CHAPTER SEVEN. Volume = 1 5 πr3 = 1 5 π 23 = 8 5 π cm3. Volume = 1 5 πr3 = 1 5 π 43 = 64 5 π cm3. Depth = 4.9t 2 = = 78.4 meters.

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1 CHAPTER SEVEN 7. SOLUTIONS 25 Solutions for Section 7. EXERCISES. Since A = x 2, the exponent is 2 and the constant of proportionality is. 2. Since P = 4x, the exponent is and the constant of proportionality is Since x = 3 V = V /3, the exponent is /3 and the constant of proportionality is. 4. The power of r is and the constant of proportionality is 2π. 5. The power of r is 2 and the constant of proportionality is 4π. 6. (a) The coefficient is 3 and the exponent is 2. (b) We have Area = 3w 2 = = 75 cm (a) The coefficient is π and the exponent is 3. 5 (b) We have Volume = 5 πr3 = 5 π 23 = 8 5 π cm3. (c) Since we know that the radius is 5 times the height, the radius is = 4 cm, so Volume = 5 πr3 = 5 π 43 = 64 5 π cm3. 8. (a) The coefficient is 4.9 and the exponent is 2. (b) After 2 seconds, the ball is at Depth = 4.9t 2 = = 9.6 meters. (c) After 4 seconds, the ball is at 9. p > 0. 0 < p <. p = 2. p < 0 3. p > Thus, the hole is 78.4 meters deep. 4. (a) We see that y is proportional to the second power of x. (b) We have Depth = 4.9t 2 = = 78.4 meters. x y = 2x ,000 2,000,000 (c) We see in the table that y increases as x gets larger.

2 26 Chapter Seven /SOLUTIONS 5. (a) We see that y is proportional to the one-half power of x. (b) We have x y = 3 x (c) We see in the table that y increases as x gets larger. 6. (a) We see that y is inversely proportional to x. (b) We have x y = /x (c) We see in the table that y decreases as x gets larger. 7. (a) We see that y is inversely proportional to the second power of x. (b) We have x y = 5/x (c) We see in the table that y decreases as x gets larger. PROBLEMS 8. (a) The graph is symmetric about the y-axis, so p is even. (b) Since the values of y get small as x gets large, p is negative. 9. (a) The graph crosses the x-axis, so p must be odd. (b) The function is defined at x = 0, so p must be positive. 20. (a) p is odd, because y changes sign when x does. (b) p is negative, because the values of f get small when x gets large. 2. As x increases, x 4 increases. So if a > b, then a 4 > b As x increases, x /4 increases. So if a > b, then a /4 > b / As x increases, x 4 decreases. So if a > b, then a 4 < b As x increases, x /4 decreases. So if a > b, then a /4 < b / (a) The graph of y = x 5 is above the graph of y = x 3 when x is greater than, and when x is between and 0. It is below when x is between 0 and, and when x is less than. (b) We have x 5 > x 3 when x > and < x < 0, and x 5 < x 3 when 0 < x < and when x <. 26. (a) The graph of y = x /2 is above the graph of y = x /4 when x is greater than, and below when x is between 0 and. (b) We have x /2 > x /4 when x > and x /2 < x /4 when 0 < x <. 27. If one graph is above another then the y-values on the first are greater than the corresponding y-values on the second. So x 4 > x 2 for x > and x 4 < x 2 for 0 < x <. 28. In Example we saw that the braking distance of an Alfa Romeo going at v mph is We are told that h(v) =.3f(v), so f(v) = 0.036v 2. h(v) = v 2 = 0.047v 2.

