2.2 Differentiation and Integration of Vector-Valued Functions

.. DIFFERENTIATION AND INTEGRATION OF VECTOR-VALUED FUNCTIONS131. Differentiation and Integration of Vector-Valued Functions Simply put, we differentiate and integrate vector functions by differentiating and integrating 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 13 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 tangent to the curve at t = a. Definition 14 The line tangent to a curve C with position vector r t at t = a is the line through r a in the direction of r a. Definition 15 Unit Tangent Vector The unit tangent vector, denoted T t is defined to be T t = r t r t.4 Remark 16 Of course, the above definition makes sense only if r t 0. The derivative is defined in terms of limits. Taking the limit of a vector function amounts to taking the limits of the component functions. Thus, we have the following theorem: Theorem 17 If r t = f t, g t, h t then r t = f t, g 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 following theorem: Theorem 18 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

13 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 19 Let r t = t, e t, sin t. Find r t and the unit tangent vector at t = 0. Then, find the equation of the tangent at t = 0. 1. Computation of r t. r t = 1, te t, cos t. Computation of the unit tangent at t = 0. First, we must find the tangent vector at t = 0. This vector is r 0. From our computation above, we see that r 0 = 1, 0, The unit tangent vector at t = 0 is T 0. T 0 = r 0 r 0 = 1, 0, = 5 1 5, 0, 5 3. Computation of the tangent line. The parametric equations of the line through r 0 = 0, 1, 0 with direction vector 1, 0, is x = t y = 1 z = t Definition 130 Smooth Curve A curve C given 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 131 Consider the curve given by r 1 t = 1 + t, t 3. Find if it is smooth on R. What about on 0,? We start by computing r t. r t = t, 3t. Wee 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.

.. DIFFERENTIATION AND INTEGRATION OF VECTOR-VALUED FUNCTIONS133 We finish with the proof of a well known result which we state as a proposition. Proposition 13 Let C be a curve given by a position vector r t. If r t = c a constant then r t r t for all t. 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 get r t r t + r t r t = 0 This says that r t r t. r t r t = 0 r t r t = 0 You may not recognize 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 tangent to the circle... Integration Definition 133 If r t = f t, g t, h t then b a b r t dt = f t dt, a b a g t dt, b a h t dt and r t dt = f t dt, g t dt, h t dt We have similar definitions for plane curves. Example 134 Let r t = cos t, sin t, which satisfies R 0 = 3,, 1. R t = r t dt 1 1 + t. Find R t = r t dt = = cos tdt, sin tdt, 1 1 + 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

134 CHAPTER. VECTOR FUNCTIONS..3 Velocity and Acceleration In this section, we look at direct applications of the derivative and the integral of a vector function. Definition 135 Velocity and Acceleration Consider an object moving along 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 is v t. Example 136 Finding the velocity and acceleration of a moving object An object is moving along 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 figure.. 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 137 Finding a Position Function by Integration A moving 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.

.. DIFFERENTIATION AND INTEGRATION OF VECTOR-VALUED FUNCTIONS135 Figure.: Motion of an object along 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 using 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 using the initial condition. r 0 = 1, 0, 0 = C1, C, C 3

136 CHAPTER. VECTOR FUNCTIONS Thus, C 1 = 1 and C = C 3 = 0. It follows that t 3 r t = 3 + t + 1, t3 t, t + t Remark 138 The above problem can be set up as a differential equation, that is an equation which involves the derivatives of an unknown function. Solving the equation amounts to finding the unknown function. We were given 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 given 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 challenging to solve. Remark 139 In general, we can recover velocity by integration when acceleration is known. Similarly, we can recover position by integration 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, integrating from t 0 to t gives t t 0 a u du = t t 0 v u du. Using the fundamental theorem of calculus, we get 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 140 When studying the motion of an object, the time at which we start tracking 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 = 0...4 Projectile Motion If you wondered how we can know the acceleration and not the other quantities, this section will hopefully answer your questions. We begin with a little bit of physics. Newton s second law of motion states that if at time t a force F t

