12.5. Differentiation of vectors. Introduction. Prerequisites. Learning Outcomes

Size: px

Start display at page:

Download "12.5. Differentiation of vectors. Introduction. Prerequisites. Learning Outcomes"

Nicholas Lee
6 years ago
Views:

1 Differentiation of vectors 12.5 Introuction The area known as vector calculus is use to moel mathematically a vast range of engineering phenomena incluing electrostatics, electromagnetic fiels, air flow aroun aircraft an heat flow in nuclear reactors. In this section we introuce briefly the ifferential calculus of vectors. Prerequisites Before starting this Section you shoul... Learning Outcomes After completing this Section you shoul be able to... 1 have a knowlege of vectors, in cartesian form 2 be able to calculate the scalar an vector proucts of two vectors 3 be able to ifferentiate an integrate scalar functions. ifferentiate an integrate vectors

2 1. Differentiation of Vectors Consier the following figure. y P r C If r represents the position vector of an object which is moving along a curve C, then the position vector will be epenent upon the time, t. We write r = r(t) toshow the epenence upon time. Suppose that the object is at the point P with position vector r at time t an at the point Q with position vector r(t + ) atthe later time t + as shown in the next figure. y x r(t) P r(t + ) PQ Q Then PQ represents the isplacement vector of the object uring the interval of time. The length of the isplacement vector represents the istance travelle while its irection gives the irection of motion. The average velocity uring the time from t to t + is efine as the isplacement vector ivie by the time interval, that is, average velocity = PQ = r(t + ) r(t) If we now take the limit as the interval of time tens to zero then the expression on the right han sie is the erivative of r with respect to t. Not surprisingly we refer to this erivative as the instantaneous velocity, v. By its very construction we see that the velocity vector is always tangential to the curve as the object moves along it. We have: v = lim 0 r(t + ) r(t) = r Now, since the x an y coorinates of the object epen upon the time, we can write the position vector r in cartesian coorinates as: r(t) = x(t)i + y(t)j x HELM (VERSION 1: March 18, 2004): Workbook Level 1 2

3 Therefore, so that, r(t + ) =x(t + )i + y(t + )j x(t + )i + y(t + )j x(t)i y(t)j v(t) = lim 0 { } x(t + ) x(t) y(t + ) y(t) = lim i + j 0 = x i + y j This is often abbreviate to v =ṙ =ẋi +ẏj using notation for erivatives with respect to time. So the velocity vector is the erivative of the position vector with respect to time. This result generalizes in an obvious way to three imensions. If, then the velocity vector is r(t) =x(t)i + y(t)j + z(t)k v =ṙ(t) =ẋ(t)i +ẏ(t)j +ż(t)k The magnitue of the velocity vector gives the spee of the object. We can efine the acceleration vector in a similar way, as the rate of change (i.e. the erivative) of the velocity with respect to the time: a = v = 2 r = r =ẍi +ÿj + zk 2 Example If w =3t 2 i + cos 2tj, fin (a) (b) (c) 2 w 2 Solution 1. (a) If w =3t 2 i+cos 2tj, then ifferentiation with respect to t yiels: (b) = (6t) 2 +( 2sin 2t) 2 = 36t 2 +4sin 2 2t (c) 2 w 2 =6i 4cos 2tj =6ti 2 sin 2tj It is possible to ifferentiate more complicate expressions involving vectors provie certain rules are ahere to. If w an z are vectors an c is a scalar, all these being functions of time t, then: 3 HELM (VERSION 1: March 18, 2004): Workbook Level 1

4 (w + z) = + + z Key Point (cw) =c + c w (w z) =w + z Example If w =3ti t 2 j an z =2t 2 i +3j, verify the results (a) + z (b) (w z) =w + z Solution Also so (a) w z =(3ti t 2 j) (2t 2 i +3j) =6t 3 3t 2. Then (w z) =18t2 6t =3i 2tj =4ti w + z = (3ti t 2 j) (4ti)+(2t 2 i +3j) (3i 2tj) = 12t 2 +6t 2 6t =18t 2 6t We have verifie + z (b) w z = 3t t 2 0 2t =(9t +2t4 )k implying (w z) =(9+8t3 )k Also, w = 3t t 2 0 4t 0 0 = 4t 3 k HELM (VERSION 1: March 18, 2004): Workbook Level 1 4

5 Solution (cont.) z = 3 2t 0 2t = (9+4t 3 )k an so, as require. w + z =4t3 k +(9+4t 3 )k =(9+8t 3 )k = (w z) Exercises 1. If r =3ti +2t 2 j + t 3 k, fin (a) r (b) 2 r 2 2. Given B = te t i + cos tjfin (a) B (b) 2 B 2 3. If r =4t 2 i +2tj 7k evaluate r an r when t =1. 4. If w = t 3 i 7tk, an z =(2+t)i + t 2 j 2k (a) fin w z, (b) fin, 5. Given r = sin ti+ cos tjfin (a) ṙ, (b) r, (c) r (c) fin, () show that + z Show also that the position vector an velocity vector are perpenicular. Answers 1. (a) 3i +4tj +3t 2 k (b) 4j +6tk 2. (a) ( te t + e t )i sin tj (b) e t (t 2)i cos tj 3. 4i +2j 7k, 8i +2j 4. (a) t(t 3 +2t ) (b) 3t 2 i 7k (c) i +2tj 5. (a) cos ti sin tj (b) sin ti cos tj (c) 1. 5 HELM (VERSION 1: March 18, 2004): Workbook Level 1

Physics 170 Week 7, Lecture 2

Physics 170 Week 7, Lecture 2 http://www.phas.ubc.ca/ goronws/170 Physics 170 203 Week 7, Lecture 2 1 Textbook Chapter 12:Section 12.2-3 Physics 170 203 Week 7, Lecture 2 2 Learning Goals: Learn about