3 7. SOLUTIONS (a) If the job pays $4 an hour, the student has to work 2400/4 = 600 hours. If the job pays $0 an hour, the student has to work 2400/0 = 240 hours. (b) As the hourly wage goes up, the number of hours the student has to work goes down. This makes sense algebraically because w is in the denominator, so as the denominator increases, the value of h decreases. In practical terms, if the wage increases, then the student has to work fewer hours to earn $2400. (c) Since h = 2400 w, we have w = 2400 h, So w is inversely proportional to h. 30. When t = 3, we have D = 6(3 2 ) = 44 feet. When t = 5, we have D = 6(5 2 ) = 400 feet. So the distance is larger after five seconds. This makes sense algebraically because 5 > 3, so 5 2 > (a) We have E = 0.5 p 3/2 = 0.5p 3/2. So E is a power function of p with exponent 3/2 and constant of proportionality 0.5. (b) When p = 0.6 dollars per kilowatt-hour, we have When p = 0.25 dollars per kilowatt-hour, we have E = 0.5 = gigawatt-hours per year. (0.6) 3/2 E = 0.5 =.2 gigawatt-hours per year. (0.25) 3/2 The electricity consumption decreases as the price increases. Since 0.6 < 0.25, so (0.6) 3/2 < (0.25) 3/2. But the p 3/2 is in the denominator, so a larger value of p means a larger denominator which results in a smaller output. 32. The surface area for a mammal of body mass 70 kilograms is larger because 70 > 60, so 70 2/3 > 60 2/ (a) The radius is graphed in Figure 7.. (b) In Figure 7., we estimate that a sphere of radius 2 cm has a volume of approximately 33.5 cm 3. r (cm) 2 r = V 33.5 V (cm 3 ) Figure The graph of x 2 x 3 in Figure 7.2 appears to be the same as the graph of x 5, not of x 6. Thus, it appears that x 2 x 3 = x 5 = x 2+3.

4 28 Chapter Seven /SOLUTIONS x 2 x 3 x x 5 x x 6 x Figure The graphs of x 4 and ( x) 4 in Figure 7.3 are not the same so x 4 ( x) 4. ( x) 4 x x 4 Figure 7.3 Solutions for Section 7.2 EXERCISES. This is a power function. 2. This is not a power function. In a power function, the exponent is a constant, not a variable. 3. This is not a power function. It is the sum of two power functions. 4. This is a power function. We can rewrite it as y = 2x This is a power function. We can rewrite it as y = 2 x3. 6. This is a power function. We can rewrite it as y = 2x We have 3 p = 3p /2. The coefficient is 3 and the exponent is /2. 8. We have 2b = (2b) /2 = 2 /2 b /2. The coefficient is 2 /2 and the exponent is /2. 9. We have 4 4 = 4 z z = /4 4z /4. The coefficient is 4 and the exponent is /4. x 0. We have 3 = x/2 x x = /3 x(/2) (/3) = x /6. The coefficient is and the exponent is /6.. Since x > 0 have 4 x = 2 4 (x 2 ) = /2 4 ( ) 3 2. We have 5 = x x 4 x 4 x 4 x = 4 x4 = 4 x3. The coefficient is /4 and the exponent is (x /2 ) = 3 25x = 3/2 25 x 3/2. The coefficient is /25 and the exponent is 3/ We have 3 x = 8/3 6 (x 6 ) = 2 /3 x = 2 2x 2. The coefficient is 2 and the exponent is We have (2 x)x 2 = 2x /2 x 2 = 2x 2+(/2) = 2x 5/2. The coefficient is 2 and the exponent is 5/2. 3 ( ) 3 /2 /2 5. We have 4 s = 4 s = 3/4 s /2 = 3/4 s /2. The coefficient is 3 /4 /2 and the exponent is /2. 4 /2 2

