Math Skills for Physics

Transcription

1 Math Skills for Physics I want all students to be as successful as possible in physics. Because physics relies heavily on math throughout the course many concepts learned in chemistry, algebra, and geometry will be necessary in physics it is important that students have the prerequisite math skills to ensure success. To this end, I am providing this optional math packet, which has several purposes: to keep math skills sharp over the summer, to give students an idea of the type of math that will be used during physics, to review concepts that may have gotten rusty over time. This packet is designed to accompany your physics textbook s math handbook (Appendix A, pp ). I suggest going through the math handbook a section at a time, reviewing each concept and completing some of the associated practice problems in this packet. Note that this packet also includes a few extra sections (called extra topics) that aren t included in the math handbook. These sections contain review of concepts of geometry that may also be needed in your study of physics. My hope is that most (if not all) of this material is review. If any of these sections contains new (or forgotten) material, take the time to work carefully through all of the problems, perhaps supplementing with additional practice problems if necessary to ensure mastery. Solutions for all problems are given in the back in case you run into trouble. You might notice that some practice problems make use of SI, scientific notation, and significant digits, even though these concepts are not explicitly reviewed in this math packet. Don t panic if you re feeling very rusty with these concepts; they will be reviewed during in Chapter 2. Because this packet is optional, I will not be collecting them or correcting them. Solutions to all problems start on page 8. Good luck and have fun with these pages! Lisa Swieson Sources: Dingrando, et al., Chemistry: Matter and Change, Glencoe/McGraw-Hill, Leff, Geometry the Easy Way, Barron s, Sobel, et al., Algebra Two with Trigonometry, Harper & Row, Zitzewitz, Physics: Principles and Problems, Glencoe/McGraw-Hill, 2002.

2 Fractions, Decimals, and Percents (pp. 737) 1. Complete the table. Simplify all fractions. Fraction 27/100 5/6 57/311 Decimal Percent 27% 55.7% 87.5% 21.8% Relative Uncertainty and Relative Error (pp ) 2. Three students calculated the density of an unknown compound, which has an accepted density of 1.59 g/cm 3. They measured the mass and volume of three separate samples and reported the densities as follows: Student ± 0.03 g/cm 3 Student ± 0.14 g/cm 3 Student ± 0.10 g/cm 3 What is each student s relative error? What is each student s relative uncertainty? Unit Operations / Dimensional Analysis (p. 740) 3. Use appropriate conversion factors to solve the following equations: a. 1.5 days = s b. 675 nm = m c mm 3 = cm 3 d km/min = km/h e. 790 km/h = m/s f cm/min = m/h g x 10 7 mm 3 = L h g/cm 3 = kg/dm 3 Properties of Exponents (p. 741) 4. Simplify. a. (5 3 ) 4 b. (2x 2 ) 3 c. (-5x 2 ) 3 (xy 4 ) 2 d. [(2a) 3 ] 2 [(½b) 3 ] 2 e. 3-4 f. a -2 (b -1 ) 3 g. (x 2 ) -1 (x -4 ) -2 h. (x 2 + 2x) -1 (x + 2) i. 81 1/2 j. 27-2/3 k. (1/16) 3/4 l. (64-1/2 ) 2/3 m. (3.1 x 10 3 ) x (1.4 x 10 4 ) n. (1.26 x 10-2 ) x (6.34 x 10-3 ) o. (5.61 x 10 6 ) (8.4 x 10 2 ) p. (7.92 x 10-5 ) (1.3 x 10 2 )

3 The Quadratic Formula (p. 742) 5. Solve. Leave answers in radical form. a. x 2 + 7x + 12 = 0 b. -12x 2 + 5x + 2 = 0 c. 3x 2 = 5 d. ½x 2/x = 2 6. Approximate each solution to two decimal places. a. x 2 5x + 2 = 0 b. 3(x 2 2x) = 5 2x Perimeter, Area, & Volume (p. 743) 7. Graph each line and find the area under the curve over the interval given. a. x = y; 0 x 4 b. y = 3.5; 1.5 x 6 c. y = ½x + 2; 1 x 5 Lines & Angles (extra topic) Review. Vertical angles are congruent. Perpendicular lines form right angles, angles with a measure of 90. If two parallel lines are cut by a transversal: alternate interior angles are congruent alternate exterior angles are congruent corresponding angles are congruent angles that are not congruent are supplementary (their measures add up to 180 ). Problem. 8. In Figure 1, what are the measures of angles a-g? Figure a b c d h f g i j e Triangles (extra topic) Review. The sum of the angles of a triangle is 180. Problem. 9. In Figure 1, the dotted line is perpendicular to the parallel lines. What are the measures of angles h-j?

