2.2 Differentiation and Integration of Vector-Valued Functions

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1 .. DIFFERENTIATION AND INTEGRATION OF VECTOR-VALUED FUNCTIONS133. Differentiation and Interation of Vector-Valued Functions Simply put, we differentiate and interate vector functions by differentiatin and interatin their component functions. Since the component functions are realvalued functions of one variable, we can use the techniques studied in calculus I and II...1 Differentiation Definition..1 Let r t be a vector function. The derivative of r with respect to t, denoted r t or d r is defined to be dt r t = lim h 0 r t + h r t h Geometrically, r a is the vector tanent to the curve at t = a. Definition.. The line tanent to a curve C with position vector r t at t = a is the line throuh r a in the direction of r a. Definition..3 Unit Tanent Vector The unit tanent vector, denoted T t is defined to be T t = r t r t.4 Remark..4 Of course, the above definition makes sense only if r t 0. The derivative is defined in terms of limits. Takin the limit of a vector function amounts to takin the limits of the component functions. Thus, we have the followin theorem: Theorem..5 If r t = f t, t, h t then r t = f t, t, h t. There is a similar result for plane curves. Since the component functions are real-valued functions of one variable, all the properties of the derivative will hold. We have the followin theorem: Theorem..6 Suppose that u and v are differentiable vector functions, c is a scalar and f is a real-valued function. Then: 1. u t ± v t = u t ± v t. c u t = c u t 3. f t u t = f t u t + f t u t 4. u t v t = u t v t + u t v t

2 134 CHAPTER. VECTOR FUNCTIONS 5. u t v t = u t v t + u t v t 6. u f t = f t u f t Example..7 Let r t = t, e t, sin t. Find r t and the unit tanent vector at t = 0. Then, find the equation of the tanent at t = Computation of r t. r t = 1, te t, cos t. Computation of the unit tanent at t = 0. First, we must find the tanent vector at t = 0. This vector is r 0. From our computation above, we see that r 0 = 1, 0, The unit tanent vector at t = 0 is T 0. T 0 = r 0 r 0 = 1, 0, = 5 1 5, 0, 5 3. Computation of the tanent line. The parametric equations of the line throuh r 0 = 0, 1, 0 with direction vector 1, 0, is Fiure 3 shows the raph of r t = its tanent at t = 0. x = t y = 1 z = t t, e t, sin t as well as the raph of

3 .. DIFFERENTIATION AND INTEGRATION OF VECTOR-VALUED FUNCTIONS135 4 z x y Graph of r t = t, e t, sin t and its tanent at t = 0. Definition..8 Smooth Curve A curve C iven by a position vector r t on an interval I is said to be smooth if the conditions below are satisfied: 1. r t is continuous.. r t 0 except possibly at the endpoints of I. Smooth curves will play an important role in the next sections. Geometrically, a curve is not smooth at points where there is a corner also called a cusp. Example..9 Consider the curve iven by r 1 t = 1 + t, t 3. Find if it is smooth on R. What about on 0,? We start by computin r t. r t = t, 3t. We see that r t is always continuous. However, r t = 0 when t = 0. Hence, this curve is not smooth on R. It is smooth on 0, since r t is continuous there and never equal to 0. We finish with the proof of a well known result which we state as a proposition. Proposition..10 Let C be a curve iven by a position vector r t. r t = c a constant then r t r t for all t. If

4 136 CHAPTER. VECTOR FUNCTIONS Proof. We know that r t = r t r t. So, we have r t r t = c. If we differentiate each side and use our rules of differentiation, we et This says that r t r t. r t r t + r t r t = 0 r t r t = 0 r t r t = 0 You may not reconize this result the way it is stated. Think of a circle. The position vector of a circle is its radius. The theorem stated in the case of a circle says that the radius of a circle is perpendicular to the tanent to the circle... Interation Definition..11 If r t = f t, t, h t then b a b r t dt = f t dt, a b a t dt, b a h t dt and r t dt = f t dt, t dt, h t dt We have similar definitions for plane curves. Example..1 Let 1 r t = cos t, sin t, 1 + t. Find R t = r t dt which satisfies R 0 = 3,, 1. R t = r t dt = = cos tdt, sin tdt, t dt 1 sin t + C 1, cos t + C, tan 1 t + C 3 Since we want R 0 = 3,, 1, we must have 3,, 1 = 0 + C 1, + C, 0 + C 3 Thus, C 1 = 3, C = 4 and C 3 = 1. It follows that 1 R t = sin t + 3, cos t 4, tan 1 t + 1