.. DIFFERENTIATION AND INTEGRATION OF VECTOR-VALUED FUNCTIONS137 acts on an object of mass m, this action will produce an acceleration a t of the object satisfying F t = m a t Thus, if we know the force acting 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 finding the trajectory of an object being thrown in the air and subject to the laws of physics. Example 141 A projectile is fired with an angle of elevation α and initial velocity v 0 as shown on the next slideif we ignore air resistance, the only force acting on the object is gravity. 1. Find the position of the object r t in terms of α and v 0.. Express the range d in terms of α. 3. Find the value of α which maximizes the range d. 4. What is the maximum height reached by the object? Answer to 1. When translating 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 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 begin by setting the axes so that y is the vertical direction and x the horizontal direction. In addition, the motion of the projectile begins at the origin. The only force the object is subject to is gravity. Since the force of gravity acts downward, we have a t = 0, g where g = 9.81m/s. We can now compute v t. Let us introduce some 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, gt + C Then, we have Thus v 0 = v0x, v 0y = C 1, C v t = v0x, gt + v 0y Elementary trigonometry tells us that v 0x = v 0 cos α v 0y = v 0 sin α

138 CHAPTER. VECTOR FUNCTIONS Thus v t = v0 cos α, gt + v 0 sin α Now, we can compute r t. r t = v t dt = v 0 cos α t + C 1, 1 gt + 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 gt + 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 writing the parametric equations of the trajectory of the object. { x = v 0 cos α t y = 1 gt + v 0 sin α t Answer to. d, the range, is the horizontal distance corresponding to y = 0. We can find d by finding the value of t for which y = 0 and plugging this value in the equation for x. y = 0 1 gt + v 0 sin α t = 0 t 1 gt + v 0 sin α = 0 This happens when either t = 0 or t = v 0 sin α. t = 0 corresponds to the g initial position. The value of t we want is t = v 0 sin α. This value gives g us d = v 0 cos α v 0 sin α g d = v 0 sin α g Answer to 3. We see that d is maximum when sin α = 1 that is when α = π 4. Answer to 4. The object reaches its maximum altitude when its vertical velocity is 0 that is when v y = gt + v 0 sin α = 0 or when t =. At that v0 sin α g

.. DIFFERENTIATION AND INTEGRATION OF VECTOR-VALUED FUNCTIONS139 Figure.3: Trajectory of a Projectile time, the altitude is v0 sin α y g = 1 g v0 sin α v0 sin α + v 0 sin α g g = 1 v 0 sin α + v 0 sin α g g = v 0 sin α g..5 Things 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 tangent vector is and be able to find it. Be able to compute the integral 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 given its position vector. Be able to find the velocity and position of a particle given its acceleration and initial velocity and position. Be able to do problems similar to the example on projectile motion.

140 CHAPTER. VECTOR FUNCTIONS..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 4 1 1 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 graph is the path of the particle, then find the particle s velocity and acceleration when t = 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 graph is the path of the particle, then find the particle s velocity and acceleration when t = ln 3. 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 along 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 along with the curve. 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. 10. 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 = 1. 1. Find the angle between the velocity and acceleration if r t = 3t + 1, 3t, t at t = 0. 13. Find the angle between the velocity and acceleration if r t = ln t + 1, tan 1 t, t + 1 at t = 0. 14. Find the equation of the line tangent to r t = sin t, t cos t, e t at t = 0.

.. DIFFERENTIATION AND INTEGRATION OF VECTOR-VALUED FUNCTIONS141 15. Find the equation of the line tangent to r t = a sin t, a cos t, bt at t = π. 16. Consider particles moving along 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 orthogonal to its velocity vector? iii. Does the particle move clockwise or counterclockwise around the circle? iv. Does the particle begin 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 0. 17. A particle moves along the top of the parabola y = x from left to right at constant speed 5 units per second. Find the velocity of the particle as it moves through,. 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 19. At time t = 0, a particle is located at the point 1,, 3. It travels in a straight 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 angle of 60. How long will it take to get 1km downrange? 1. A projectile is fired with an initial speed of 500m/s at an angle of elevation of 45. a When and how far away will the projectile strike? b How high overhead will the projectile be when it is 5km downrange? c What is the greatest height reached by the projectile?