4 Similar and Congruent Triangles (extra topic) Review. Two triangles are similar if any one of these conditions is true: two angles are congruent (AA) the three sides are in proportion two sides are in proportion and the angles between them are congruent Remember that the converse is also true: if two triangles are similar, their sides will be in proportion and their angles will be congruent. A line parallel to one side of a triangle and intersecting the other two sides divides these sides proportionally and forms a smaller, similar triangle. (The converse is also true.) Two triangles are congruent if any one of these conditions is true: the three sides are all congruent (SSS) two angles and one side are congruent (ASA or AAS) two sides and the included angle are congruent (SAS) Remember that the converse is also true: if two triangles are congruent, their sides will be congruent and their angles will be congruent. An altitude drawn to the base of an isosceles triangle separates it into 2 congruent right triangles. Two right triangles are congruent if any one of these conditions is true: the two legs are congruent (LL) [which is like SAS] one leg and one acute angle are congruent (LA) [which is like AAS or ASA] the hypotenuse and one acute angle are congruent (HA) [which is like AAS] the hypotenuse and one leg are congruent (HL) An altitude drawn to the hypotenuse of a right triangle separates it into two triangles that are similar to each other and to the original triangle. 10. In Figure 1, draw a line perpendicular to the transversal through the point and intersecting both parallel lines. What is true about the two triangles you formed? Defend your conclusion. 11. In the figure, line segment EF is parallel to line segment RT. a. SE = 8, ER = 6, FT = 15, SF =. b. SF = 4, ST = 12, SR = 27, SE =. c. SE = 6, ER = 4, ST = 20, FT =. d. SE = 9, ST = 42, ER = 12, FT =. 12. For each case listed below, determine whether AB is parallel to KJ. a. KA = 2, AL = 5, JB = 6, BL = 15. b. KL = 8, AL = 3, JB = 10, JL = 16. c. KA = 9, AL = 5, JB = 15, BL = 10.

5 Pythagorean Theorem (p. 744) 13. Find the value of x in each shape. Round to the nearest tenth. a. b. c. d. 14. Find the values of r, s, and t in each shape. Leave answer in radical form. a. b. c. 15 Special Triangles (p. 744) 15. What is the measure of each angle in the triangles in 14a? 16. Find the values of x, y, and z (if labeled) in each shape. Leave answer in radical form. a. b. c.

6 Trigonometry (p. 745) 17. In each shape, find the values of x and y (if labeled) to the nearest tenth. a. b. c. d. 18. In each right triangle, find the measure of the marked angle to the nearest tenth. a. b. c. Law of Cosines and Law of Sines (pp ) 19. Refer to the figure to find the indicated measurement to the nearest tenth. a. AC = 8, BC = 6 2, C = 45, AB =. b. AC = 7, BC = 5 2, AB = 29, C =. c. A = 60, B = 45, BC = 9, AC =. 20. Refer to the figure to find the indicated measurement(s) to the nearest tenth. a. AC = 5.6, AB = 3.6, A = 20, BC =, C =. b. AB = 1.6, BC = 1.2, B = 105, AC =, C =. c. A = 30, C = 45, AB = 10, BC =. d. B = 120, C = 30, AB = 10, AC =.

