Introduction to Vector Functions

Transcription

1 Introduction to Vector Functions Differentiation and Integration Philippe B. Laval KSU Today Philippe B. Laval (KSU) Vector Functions Today 1 / 14

2 Introduction In this section, we study the differentiation and integration of vector functions. Simply put, we differentiate and integrate vector functions by differentiating and integrating their component functions. Since the component functions are real-valued functions of one variable, we can use the techniques studied in calculus I and II. We then look at applications. In the case a vector function represents the path followed by a moving object, we see how differentiation and integration relate position, velocity and acceleration. We finish by deriving the equations of an object moving in space, subject to an initial force and gravity. Philippe B. Laval (KSU) Vector Functions Today 2 / 14

3 Differentiation Definition 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 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). Philippe B. Laval (KSU) Vector Functions Today 3 / 14

4 Differentiation Definition (Unit Tangent Vector) The unit tangent vector, denoted T (t) is defined to be T (t) = r (t) r (t) 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 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. Philippe B. Laval (KSU) Vector Functions Today 4 / 14

5 Differentiation 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 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) ( f (t) ) u (t) = f (t) u (t) + f (t) u (t) ( u (t) ) v (t) = u (t) v (t) + u (t) v (t) ( u (t) ) v (t) = u (t) v (t) + u (t) v (t) ( u ) (f (t)) = f (t) u (f (t)) Philippe B. Laval (KSU) Vector Functions Today 5 / 14

6 Differentiation Example Let r (t) = t, e t2, sin 2t. Find r (t) and the unit tangent vector at t = 0. Then, find the equation of the tangent at t = 0. Definition (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. 2 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. Philippe B. Laval (KSU) Vector Functions Today 6 / 14

7 Differentiation Example Consider the curve given by r 1 (t) = 1 + t 2, t 3. Find if it is smooth on R. What about on (0, )? We finish with the proof of a well known result which we state as a theorem. Theorem 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. 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. Philippe B. Laval (KSU) Vector Functions Today 7 / 14

8 Integration Definition If r (t) = f (t), g (t), h (t) then b a r (t) dt = b a b b f (t) dt, g (t) dt, h (t) dt a a and r (t) dt = f (t) dt, g (t) dt, h (t) dt We have similar definitions for plane curves. Example Let 1 r (t) = cos 2t, 2 sin t, 1 + t 2. Find R (t) = r (t) dt which satisfies R (0) = 3, 2, 1. Philippe B. Laval (KSU) Vector Functions Today 8 / 14

9 Velocity and Acceleration In this section, we look at direct applications of the derivative and the integral of a vector function. Definition (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) (1) 2 The acceleration of the object, denoted a (t), is defined to be a (t) = v (t) = r (t) (2) 3 The speed of the object, denoted v (t), is the magnitude of the velocity, that is v (t) = v (t) Philippe B. Laval (KSU) Vector Functions Today 9 / 14

10 Velocity and Acceleration Example (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. 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 (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. Philippe B. Laval (KSU) Vector Functions Today 10 / 14

11 Velocity and Acceleration Figure: Motion of an object along r (t) = t, t 3, 3t Philippe B. Laval (KSU) Vector Functions Today 11 / 14

12 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) 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. Philippe B. Laval (KSU) Vector Functions Today 12 / 14

13 Projectile Motion - Example A projectile is fired with an angle of elevation α and initial velocity v 0. If 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. 2 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? Figure: Trajectory of a Projectile Philippe B. Laval (KSU) Vector Functions Today 13 / 14

14 Exercises Review the notions of differentiation and integration from Calculus I and II. See the problems at the end of my notes on vector functions: differentiation and integration. Philippe B. Laval (KSU) Vector Functions Today 14 / 14