14 CHAPTER. VECTOR FUNCTIONS..7 Answers 1. Evaluate 1 0 t 3, 7, t + 1 dt 1 t 3, 7, t + 1 1 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 4 1 1 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 graph is the path of the particle, then find the particle s velocity and acceleration when t = 1. y = x 1 + 1 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 graph is the path of the particle, then find the particle s velocity and acceleration when t = ln 3. y = 9 x

.. DIFFERENTIATION AND INTEGRATION OF VECTOR-VALUED FUNCTIONS143 v t = e t, 4 9 et Thus and v ln 3 = 3, 4 a t = e t, 8 9 et So 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 along with the curve. π r = 4, blue r π = 1, 0 red v t = cos t, sin t So π v = 4, light blue π v = 0, 1 light red a t = sin t, cos t π a = 4, cyan π a = 1, 0 pink

144 CHAPTER. VECTOR FUNCTIONS y 1.0 0.8 0.6 0.4 0. 1.0 0.8 0.6 0.4 0. 0. 0.4 0.6 0.8 1.0 1. 1.4 0. x 0.4 0.6 0.8 1.0 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 along with the curve. r π = π, blue r 3π = 3π + 1, 1 red v t = 1 cos t, sin t So v π =, 0 light blue v 3π = 1, 1 light red a t = sin t, cos t a π = 0, 1 cyan a 3π = 1, 0 pink

.. DIFFERENTIATION AND INTEGRATION OF VECTOR-VALUED FUNCTIONS145 y 1 0 1 3 4 5 6 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 = 3 10. 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

146 CHAPTER. VECTOR FUNCTIONS 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 = 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 = 7 1. Find the angle between the velocity and acceleration if r t = 3t + 1, 3t, t at t = 0. Let θ be the angle in question. Then θ = cos 1 v 0 a 0 v 0 a 0 = π 13. Find the angle between the velocity and acceleration if r t = ln t + 1, tan 1 t, t + 1 at t = 0. Let θ be the angle in question then θ = cos 1 v 0 a 0 v 0 a 0 = π 14. Find the equation of the line tangent to r t = sin t, t cos t, e t at t = 0. The equation of the tangent is x = t y = 1 z = 1 + t 15. Find the equation of the line tangent to r t = a sin t, a cos t, bt at t = π. The equation of the tangent is x = at y = a z = πb + bt

.. DIFFERENTIATION AND INTEGRATION OF VECTOR-VALUED FUNCTIONS147 16. Consider particles moving along 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 orthogonal to its velocity vector? iii. Does the particle move clockwise or counterclockwise around the circle? iv. Does the particle begin 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 iii. counterclockwise iv. yes 17. A particle moves along the top of the parabola y = x from left to right at constant speed 5 units per second. Find the velocity of the particle as it moves through,. 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, 5.

148 CHAPTER. VECTOR FUNCTIONS 18. Solve for r t in { d r dt = t, t, t. r 0 = 1,, 3 r t = t + 1, t t +, + 3 19. At time t = 0, a particle is located at the point 1,, 3. It travels in a straight 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 finding r t given a t = 3, 1, 1, v 0 = and r 0 = 1,, 3. v t = 3t + 6, t, t + 11 11 11 3t r t = + 6 t + 1, t 11 t +, t 11 + t + 3 11 0. A projectile is fired at a speed of 840m/s at an angle of 60. How long will it take to get 1km downrange? t = 50s 1. A projectile is fired with an initial speed of 500m/s at an angle of elevation of 45. a When and how far away will the projectile strike? t 7.1 x 7.1 5484m b How high overhead will the projectile be when it is 5km downrange? First, we find t 1 such that x t 1 = 5000, then compute y t 1. t = 10 14.14 Hence, the height is y 14.14 4018m c What is the greatest height reached by the projectile? The maximum height is given by y max 6371m