7 Mixed Review In physics, you won t usually be given a labeled diagram and asked to solve for x. Nor will you be told which tool (Pythagorean Theorem, Law of Sines, etc.) you ll need to solve a problem. Instead, you will need to translate a word problem into a diagram yourself and then figure out how to solve it. Below are some problems similar to those you will learn to solve in physics. First draw the diagram and check it, then solve the problem. 21. The diagonals of a rhombus measure 12.0 and 16.0 cm. What are the measures of the angles? 22. Jane hangs her purse on a hook and notices that the strap is bent at a 72 angle. If the purse itself is 10 cm below the doorknob, how long is the purse strap? How wide is the purse? 23. Paul walks 11 blocks south, then turns and walks 16 blocks east. If he decides to return home following a straight-line path, how far must he walk? What angle does his return path make with his original route? 24. A boat travels east across a river at 2.5 m/s. The river is flowing south at 1.8 m/s. What direction will the boat end up traveling? What is the boat s net speed? 25. A 2.1-m long ladder leans against a house, making an angle of 58 with the ground. How high off the ground is the top of the ladder? 26. Two cars each travel 72 miles at constant rates. One car travels 6 miles per hour faster than the other and arrives 10 minutes before the other arrives. Find the rates of speed of the two cars. 27. The wire used to hang a crooked picture forms an obtuse (129 ) angle. If one side of the wire measures 32 cm and the other side measures 34 cm, how far apart are the ends of the wire? If someone were to straighten the picture, what angle would the wire make? 28. Two cars start at the intersection of two straight highways. One travels at an average rate of 44 mph, and the other at 52 mph. If the angle between their paths measures 38, how far apart are the cars 45 minutes later? 29. Two guy wires that support a telephone pole stretch from the top of the pole to points on the ground on opposite sides of the pole. The wires form angles of 62 and 69 with the ground, and the distance between the points on the ground is 46 feet. Find the length of each guy wire. 30. A plane takes off from the ground and travels in a straight line upward for m. At that instant, the plane s altimeter reads m. At what angle has the plane risen with respect to the ground?

8 Solutions Fractions, Decimals, and Percents 1. Fraction 27/100 13/40 557/1000 5/6 23/80 7/8 57/ /500 Decimal Percent 27% 32.5% 55.7% 83.3% 28.75% 87.5% 18.3% 21.8% Relative Uncertainty and Relative Error 2a. Relative error Student x 100% = 1.258% 1.26% Student x 100% = 5.031% 5.03% Student x 100% = 6.918% 6.92% 2b. Relative uncertainty Student cm 1.57 cm x 100% = 1.911% 2% Student cm 1.51 cm x 100% = 9.272% 9.3% Student cm 1.70 cm x 100% = 5.882% 5.9% Unit Operations / Dimensional Analysis 3. a. 1.5 days x (24 h/1 day) x (60 min/1 h) x (60 s/1 min) = 129,600 s 130,000 s b. 675 nm x (1 m/10 9 nm) = 6.75 x 10-7 m c mm 3 x (1 cm/10 mm) 3 = 3.72 cm 3 d km/min x (60 min/1 h) = km/h e. 790 km/h x (1000 m/1 km) x (1 h/3600 s) = m/s 220 m/s f cm/min x (1 m/100 cm) x (60 min/1 h)= m/h 3.59 m/h g x 10 7 mm 3 x (1 cm/10 mm) 3 x (1 L/1000 cm 3 ) = 35.4 L h g/cm 3 x (1 kg/1000 g) x (10 cm/1 dm) 3 = 1.03 kg/dm 3 Properties of Exponents 4. a. (5 3 ) 4 = 5 (3 4) = 5 12 = 244,140,625 b. (2x 2 ) 3 = (2 3 )(x (2 3) ) = 8x 6 c. (-5x 2 ) 3 (xy 4 ) 2 = (-5 3 )(x (2 3) )(x 2 )(y (4 2) ) = -125x 6 x 2 y 8 = -125x 8 y 8 d. [(2a) 3 ] 2 [(½b) 3 ] 2 = [(2 3 )(a 3 )] 2 [(½) 3 (b) 3 ] 2 = (8a 3 ) 2 [(1/8)b 3 ] 2 = (64a 6 )(1/64)b 6 = a 6 b 6