5 7.2 SOLUTIONS We have 4t3 = ( (4t 3 ) /2) /2 = ( 4 /2 t 3/2) /2 = ( 2t 3/2 ) /2 = 2 /2 t 3/4. The coefficient is 2 /2 and the exponent is 3/4. 7. We have so k = 2/3 and p = /2. 8. We have so k = /3 and p = We have We see that k = 49 and p = We have 49 Thus k = 7 and p = 5/2. 2. We have Thus k = 27 and p = x = (2/3)x /2, 3x 2 = (/3)x 2, (7x 3 ) 2 = 7 2 (x 3 ) 2 = 49x 6. x 5 = 49 x 5 = 7 x 5/2 = 7x 5/2. (3x 2 ) 3 = 3 3 (x 2 ) 3 = 27x We have ( 2 x ) 3 = 23 (x /2 ) 3 = 8 x 3/2 = 8x 3/2. Thus k = 8 and p = 3/ This represents proportionality to a power: Thus k = /5 and p =. y = 5x = 5 x. 24. Since x 2 and 3x 2 are terms with the same power of x, we begin by combining like terms. We have We see that k = 6 and p = 4. (x 2 + 3x 2 ) 2 = (4x 2 ) 2 = 4 2 (x 2 ) 2 = 6x Since 3 and x 2 are not like terms, we cannot combine them. It is not possible to write this expression in the form kx p. 26. We have 9x5 = 9 x 5 = 3(x 5 ) /2 = 3x 5/2. We see that k = 3 and p = 5/ We have ( ) = x 2 3 ( x) = 3 8x = 3/2 8 x 3/2. We see that k = /8 and p = 3/ Since x 2 and 4x are not like terms, we cannot combine them. It is not possible to write this expression in the form kx p. 29. We have ( ( 2x) 2 ) 3 = ( 2x( 2x)) 3 the power is p = 6, and the coefficient is k = 64. = ( 4x 2) 3 = 4x 2 4x 2 4x 2 = 64x 6,

6 220 Chapter Seven /SOLUTIONS x x/8 = 3 8 so k = /2, p = /3. = x/3 2 = 2 x/3, 3. Writing this as x 3 3 = 3 x3, we have k = /3 and p = 3. If a > b, then a 3 > b 3 and k is positive. So f(a) > f(b). 32. Writing this as 5 t = 5 7 t = 5 t 7, 7 we have k = 5, p = 7. If a > b, then a 7 < b 7 and k is positive. So f(a) < f(b) r = r = 2 r /2 = 2r /2, so k = 2, p = /2. If a > b, then a /2 < b /2 and k is positive. So f(a) < f(b). 3t 2 ( 2t 3) 5 = 3t 2 ( 2) 5 ( t 3) 5 = 3t 2 ( 32)t 3 5 = 96t 2+5 = 96t 7, so k = 96, p = 7. If a > b, then a 7 > b 7 and k is positive. So f(a) > f(b). 35. Writing this as 5 ( ( 2v) 2) 3 = 5 ( ( 2) 2 v 2) 3 = 320v 6, = 5(4) 3 (v 2 ) 3 = 5 64 v 6 we have k = 320, p = 6. If a > b, then a 6 > b 6. But k is negative and so f(a) < f(b). 36. Writing this as 2 3 x x 3 = 2x 3 + 3x 3 9 = 2x 3 + 3x 3 = 5x 3, we have k = 5, p = /3. If a > b, then a /3 > b /3 and k is positive. So f(a) > f(b). PROBLEMS 37. This is a power function with exponent so the graph is a line and graph (III) fits best. 38. Writing this as N = ka /3, we see that this is a power function with exponent /3, so graph (IV) fits best.