5 .. DIFFERENTIATION AND INTEGRATION OF VECTOR-VALUED FUNCTIONS Velocity and Acceleration In this section, we look at direct applications of the derivative and the interal of a vector function. Definition..13 Velocity and Acceleration Consider an object movin alon C, a smooth curve, twice differentiable, with position vector r t. 1. The velocity of the object, denoted v t, is defined to be v t = r t.5. The acceleration of the object, denoted a t, is defined to be a t = v t = r t.6 3. The speed of the object, denoted v t, is the manitude of the velocity, that is v t. Example..14 Findin the velocity and acceleration of a movin object An object is movin alon the curve r t = t, t 3, 3t for t 0. Find v t, a t and sketch the trajectory of the object as well as the velocity and acceleration when t = 1. Velocity: Acceleration: v t = 1, 3t, 3 a t = 0, 6t, 0 Sketch: When t = 1, we have v 1 = 1, 3, 3 and a 1 = 0, 6, 0 The trajectory of the object as well as the velocity and acceleration at t = 1 are shown in fiure.. In many applications, we do not know the position function. Instead, we know the acceleration and we must find the velocity and position function. The next example illustrates this. Example..15 Findin a Position Function by Interation A movin particle starts at position r 0 = 1, 0, 0 with initial velocity v 0 = 1, 1, 1. Its acceleration is a t = 4t, 6t, 1. Find its velocity and position function at time t.

6 138 CHAPTER. VECTOR FUNCTIONS Fiure.: Motion of an object alon r t = t, t 3, 3t Velocity: Since a t = v t, it follows that v t = a t dt Thus v t = t + C 1, 3t + C, t + C 3 We find the constants by usin the initial condition. v 0 = 1, 1, 1 = C1, C, C 3 Thus, C 1 = 1, C = 1 and C 3 = 1. It follows that v t = t + 1, 3t 1, t + 1 Position Function: Since v t = r t, it follows that r t = v t dt t 3 = 3 + t + C 1, t 3 t + C, t + t + C 3 We find the constants by usin the initial condition. r 0 = 1, 0, 0 = C1, C, C 3

7 .. DIFFERENTIATION AND INTEGRATION OF VECTOR-VALUED FUNCTIONS139 Thus, C 1 = 1 and C = C 3 = 0. It follows that t 3 r t = 3 + t + 1, t3 t, t + t Remark..16 The above problem can be set up as a differential equation, that is an equation which involves the derivatives of an unknown function. Solvin the equation amounts to findin the unknown function. We were iven the acceleration a t and we had to find the position function r t. Since a t = r t, the above problem could have been stated as: Find r t iven that r t = 4t, 6t, 1, r 0 = 1, 0, 0 and v 0 = 1, 1, 1. The last two conditions are called the initial conditions because they are the conditions when t = 0. Of course, we would solve the problem the same way in this particular example. However, many diff erential equations which appear in applied mathematics are far more challenin to solve. Remark..17 In eneral, we can recover velocity by interation when acceleration is known. Similarly, we can recover position by interation when velocity is known. More precisely, if a t 0 and v t 0 are known, then t v t = v t0 + a u du t 0 This can be derived from the definition of acceleration. Since a t = v t, interatin from t 0 to t ives t t 0 a u du = t t 0 v u du. Usin the fundamental theorem of calculus, we et t t 0 a u du = v t v t 0 hence the result. Similarly, if v t and r t 0 are known, then t r t = r t0 + v u du t 0 This can be proven the same way as the formula for velocity. Remark..18 When studyin the motion of an object, the time at which we start trackin the object is usually set to 0. The acceleration at time t = 0 is called initial acceleration and denoted a 0. Similarly, the velocity at time t = 0 is called initial velocity and usually denoted v 0. The position at time t = 0 is called initial position and usually denoted r 0. The term initial means at time t = Projectile Motion If you wondered how we can know the acceleration and not the other quantities, this section will hopefully answer your questions. We bein with a little bit of physics. Newton s second law of motion states that if at time t a force F t