9 e. 3-4 = 1/3 4 = 1/81 f. a -2 (b -1 ) 3 = (1/a 2 )(b -3 ) = (1/a 2 )(1/b 3 ) = 1/a 2 b 3 g. (x 2 ) -1 (x -4 ) -2 = (x (2-1) )(x (-4-2) ) = (x -2 )(x 8 ) = x 6 h. (x 2 + 2x) -1 (x + 2) = (x + 2) / x(x + 2) = 1/x i. 81 1/2 = 9 j. 27-2/3 = 1/(27) 2/3 = 1/(3 2 ) = 1/9 k. (1/16) 3/4 = (1/2) 3 = 1/8 l. (64-1/2 ) 2/3 = 64-2/6 = (1/64) 1/3 = 1/4 m. (3.1 x 10 3 ) x (1.4 x 10 4 ) = (3.1 x 1.4) x (10 3 x 10 4 ) = 4.34 x 10 (3+4) = 4.3 x 10 7 n. (1.26 x 10-2 ) x (6.34 x 10-3 ) = (1.26 x 6.34) x (10-2 x 10-3 ) = x 10 (-2+-3) = 7.99 x 10-5 o. (5.61 x 10 6 ) (8.4 x 10 2 ) = ( ) x ( ) = x 10 (6-2) = 0.67 x 10 4 p. (7.92 x 10-5 ) (1.3 x 10 2 ) = ( ) x ( ) = x 10 (-5-2) = 6.1 x 10-7 The Quadratic Formula 5. a. x 2 + 7x + 12 = 0 b. -12x 2 + 5x + 2 = 0 (mult. both sides by -1) (x + 3)(x + 4) = 0 12x 2 5x 2 = 0 x = -3 or -4 (3x 2)(4x + 1) = 0 3x = 2 or 4x = -1 x = 2/3 or -1/4 c. 3x 2 = 5 d. ½x 2/x = 2 (mult. both sides by 2x) 3x 2 + 0x 5 = 0 x 2 4 = 4x x 2 4x 4 = 0 x = -0 ± 0 2 4(3)(-5) x = -(-4) ± (-4) 2 4(1)(-4) 2(3) 2(1) x = ± ( 60)/6 x = ± 15/3 x = (4 ± 32)/2 x = 2 ± a. x 2 5x + 2 = 0 b. 3(x 2-2x) = 5 2x 3x 2 4x 5 = 0 x = -(-5) ± (-5) 2 4(1)(2) x = -(-4) ± (-4) 2 4(3)(-5) 2(1) 2(3) x = (5 ± 17)/2 x = (4 ± 76)/6 x = 4.56 or 0.44 x = 2.12 or Perimeter, Area, & Volume 7. a. x = y; 0 x 4 area of Δ = ½bh b = 4; h = 4 A = ½(4)(4) = 8

10 b. y = 3.5; 1.5 x 6 area of = bh b = 4.5; h = 3.5 A = (4.5)(3.5) = c. y = ½x + 2; 1 x 5 area of = ½(b 1 + b 2 )h b 1 = 2.5; b 2 = 4.5; h = 4 A = ½( )(4) = 14 Lines & Angles 8. Angles a, b, e, and f measure 40. Angles c, d, and g measure 140. Triangles 9. h = 90, i = 50, j = 130 Similar and Congruent Triangles 10. The triangles are similar right triangles. The perpendicular line forms right angles, which are always congruent. Angles b and e are also congruent (alternate interior angles ). According to AA, the triangles are similar. Figure 1. d h f g i j e 140 a b c 11. a SE = ---- SF b SE = ---- SF ER FT SR ST (SE)(FT) = (ER)(SF) (SE)(ST) = (SR)(SF) (8)(15) = (6)(SF) (SE)(12) = (27)(4) 120 = 6(SF) 12(SE) = 108 SF = 20 SE = 9 c SR = ---- ST d ER = ---- FT ER FT SR ST (SR)(FT) = (ER)(ST) (ER)(ST) = (SR)(FT) (6 + 4)(FT) = (4)(20) (12)(42) = (9 + 12)(FT) 10(FT) = = 21(FT) FT = 8 FT = 24 KA? JB KA? JB KA? JB 12. a = ---- b = ---- c = ---- AL BL KL JL AL BL (KA)(BL) = (AL)(JB) (KA)(JL) = (KL)(JB) (KA)(BL) = (AL)(JB) (2)(15) = (5)(6) (8-3)(16) = (8)(10) (9)(10) = (5)(15) 30 = 30 YES 80 = 80 YES NO

11 Pythagorean Theorem 13. a. c 2 = a 2 + b 2 b. first find s c = x 2 d = x = x 2 s 2 = x 2 = x 2 = x 2 = 160,000 90,000 s = 3 x = 64 x = 576 x = 70,000 5 = 3 x = 8.0 x = 24.0 x = x 4 3x = 20 a. n 13b. x = a. 4 = t 4 = r 16 2 = r 2 + s 2 t 12 r 16 s 2 = t 2 = 48 r 2 = 64 s = 192 t = 4 3 r = 8 s = 8 3 s b. s 2 = = 10 r 2 = s 2 + t 2 s 2 = t r 2 = s = 5 5 5t = 50 5 r = 625 t = 10 5 r = 25 c = t 2 15 = r 25 = s + r t 2 = s = 25 9 t = r = 225 s = 16 t = 20 r = 9 Special Triangles 15. Because the hypotenuse is twice as long as the short leg, and the long leg = short leg x 3, all triangles in 14a are triangles. 16. a. x = (4/2) 3 y = 2 x 4 y 2 = z 2 x = 2 3 y = 8 z 2 = z = 48 z = 4 3 b. x = 12/ 2 x = 6 2 c. y = 8/2 8 2 = y 2 + x 2 z 2 = (14 y) 2 + x 2 y = 4 x 2 = z 2 = x = 48 z = 148 x = 4 3 z = 2 37