7 39. Energy E is a power function of v with exponent 3, so graph (I) fits best. 40. Strength S is a power function of h with exponent 2, so graph (I) fits best. 7.2 SOLUTIONS Writing this as v = dt. we see that v is a power function of t with exponent. Since the exponent is negative, graph (II) fits best. 42. We see that S is a power function of B with exponent 2/3, so graph (IV) fits best. 43. Writing this as N = AL 2, we see that N is a power function of L with exponent 2. Since the exponent is negative, graph (II) fits best. 44. Since T = mb /4, we see that T is a power function of B with exponent /4. Graph (IV) fits best. 45. We see that W is a power function of L with exponent 3, so graph (I) fits best. 46. Since s = (0.0h 0.75 ) w 0.25, we see that s is a power function of w with exponent Graph (IV) fits best. 47. The judged loudness J is a power function of L with exponent 0.3 so graph (IV) fits best. 48. This is a power function with exponent so the graph is a line and graph (III) fits best. 49. Negative. Since the expression does not equal zero, and since the square of any non-zero number is positive, we have (b ) 2 = (A positive number) = A negative number. 50. Negative. The expression does not equal zero, so c 0. Since c is defined, c must be positive, so c itself must be negative. Because the square root of a positive number is positive, this means c c = (A negative number) (A positive number) = A negative number. 5. Positive. Since r 2 > 0 for any value of r, we have ( + r 2 ) 2 = ( + A positive number) 2 = (A number larger than ) 2 = (A number larger than ) = A positive number. 52. Negative. Note that neither v nor w equal 0, for otherwise the expression equals zero. This means v is negative, for otherwise v is undefined. Since v is negative, w must also be negative, for otherwise vw is undefined. We know that the square root of any positive number is positive, so we have w v + v vw = A negative number A positive number + A negative number A positive number = A negative number + A negative number = A negative number. 53. (a) The term in l and the term in w cannot be combined, so we cannot write this expression as a power function. (b) Since l = 3w, we have P = 2 3w + 2w = 8w. 54. (a) We have So P is a power function of w, with k = 8, and p =. A = l 2 + π ( ) l 2 = l 2 + π l2 = ( + π ) l 2. 8 So A is a power function of l, with k = + π/8 and p = 2. (b) The area is larger when l = 2.5 feet because (2.5) 2 > (.5) 2 and the coefficient k is positive.

8 222 Chapter Seven /SOLUTIONS 55. Rewriting this as s = we see that the coefficient is 2/3 /4 and the exponent is / Remains unchanged, since A = 3 A = 3 /4 A/2, (ax) 2 a 2 = a2 x 2 a 2 = x Increases, since (ax) 2 + a 2 = a 2 (x 2 + ). 58. Decreases, since 59. Increases. 60. Increases, since 6. Decreases, since (ax) 2 = a2 x 2 = x2 a 3 a 3 a. a x = ax. x + a = x + a. 62. Remains unchanged, since a 0 x = x. 63. Remains unchanged, since x + a 0 = x Increases, since 65. Decreases, since a 4 (ax) 3 = a4 a 3 x 3 = a x 3. (ax) /3 a = a/3 x /3 a /2 = x/3 a / For investment A, P = $7000, r = 4%, and t = 5 years, so the value of the certificate will be 7000( ) 5 = For investment B, P = $6500, r = 6% and t = 7 so the value of the certificate will be 6500( ) 7 = Therefore the second investment is worth more. 67. For investment A, P = $0,000, r = 2%, and t = 0 years, so the value of the certificate will be $0,000( ) 0 = 2, For investment B, P = $5000, r = 4%, and t = 0 years, so the value of the certificate will be 5000( ) 0 = 740 = Therefore the first investment is worth more.

9 68. For investment A, P = $0,000, r = 3%, and t = 30 years, so the value of the certificate will be 0,000( ) 30 = 24, For investment B, P = $5,000, r = 4%, and t = 5 years, so the value of the certificate will be Therefore the second investment is worth more. 5,000( ) 5 = 27, (a) We have x = 2% = 0.02, so Population = 5( ) 20 = , so the town has 22,289 people in 20 years. (b) We have x = 7% = 0.07, so Population = 5( ) 20 = , so the town has 58,045 people in 20 years. (c) We have x = 5% = 0.05, so so the town has 5,377 people in 20 years. Population = 5( 0.05) 20 = 5.377, 7.3 SOLUTIONS (a) Since r is expressed in thousands of miles, we express the other distances in the same units. The distance from the center to the surface is 4 thousand miles, and the distance from the surface to the astronaut is h miles, or h/000 thousand miles. So the total distance from the center to the astronaut is r = 4 + h 000. (b) Since w = 2880/r 2, we have w = 2880 (4 + h/000) 2. Solutions for Section 7.3 EXERCISES. We raise both sides to the /3 power to obtain: x = 50 /3 = We first solve for x 2, then take the square root of both sides. Remember to include both the positive and negative roots: 2x 2 = 8.6 x 2 = 8.6/2 = 4.3 x = ± 4.3 = ± There are two solutions: x = and x = We raise both sides to the 2 power: 4 = x /2 4 2 = (x /2 ) 2 = x = x x = 4 2 = /(4 2 ) = /6. 4. We solve for w 3 and then raise both sides to the /3 power: 4w 3 = 7 w 3 = 7/4 =.75 w = (.75) /3 =.205.