8 140 CHAPTER. VECTOR FUNCTIONS acts on an object of mass m, this action will produce an acceleration a t of the object satisfyin F t = m a t.7 Thus, if we know the force actin on the object, we can find the acceleration. We will then be in the situation of the previous example. We illustrate this with a classical problem in physics, the problem of findin the trajectory of an object bein thrown in the air and subject to the laws of physics. Example..19 A projectile is fired with an anle of elevation α and initial velocity v 0 as shown on fiure.3. If we inore air resistance and assume there is no wind, the only force actin on the object is ravity. 1. Find the position of the object r t in terms of α and v 0.. Express the rane d in terms of α. 3. Find the value of α which maximizes the rane d. 4. What is the maximum heiht reached by the object? Answer to 1. When translatin a real world problem into mathematics so we can solve it, we will have to introduce a coordinate system so we can measure distances and position of objects. This step is crucial. We want to do it in a way that makes all the equations we will derive as simple as possible. We bein by settin the axes so that y is the vertical direction and x the horizontal direction. In addition, the motion of the projectile beins at the oriin. From physics, we can find the ravity vector. Let us introduce some notation. Suppose that a t = a x, a y. Let m be the mass of the object and F t = F x, F y be the force actin on the object at time t > 0. From Newton s second law of motion, F t = m a t in other words, F x, F y = m a x, a y in other words { Fx = ma x F y = ma y.8 The object will move in both the x and y directions. In the x direction, no force is actin on the object from our assumptions, hence F x = 0. From equation.8, it follows that a x = 0. In the y direction, aain from our assumptions, the only force actin on the object comes from ravity. Physics tells us this force is F y = m where = 9.81m/s. The minus sin comes from the fact that the positive direction is upward and ravity pulls the object downward, that is in the neative direction. From equation.8, it follows that F y = ma y = m hence a y =. Combinin what we found for a x and a y, we have a t = 0,

9 .. DIFFERENTIATION AND INTEGRATION OF VECTOR-VALUED FUNCTIONS141 We can now compute v t. Let us introduce more notation, let v 0 = v 0x, v 0y let v t = v x, v y and let v 0 = v 0. Now, by definition, v t = a t dt = C 1, t + C Then, we have Thus v 0 = v0x, v 0y = C 1, C v t = v0x, t + v 0y Elementary trionometry tells us that v 0x = v 0 cos α v 0y = v 0 sin α Thus v t = v0 cos α, t + v 0 sin α Now, we can compute r t. r t = v t dt = v 0 cos α t + C 1, 1 t + v 0 sin α t + C Since r 0 = 0, 0, both C 1 and C are 0 hence r t = v 0 cos α t, 1 t + v 0 sin α t So, we can see that the motion in the x-direction and in the y-direction follow different laws. This can be made more obvious by writin the parametric equations of the trajectory of the object. { x = x t = v 0 cos α t y = y t = 1 t + v 0 sin α t Answer to. d, the rane, is the horizontal distance correspondin to y = 0. We can find d by findin the value of t for which y = 0 and pluin this value in the equation for x. y = 0 1 t + v 0 sin α t = 0 t 1 t + v 0 sin α = 0

