 the acceleration due to gravity i.e., about 11 feet per sec per sec! Drag must also be considered when Magnus force is considered, giving us an approximate acceleration of a = h11; 32; 32i 13

Use this information to construct the trajectory of such a baseball given that r = h; 5; 6i and v = h; 132; i (i.e., ball 6 feet above the ground traveling 9 mph at 5 feet from the back of home plate). 38. Magnus Force Theory: In reality, the drag, magnus, and gravitational forces lead to an acceleration vector of the form a = KC D vv KC L v (v!) g where K is a constant equalt to about 5:44 1 3 per foot, v is the magnitude of the velocity v vector, g is the acceleration due to gravity,! is the unit vector parallel to the axis of rotation of the ball, and C D and C L are the drag and lift constants, respectively. Explain why a cannot be constant (i.e., the same at all times), and use this to explain why the Gameday curves are only statistical approximations and not actual physical trajectories. (more with Magnus force in later sections!) 39. When r = h; ; i ; then projectile motion begins at the origin. Show that if r = h; ; i ; then t imp = 2t max : 4. * A Basketball s Initial Velocity: An NBA player shoots a 3-pointer from 24 feet ( 3 point line is at 23 9 ), and 1.4 seconds elapse before it passes through the hoop. If we assume that the player released the ball from a height of 8 feet, then what is the initial velocity of the ball? And what is the ball s maximum altitude? 41. Bezier Curves: The Bezier curve determined by the 4 points P (x ; y ) ; P 1 (x 1 ; y 1 ) ; P 2 (x 2 ; y 2 ) ; and P 3 (x 3 ; y 3 ) is the curve with endpoints P and P 3!! which is tangent to the vectors P P 1 and P 2 P 3 at the points P and P 3 ; respec- 14

tively. In this exercise, we show that a Bezier curve is the graph of the vector valued function r (t) = hx ; y i t 3 + 3 hx 1 ; y 1 i t 2 (1 t) + 3 hx 2 ; y 2 i t (1 t) 2 + hx 3 ; y 3 i (1 t) 3 for t in [; 1] : 1. (a) Compute r () and r (1) : (b) Compute v (t) ; and then compute v () and v (1) : (c) Explain how (a) and (b) relate to the gure shown above. 42. Bezier Curves: Construct the Bezier curve using the parameterization in exercise 39 using the 4 points P (; ) ; P 1 (; 1) ; P 2 (1; 1) ; and P 3 (1; ) ; and!! then sketch the result along with the vectors P P 1 and P 2 P 3 : 43. Write to Learn: Bezier curves are often used to connect a sequence of points (position vectors) P ; P 1 ; : : : ; P n with a curve subject to a set of control 15

points (position vectors) Q ; : : : ; Q n 1 ; R ; : : : ; R n 1 : Write short essay in which you show that for each j = 1; : : : ; n, the curve r j (t) = P j 1 t 3 + 3R j 1 t 2 (1 t) + 3Q j 1 t (1 t) 2 + P j (1 t) 3 for t in [; 1] satis es r j () = P j 1 and r j (1) = P j, as well as v j () = 3 (R j 1 P j ) ; and v j (1) = 3 (P j 1 Q j 1 ) What conditions on P j ; R j 1 and Q j, j = ; : : : ; n 1; are necessary for the collection of "pieces" r 1 (t) ; : : : ; r n (t) to form a smooth curve connecting all the points P ; P 1 ; P 2 ; : : : ; P n. 44.Write to Learn: Suppose that r (t) = hf (t) ; g (t) ; h (t)i ; t in [a; b] ; parametrizes a curve C; and suppose that is a di erentiable, 1-1 function. Write a short essay explaining why r (t) = hf ( (u)) ; g ( (u)) ; h ( (u))i ; u in [c; d] parameterizes the same curve C when c = 1 (a) and d = 1 (b) : What is the velocity vector for the new parameterization? How is it related to the velocity vector for the original parameterization? 16