10 224 Chapter Seven /SOLUTIONS 5. When we solve for z 2, we obtain z 2 = 5. If we take the square root of both sides, we try to evaluate 5 which has no real solutions. This equation has no solutions. 6. We solve for b 4 and then raise both sides to the /4 power. Since we are finding an even root, we need to remember to include both the positive and negative roots. 2b 4 = 92 There are two solutions: b = and b = We solve for a and then square both sides: b 4 = 92/2 = 46 b = ±46 /4 = ± a = 9 a = 9 2 = We solve for 3 x and then raise both sides to the 3 rd power. 3 3 x = 5 3 x = 5/3 = 5 x = 5 3 = y 2 = y 2 = 2 y = y = 9 2y = 8 2y = 82 y = 4.. 3x 2 + = 0 3x 2 = 9 3x 2 = 8 3x = 83 x = We solve for the power (x + ) 2 and then take the square root of both sides. Since we are taking an even root, we need to remember to include both the positive and negative roots. (x + ) 2 = 25 x + = ± 25 = ±5. We have x + = 5 which gives x = 4 and also x + = 5 which gives x = 6. There are two solutions: x = 4 and x = 6.

11 7.3 SOLUTIONS We solve for the power (3c 2) 3, then take the cube root of both sides and then solve for c. (3c 2) 3 = 50 3c 2 = (50) /3 = c = 7.33 c = 7.33/3 = We rewrite the equation with positive exponents. Then we take the cube root of both sides and solve it for x. 2x = 54x 2 2x = 54 x 2 2x 3 = 54 x 3 = 27 x = (27) /3 = We solve for p 5 then take the fifth root of both sides. 2p = 0 2p 5 = 64 p 5 = 32 p = ( 32) /5 p = First we put the powers of t on one side and the constants on the other. Then we take the fourth root of both sides. 4 t3 = 4 t 4 t4 = 4 t 4 = 6 t = ±(6 /4 ) t = ± L 2 = 0 6 = L 2 6L 2 = L 2 = 6 L = ± 4.

12 226 Chapter Seven /SOLUTIONS 8. We can solve this equation by squaring both sides. r = 3 r = 69 r 2 = 25 r = ±5. Because we squared the original equation (which is not a reversible step) we need to check that the values we got r = 5 and r = 5 actually satisfy r = 3. They do, so they are both solutions. 9. We can solve this equation by cubing both sides. 3 = 3 x ( ) 3 3 = ( 3) 3 x x = 27 x = First we combine similar terms and solve for x. We then square both sides. 2. We can solve this equation by squaring both sides. 4 x 2 x = 2 3 x 2 x = 2 3 x x = 3 x = 3 x 3 = x x = 3 2 = 9. z 2 = 5π 44 = z 5π 720π = z. Because we squared the original equation (which is not a reversible step) we need to check that the value we got, z = z 720π, actually satisfies 2 =. It does, so it is a solution. 5π 22. We divide both sides by k and then take the square root of both sides. Since we are taking an even root, we have to remember to include both the positive and negative roots. y = kx 2 y k = x2 x = ± y k.