10 14 CHAPTER. VECTOR FUNCTIONS Fiure.3: Trajectory of a Projectile This happens when either t = 0 or t = v 0 sin α. t = 0 corresponds to the initial position. The value of t we want is t = v 0 sin α. This value ives us v0 sin α d = x = v 0 cos α v 0 sin α d = v 0 sin α Answer to 3. We see that d is maximum when sin α = 1 that is when α = π or α = π 4. Answer to 4. The object reaches its maximum altitude when its vertical velocity is 0 that is when v y = t + v 0 sin α = 0 or when t = v0 sin α. At that time, the altitude is v0 sin α y = 1 v0 sin α v0 sin α + v 0 sin α = 1 v 0 sin α + v 0 sin α = v 0 sin α

11 .. DIFFERENTIATION AND INTEGRATION OF VECTOR-VALUED FUNCTIONS Thins to Know Be able to tell if and where a curve is smooth. Be able to compute the derivative of a space curve. Know what the unit tanent vector is and be able to find it. Be able to compute the interal of a space curve. Know the relationship between the position vector, the velocity and acceleration of a particle in motion. Be able to find the velocity and acceleration of a particle iven its position vector. Be able to find the velocity and position of a particle iven its acceleration and initial velocity and position. Be able to do problems similar to the example on projectile motion...6 Problems 1. Evaluate 1 0 t 3, 7, t + 1 dt. Evaluate π 4 sin t, 1 + cos t, sec t dt π 4 3. Evaluate t, 1 5 t, 1 t dt 4. Determine if the curves below are smooth on R. a r t = t 3, t 4, t 5 b r t = t 3 + t, t 4, t 5 5. r t = t + 1, t + 1 is the position of a particle in the xy-plane at time t. Find an equation in x and y whose raph is the path of the particle, then find the particle s velocity and acceleration when t = r t = e t, 9 et is the position of a particle in the xy-plane at time t. Find an equation in x and y whose raph is the path of the particle, then find the particle s velocity and acceleration when t = ln Motion on a circle x + y = 1. r t = sin t, cos t is the position of a particle in the xy-plane at time t. Find the particle s velocity and acceleration when t = π 4 and t = π and plot them as vectors alon with the curve. 8. Motion on a cycloid. r t = t sin t, 1 cos t is the position of a particle in the xy-plane at time t. Find the particle s velocity and acceleration when t = π and t = 3π and plot them as vectors alon with the curve.

12 144 CHAPTER. VECTOR FUNCTIONS 9. r t = t + 1, t 1, t is the position of a particle in space at time t. Find the particle s velocity, acceleration and speed when t = r t = cos t, 3 sin t, 4t is the position of a particle in space at time t. Find the particle s velocity, acceleration and speed when t = π. 11. r t = ln t + 1, t, t is the position of a particle in space at time t. Find the particle s velocity, acceleration and speed when t = Find the anle between the velocity and acceleration if r t = 3t + 1, 3t, t at t = Find the anle between the velocity and acceleration if r t = ln t + 1, tan 1 t, t + 1 at t = Find the equation of the line tanent to r t = sin t, t cos t, e t at t = Find the equation of the line tanent to r t = a sin t, a cos t, bt at t = π. 16. Consider particles movin alon the unit circle x + y = 1. Answer the questions below for each of the particle a e. i. Does the particle have constant speed? If so, what is it? ii. Is the particles s acceleration always orthoonal to its velocity vector? iii. Does the particle move clockwise or counterclockwise around the circle? iv. Does the particle bein at the point 1, 0? a r t = cos t, sin t, t 0. b r t = cos t, sin t, t 0. c r t = cos t π, sin t π, t 0. d r t = cos t, sin t, t 0. e r t = cos t, sin t, t A particle moves alon the top of the parabola y = x from left to riht at constant speed 5 units per second. Find the velocity of the particle as it moves throuh,. hint: You first need to find the parametric equations { of the curve. Recall, if y = f x then the parametric equations x = at are where a is a constant. In this case, first write x = f y y = f at then use a similar trick for the parametric equations. 18. Solve for r t in { d r dt = t, t, t. r 0 = 1,, 3

