Math 20C. Lecture Examples.

Transcription

1 Math 20C. Lecture Eamples. (8//08) Section 2.. Vectors in the plane Definition A ector represents a nonnegatie number and, if the number is not zero, a direction. The number associated ith the ector is called its length or magnitude and is denoted. The ector of zero length is the zero ector and is denoted O. It has no direction. Vectors are generall denoted b bold-faced letters, like the in Definition, or b letters ith arros oer them, as in the smbol PQ. Vectors are represented in draings b arros, here in each case the length of the arro is the magnitude of the ector, measured ith a scale that might or might not be the scale used on the coordinate aes. The direction of the arro is the direction associated ith the ector (Figure ). The same ector can be represented b different arros in different locations (Figure 2), proided that the different arros are parallel, hae the same lengths, and point in the same direction. FIGURE FIGURE 2 If a ector is represented b an arro in an -plane as in Figure 3, then the - and -components of the ector are the changes in the - and -coordinates from the base to the tip of the arro, measured ith the scale used for measuring the length of the arro. If the -component of is a and the -component is b, as shon in Figure 6 ith positie a and b, e rite = a,b. The length of a ector can be calculated from its - and -components b using the Pthagorean Theorem: = a,b = a 2 + b 2. = a,b a b FIGURE 3 Lecture notes to accompan Section 2. of Calculus, Earl Transcendentals b Rogaski.

2 Math 20C. Lecture Eamples. (8//08) Section 2., p. 2 A nonzero ector in an -plane can also be described b giing its length and its angle of inclination, hich is an angle θ from the positie -direction to the ector (Figure 4). Then the components of the ector are gien b = cos θ,sin θ = cos θ, sin θ. This formula is a consequence of the definition of the sine and cosine functions. The zero ector has zero components: O = 0, 0. = cos θ, sin θ θ FIGURE 4 Eample Find the - and -components of the ector u of length 0 ith angle of inclination 5 6 π. Anser: Figure Aa u = 5 3, 5 Figure Ab u 0 θ = 5 6 π u = 5 3, θ = 5 6 π 5 3 Figure Aa Figure Ab Eample 2 Find an angle of inclination of the ector = 3,4. Anser: Figure A2. θ = tan ( 4 3 ) 4 = 3,4 θ Figure A2 3

3 Section 2., p. 3 Math 20C. Lecture Eamples. (8//08) Adding ectors and multipling ectors b numbers Definition 2 For an ectors = a,b and = c,d and an number λ, + = a,b + c,d = a + c,b + d () λ = λ a, b = λa, λb. (2) Equation () has to geometric interpretations that are illustrated in Figures 5 and 6 for ectors ith positie components: If e place the base of at the tip of, then the ector + is gien b the arro from the base of to the tip of, as shon in Figure 8. If e place the bases of and together, as in Figure 9 and complete the parallelogram ith those ectors as sides, then + is represented b the arro ith base at the bases of and that goes along the diagonal of the parallelogram to the opposite erte. + d + b a c FIGURE 5 FIGURE 6 Equation (2) is illustrated in Figure 7. Multipling the ector b a positie number λ ields a ector ith the same direction as hose length is λ multiplied b the length of. Multipling b a negatie number µ ields a ector ith the opposite direction as hose length is µ multiplied b the length of. λ (λ > 0) µ (µ < 0) FIGURE 7

4 Math 20C. Lecture Eamples. (8//08) Section 2., p. 4 You can think of the difference of to ectors as the ector hich hen added to gies (Figure 8). It the bases of and are at the same point, then is the ector from the tip of to the top of. FIGURE 8 Eample 3 Calculate (a) + and (b) for = 4, and =, 3. Then dra the four ectors. Anser: (a) + = 5, 4 Figure A3a (b) = 3, 2 Figure A3b Figure A3a Figure A3b Eample 4 Write 3 4, 2 0, 5 in the form a, b. Anser: 3 4, 2 0, 5 = 8, 7

5 Section 2., p. 5 Math 20C. Lecture Eamples. (8//08) Displacement and position ectors The displacement ector from one point P( 0, 0 ) to a second point Q(, ) is denoted PQ and can be represented b an arro ith base at P and tip at Q. Its components are obtained b subtracting the coordinates of P from the coordinates of Q: (See Figure 9.) PQ = 0, 0. Points can be located in an -plane b using their position ectors. The position ector of the point P(,) is the displacement ector OP OP =, from the origin to the point (Figure 0). The components of the position ector are the coordinates of the tip of the ector. PQ Q(, ) 0 OP P(,) P( 0, 0 ) 0 PQ = 0, 0 OP =, FIGURE 9 FIGURE 0 Eample 5 Three ertices of the parallelogram PRSQ in Figure are P = (3, 4),Q = (7,8), and R = (2, 2). What are the coordinates of S? Q(7, 8) S P(3, 4) FIGURE R(2, 2) Anser: S = (6, 6)

6 Math 20C. Lecture Eamples. (8//08) Section 2., p. 6 Unit ectors Vectors of length are called unit ectors. For an nonzero ector, the unit ector ith the same direction as is e =. Eample 6 Gie the unit ector e and the ector of length 5 ith the same direction as = 3,2. 3, 2 5, 0 Anser: e = = 3 3 The unit ectors,0 and 0, in the directions of the positie - and -aes in Figure 2 are denoted i and j, respectiel. These ectors can be used in place of angular brackets to epress a ector in terms of its -and -components b riting (Figure 3) = a,b = a,0 + b 0, = ai + bj. bj = ai + bj j i ai FIGURE 2 FIGURE 3 Eample 7 Epress 3(4i j) 2(0i 5j) in the form ai + bj. Anser: 3(4i j) 2(0i 5j) = 8i + 7j. (Eample 4 is the same calculation ith bracket notation for the ectors.) Sums of force ectors It is an empirical fact that to forces F and G applied at the same point P on an object hae the same effect as their sum F + G (Figure 4). Because of this, the sum is called the resultant of F and G. F + G G F P FIGURE 4

7 Section 2., p. 7 Math 20C. Lecture Eamples. (8//08) Eample 8 One man is lifting a boulder ith a rod hile another is pulling it ith a rope as in Figure 3. (a) Find the - and -components of the to force ectors, ith the usual orientation of aes. (b) Find the resultant of the to forces and the approimate decimal alues of its magnitude and angle of inclination FIGURE 5 Interactie Eamples Anser: (a) [Force eerted b the man ith the rod] = F = 300 [Force eerted b the man ith the rope] = G = 50 cos( 7 π) (, sin 7 π) pounds. 8 8 cos( π), sin( π) pounds. 9 9 ( 7. = 244, 333 pounds (b) [Resultant] = 300 cos( 7 π) + 50 cos( π), 300 sin π) + 50 sin( π) [Magnitude of the combined force =. 43 pounds [Angle of inclination] =. ( tan 333.= 244) 0.94 radians. Work the folloing Interactie Eamples on Shenk s eb page, http//.math.ucsd.edu/ ashenk/: Section 2.: Eamples, 6, 7 The chapter and section numbers on Shenk s eb site refer to his calculus manuscript and not to the chapters and sections of the tetbook for the course.