13 7.3 SOLUTIONS We solve for r 2 and then take the square root of both sides. Since we are taking an even root, we have to remember to include both the positive and negative roots. A = 2 πr2 r 2 = 2A π 2A r = ± π. If we knew that r had to be non-negative (for example, if r represented the radius of a circle), then we would only include the positive root. 24. We solve for D 3 and then raise both sides to the /3 power. L = kb 2 D 3 D 3 = L kb ( 2 ) L /3 D =. kb If x 0, first we simplfy the right side of the equation then divide by y 2 in order to solve for x. We then take the square root of both sides. y 2 x 2 = (3y 2 ) 2 y 2 x 2 = 9y 4 x 2 = 9y 2 x = ± 9y 2 = ±3y. 26. We begin by squaring both sides of the equation in order to eliminate the radical. x w = 4π t ( ) x w 2 = 6π 2 t tw 2 6π = x. 2 Because we squared the original equation (which is not a reversible step) we need to check that the value we got, x tw 2 /6π 2, actually satisfies w = 4π. It does, so it is a solution. t 27. (i), because the cube root of a positive is positive. 28. (i), because the fifth root of a positive is positive. 29. (iv), because positive numbers have two square roots. 30. (iii), because the square root of 0 is (vi), because no real number squared is negative. 32. (ii), because the cube root of a negative is negative. 33. (vi), because even powers of real numbers are not negative. 34. (ii), because an odd power of x is negative only if x is negative. 35. (i), because the cube root of x is positive only if x is positive. 36. (ii), because the cube root of x is negative only if x is negative. 37. (i), because the square root of x is positive only if x is positive. 38. (vi), because the square root of x is non-negative. Note that this equation is different from the equation x 2 = 44, which has two solutions, written 44 /2 and (44 /2 ).

14 228 Chapter Seven /SOLUTIONS 39. (i), because only a positive number to an odd power is positive. 40. (iv), because both positives and negatives raised to even powers are positive. 4. (ii), because only negatives have negative odd powers. 42. (vi), because no even power is negative. 43. The second equation results from squaring both sides of the first equation. We solve the second equation x + 4 = (x 2) 2 x + 4 = x 2 4x + 4 5x x 2 = 0 x(5 x) = 0 so x = 5 and x = 0 are solutions. Checking both of these solutions in the first equation gives = = 3 and 3 = = = The extraneous solution, x = 0, was introduced by the squaring operation. 44. The second equation results from squaring both sides of the first equation. We solve the second equation (t + ) 2 = t 2 + 2t + = t 2 + 2t = 0 t(t + 2) = 0 so t = 0 and t = 2 are solutions. Checking both these solutions in the first equation gives and 0 + = = 2 + =. The extraneous solution, t = 2, was introduced by the squaring operation. 45. The second equation results from dividing both sides of the first equation by the variable r. To check whether the two equations have the same solutions, we solve the first equation r 2 + 3r = 7r r 2 4r = 0 r(r 4) = 0 so r = 4 and r = 0 are solutions. Checking both solutions in the second equation gives = 7 7 = 7

15 7.3 SOLUTIONS 229 and = The two equations are not equivalent because they do not have the same solutions. The solution r = 0 was lost in the operation of dividing by the variable r. 46. The second equation results from improperly taking the square root of both sides. To check whether the two equations have the same solutions, we note that the first equation has x = ±2 as solutions but the second equation has only x = 2 as a solution. The two equations are not equivalent and x = 2 was lost in the operation. The correct result of taking the square root of both sides of the first equation would have yielded 2x = ±4 and preserved both solutions. 47. The second equation results from multiplying both sides of the first equation by the variable p. Solving the second equation we find, 3p = p 2p = 0 p = 0. However, p = 0 is not a solution of the first equation because the denominators are zero for p = 0, thus p = 0 is an extraneous solution. 48. The second equation results from multiplying both sides of the first equation by x +. Solving the second equation, we find, 2x = x + 2 2x = x x =. However, x = is not a solution of the first equation because the denominators are zero for x =, thus x = is an extraneous solution. PROBLEMS 49. Both solutions must be positive, for if x = 0, the equation clearly does not work, and if x < 0, then making the value of the left-hand side less than 5. x 8 x = A negative number A positive number }{{}}{{} x 8 x = A negative number, 50. Since the volume, V is 3π, and since V = 2 3 πr2, we have V = 3π = 2 3 πr π = πr2 9 2 π = πr2 π 9 2 π = π πr2 9 2 = r2 ± 9 2 = r 2