13 .. DIFFERENTIATION AND INTEGRATION OF VECTOR-VALUED FUNCTIONS At time t = 0, a particle is located at the point 1,, 3. It travels in a straiht line to the point 4, 1, 4, has speed at 1,, 3 and constant acceleration 3, 1, 1. Find the position vector r t. 0. A projectile is fired at a speed of 840m/s at an anle of 60. How lon will it take to et 1km downrane? 1. A projectile is fired with an initial speed of 500m/s at an anle of elevation of 45. a When and how far away will the projectile strike? b How hih overhead will the projectile be when it is 5km downrane? c What is the reatest heiht reached by the projectile?. Consider the helix iven by r t = cos t, sin t, t. Prove that r t r t two different ways...7 Answers 1. Evaluate 1 0 t 3, 7, t + 1 dt 1 t 3, 7, t dt = 4, 7, 3. Evaluate π 4 sin t, 1 + cos t, sec t dt π 4 3. Evaluate 4 1 π 4 π 4 0 sin t, 1 + cos t, sec t dt = 0, π +, 1 t, 1 5 t, 1 t dt t, 1 5 t, 1 dt = ln 4, ln 4, ln t 4. Determine if the curves below are smooth on R. a r t = t 3, t 4, t 5 Not smooth. b r t = t 3 + t, t 4, t 5 Smooth. 5. r t = t + 1, t + 1 is the position of a particle in the xy-plane at time t. Find an equation in x and y whose raph is the path of the particle, then find the particle s velocity and acceleration when t = 1. y = x 1 + 1

14 146 CHAPTER. VECTOR FUNCTIONS v t = 1, t Thus v 1 = 1, and a t = 0, = a 1 6. r t = e t, 9 et is the position of a particle in the xy-plane at time t. Find an equation in x and y whose raph is the path of the particle, then find the particle s velocity and acceleration when t = ln 3. y = 9 x v t = e t, 4 9 et Thus and So v ln 3 = 3, 4 a t = e t, 8 9 et a ln 3 = 3, 8 7. Motion on a circle x + y = 1. r t = sin t, cos t is the position of a particle in the xy-plane at time t. Find the particle s velocity and acceleration when t = π 4 and t = π and plot them as vectors alon with the curve. π r = 4, blue π r = 1, 0 red v t = cos t, sin t So π v = 4, liht blue π v = 0, 1 liht red a t = sin t, cos t π a = 4, cyan π a = 1, 0 pink

15 .. DIFFERENTIATION AND INTEGRATION OF VECTOR-VALUED FUNCTIONS147 y x Motion on a cycloid. r t = t sin t, 1 cos t is the position of a particle in the xy-plane at time t. Find the particle s velocity and acceleration when t = π and t = 3π and plot them as vectors alon with the curve. r π = π, blue r 3π = 3π + 1, 1 red v t = 1 cos t, sin t So v π =, 0 liht blue v 3π = 1, 1 liht red a t = sin t, cos t a π = 0, 1 cyan a 3π = 1, 0 pink

16 148 CHAPTER. VECTOR FUNCTIONS y x 1 9. r t = t + 1, t 1, t is the position of a particle in space at time t. Find the particle s velocity, acceleration and speed when t = 1. v t = 1, t, a t = 0,, 0 v t = 5 + 4t v 1 = 1,, a 1 = 0,, 0 v 1 = r t = cos t, 3 sin t, 4t is the position of a particle in space at time t. Find the particle s velocity, acceleration and speed when t = π. v t = sin t, 3 cos t, 4 a t = cos t, 3 sin t, 0 v t = 4 sin t + 9 cos t + 16 π v =, 0, 4 π a = 0, 3, 0 π v = 0