12 Trigonometry 17. a. cos40 = x/20 b. tan75 = 28/x c. sin40 = x/20 d. sin38 = y/ = x/ = 28/x = x/ = y/10 x = 20(0.7660) x = 28/3.732 x = 20(0.6428) y = 10(0.6157) x = 15.3 x = 7.5 x = 12.9 y = = x 2 + (y + s) 2 cos38 = s/10 c. d. (y + s) 2 = = s/10 (y + s) = s = 10(0.7880) s (y + s) = 15.3 s = 7.88 tan54 = x/s x = s = 12.9/ x = 21.9 s = 8.72 y = s y = a. tanx = 5/12 = b. cosx = 300/400 = c. sinx = 4/5 = x = tan x = cos x = sin x = 22.6 x = 41.4 x = 53.1 Law of Cosines and Law of Sines 19. a. AB 2 = (6 2) (6 2)(8)cos45 b. ( 29) 2 = (5 2) (5 2)(7)cosC AB 2 = (96 2)(0.7071) 29 = (70 2)cosC AB 2 = 40 cosc = -70/70 2 AB = 6.3 cosc = C = 45.0 sina sinb c = BC AC (sina)(ac) = (BC)(sinB) (0.8660)(AC) = (9)(0.7071) AC = a. BC 2 = (5.6) 2 + (3.6) 2-2(5.6)(3.6)cos20 b. AC 2 = (1.6) 2 + (1.2) 2 2(1.6)(1.2)cos105 BC 2 = (40.32)(0.9397) AC 2 = (3.84)( ) BC 2 = AC2 = BC = 2.5 AC = sina = sinc sinb = sinc BC AB AC AB (sina)(ab) = (BC)(sinC) (sinb)(ab) = (AC)(sinC) (0.3420)(3.6) = (2.5)(sinC) (0.9659)(1.6) = (2.2)(sinC) sinc = sinc = C = 29.5 C = 44.6 c sina = sinc sinb sinc d = BC AB AC AB (sina)(ab) = (BC)(sinC) (sinb)(ab) = (AC)(sinC) (0.500)(10) = (BC)(0.7071) (0.8660)(10) = (AC)(0.500) BC = 7.1 AC = 17.3

13 Mixed Review 21. tanx = 6/8 = x = tan x = 36.9 A = C = 2(36.9 ) = 73.7 B = D = = cos36 = 10/x x = 10/ x = length of strap = 2(12.36) = 24.7 cm tan36 = (½w)/10 10(0.7265) = ½w w = 14.5 cm 23. x 2 = a 2 + b 2 11 x x 2 = x = 377 x = blocks y tany = 11/16 = y = tan = tany = 1.8/2.5 = x y = tan = 35.8 south of east x 2 = y x = x = m/s 25. sin58 = x/2.1 x = 2.1(0.8480) x = m 26. distance = rate x time rt = 72 so t = 72/r 72 r t (r + 6)(t 1/6) = 72 so (r + 6)(72/r 1/6) = r/6 +432/r 1 = 72 [72s cancel, mult. by -6r] 72 r + 6 t 1/6* r 2 + 6r 2592 = 0 (r - 48)(r + 54) = 0 note that, since rate is in mph, r = 48 or -54. time must be in hours Rate can t be negative, so r = 48 mph.

14 27. x 2 = (32) 2 + (34) 2 2(32)(34)cos x x 2 = (2176)( ) x 2 = x = x 10 cm z If the picture were straight, wires would be same length Draw the altitude to form congruent right triangles sin(z/2) = 30/33 = z/2 = sin = z = minutes = ¾ hour so d 1 = ¾(44) = 33 miles and d 2 = ¾(52) = 39 miles x 2 = (33) 2 + (39) 2 2(33)(39)cos38 x 2 = (2574)(0.7880) x 2 = x = miles 29. At the top, the wires form an angle of = 49 sin = sin x x(sin49 ) = 46(sin69 ) x(0.7547) = 46(0.9336) x = feet sin49 sin = y y(sin49 ) = 46(sin62 ) y(0.7547) = 46(0.8829) y = feet 30. sinx = opp/hyp sinx = 290.0/ = x = sin = x = 16.9