Introduction to Vector Functions Limits and Continuity Philippe B Laval KSU Spring 2012 Philippe B Laval (KSU) Introduction to Vector Functions Spring 2012 1 / 14
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) Introduction to Vector Functions Spring 2012 2 / 14
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) Introduction to Vector Functions Spring 2012 3 / 14
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) Introduction to Vector Functions Spring 2012 4 / 14
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) 2 (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) 5 ( u (t) v (t)) = u (t) v (t) + u (t) v (t) 6 ( u (f (t))) = f (t) u (f (t)) Philippe B Laval (KSU) Introduction to Vector Functions Spring 2012 5 / 14
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) Introduction to Vector Functions Spring 2012 6 / 14
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) Introduction to Vector Functions Spring 2012 7 / 14
Integration Definition If r (t) = f (t), g (t), h (t) then ˆ b ˆ b ˆ b ˆ b r (t) dt = f (t) dt, g (t) dt, h (t) dt a a a a and ˆ r (t) dt = ˆ ˆ f (t) dt, ˆ g (t) dt, h (t) dt We have similar definitions for plane curves Example Let r (t) = 1 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) Introduction to Vector Functions Spring 2012 8 / 14
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 3 The speed of the object is a (t) = v (t) = r (t) (2) v (t) = v (t) Philippe B Laval (KSU) Introduction to Vector Functions Spring 2012 9 / 14
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) Introduction to Vector Functions Spring 2012 10 / 14
Velocity and Acceleration Figure: Motion of an object along r (t) = t, t 3, 3t Philippe B Laval (KSU) Introduction to Vector Functions Spring 2012 11 / 14
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) Introduction to Vector Functions Spring 2012 12 / 14
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) Introduction to Vector Functions Spring 2012 13 / 14
Exercises Review the notions of differentiation and integration from Calculus I and II See the problems at the end of section 22 in my notes on differentiation and integration of vector functions Philippe B Laval (KSU) Introduction to Vector Functions Spring 2012 14 / 14