17 .. DIFFERENTIATION AND INTEGRATION OF VECTOR-VALUED FUNCTIONS r t = ln t + 1, t, t is the position of a particle in space at time t. Find the particle s velocity, acceleration and speed when t = 1. v t = t + 1, t, t a t = t + 1,, 1 v t = 5t + 4 t + 1 v 1 = 1,, 1 a 1 = 1,, 1 v 1 = 6 1. Find the anle between the velocity and acceleration if r t = 3t + 1, 3t, t at t = 0. Let θ be the anle in question. Then θ = cos 1 v 0 a 0 v 0 a 0 = π 13. Find the anle between the velocity and acceleration if r t = ln t + 1, tan 1 t, t + 1 at t = 0. Let θ be the anle in question then θ = cos 1 v 0 a 0 v 0 a 0 = π 14. Find the equation of the line tanent to r t = sin t, t cos t, e t at t = 0. The equation of the tanent is x = t y = 1 z = 1 + t

18 150 CHAPTER. VECTOR FUNCTIONS 15. Find the equation of the line tanent to r t = a sin t, a cos t, bt at t = π. The equation of the tanent is x = at y = a z = πb + bt 16. Consider particles movin alon the unit circle x + y = 1. Answer the questions below for each of the particle a e. i. Does the particle have constant speed? If so, what is it? ii. Is the particles s acceleration always orthoonal to its velocity vector? iii. Does the particle move clockwise or counterclockwise around the circle? iv. Does the particle bein at the point 1, 0? a r t = cos t, sin t, t 0. i. yes, 1 ii. yes iii. counterclockwise iv. yes b r t = cos t, sin t, t 0. i. yes, ii. yes iii. counterclockwise iv. yes c r t = cos t π, sin t π, t 0. i. yes, 1 ii. yes iii. counterclockwise iv. no, 0, 1 d r t = cos t, sin t, t 0. i. yes, 1 ii. yes iii. clockwise iv. yes e r t = cos t, sin t, t 0. i. no ii. no

19 .. DIFFERENTIATION AND INTEGRATION OF VECTOR-VALUED FUNCTIONS151 iii. counterclockwise iv. yes 17. A particle moves alon the top of the parabola y = x from left to riht at constant speed 5 units per second. Find the velocity of the particle as it moves throuh,. hint: You first need to find the parametric equations { of the curve. Recall, if y = f x then the parametric equations x = at are where a is a constant. In this case, first write x = f y y = f at then use a similar trick for the parametric equations. The velocity is 5, Solve for r t in { d r dt = t, t, t. r 0 = 1,, 3 r t = t + 1, t t +, At time t = 0, a particle is located at the point 1,, 3. It travels in a straiht line to the point 4, 1, 4, has speed at 1,, 3 and constant acceleration 3, 1, 1. Find the position vector r t. This amounts to findin r t iven a t = 3, 1, 1, v 0 = and r 0 = 1,, 3. v t = 3t + 6, t, t t r t = + 6 t + 1, t 11 t +, t 11 + t A projectile is fired at a speed of 840m/s at an anle of 60. How lon will it take to et 1km downrane? t = 50s 1. A projectile is fired with an initial speed of 500m/s at an anle of elevation of 45. a When and how far away will the projectile strike? t 7.1 x m b How hih overhead will the projectile be when it is 5km downrane? First, we find t 1 such that x t 1 = 5000, then compute y t 1. t =

20 15 CHAPTER. VECTOR FUNCTIONS Hence, the heiht is y m c What is the reatest heiht reached by the projectile? The maximum heiht is iven by y max 6371m. Consider the helix iven by r t = cos t, sin t, t. Prove that r t r t two different ways. hint: you can compute both vectors and show they are perpendicular or you can use proposition..10.

21 Biblioraphy [1] Joel Hass, Maurice D. Weir, and Geore B. Thomas, University calculus: Early transcendentals, Pearson Addison-Wesley, 01. [] James Stewart, Calculus, Cenae Learnin, 011. [3] Michael Sullivan and Kathleen Miranda, Calculus: Early transcendentals, Macmillan Hiher Education,

2.5 Velocity and Acceleration

82 CHAPTER 2. VECTOR FUNCTIONS 2.5 Velocity and Acceleration In this section, we study the motion of an object alon a space curve. In other words, as the object moves with time, its trajectory follows