16 230 Chapter Seven /SOLUTIONS ± 3 2 = r. Since a cone s radius cannot be negative, r = 3/( 2) = Since the volume, V, is 27 cm 3 and V = s 3, we have The side length is 3 cm for a cube of volume 27 cm 3. V = 27 = s = 3 s 3 3 = s. 52. The solution to the equation represents the price charged when the electricity consumption is 2 gigawatt-hours per year. To find the solution, we solve for p: 2 = 0.5 p 3/2 Check that 0.8 is a solution to the original equation. 53. (a) We have We substitute S = 2,000 and solve for M: 2p 3/2 = 0.5 p 3/2 = ( 0.5 ) 2/3 p = = 0.8 dollars per kilowatt-hour. 2 S = 095M 2/3. 2,000 = 095M 2/3 2, = M2/3 ( ) 2,000 3/2 M = = kg. 095 The body mass is about 84 kg. (b) The solution represents the body mass of a human with surface area 30,000 cm 2. (c) We have so The formula is M = S = 095M 2/3 095 S = M2/3, ( ) 3/2 ( ) 3/2 095 S = S 3/2 = S 3/ M = S 3/ Because r > 0, we will ignore any negative solutions for r. (a) We have (b) We have r = 2π C. r = π A.

17 7.3 SOLUTIONS 23 (c) We have (d) We have (e) We have r = ( 3 4π V ) /3. V r = πh. 3 r = πh V. 55. (a) For the sphere and the cone to have the same volume, we want 4 3 πr3 = 3 πr2 h. (b) We solve the equation in part (a) by canceling (/3)πr 2 from both sides, giving h = 4r. Thus, the cone should be twice as high as it is wide. 56. (a) When r = 0, we have A = P. This makes sense, because if the interest rate is 0%, there is no interest being added, and the amount remains the same as the initial amount, P. (b) If the interest rate is 5%, then A = P(.05) 2 = P(.025). If the interest rate is 6%, then A = P(.06) 2 = P(.236). So if the interest rate is between 5% and 6%, then the percentage growth in the amount after 2 years will be between 0.25% and 2.36%. (c) A = P(+r) 2, so A/P = (+r) 2. Taking the square root of both sides, we get A/P = +r, or r = A/P. This form of the equation might be useful if you have a goal for the amount of money that you would like to have saved at the end of 2 years, and you want to find out at what interest rate you would need to invest your principal in order to achieve that goal. (d) If the principal amount increases by 25% by the end of 2 years, then A will be at least.25p. Using this value of A in the equation r = A/P, we get r =.25P/P =.25 = 0.8. So r must be.8%. 57. After t years, the balance B is given by ( B = P + r ) t, 00 where P is the initial deposit. We know that B = 2P when t = 0, so ( 2P = P + r ) Dividing by P gives Solving for r, we take roots Thus the interest rate is 7.77%. 2 = ( + r ) r 00 = 2/0 r 00 = 2/0 r = 00(2 /0 ) = For x to be meaningful, we need x 0, in which case x + 5 x 0, which cannot be negative. 59. (a) If t = 0 then t 3 = 0, so A = 0. (b) If t > 0 then t 3 > 0, so A > 0. (c) If t < 0 then t 3 < 0, so A < (a) If t = 0 then t 4 = 0, so A = 0. (b) If t > 0 then t 4 > 0, so A > 0. (c) If t < 0 then t 4 > 0, so A > 0.

18 232 Chapter Seven /SOLUTIONS 6. (a) If t = 0 then ( t) 3 = 0, so A = 0. (b) If t > 0 then ( t) 3 = ( ) 3 t 3 = t 3 < 0, so A < 0. (c) If t < 0 then ( t) 3 = ( ) 3 t 3 = t 3 > 0, so A > (a) If t = 0 then ( t) 4 = 0 4 = 0, so A = 0. (b) If t > 0 then ( t) 4 = ( ) 4 t 4 = t 4 > 0, so A > 0. (c) If t < 0 then ( t) 4 = ( ) 4 t 4 = t 4 > 0, so A > (a) If t = 0 then t 3 = 0, so A 2 = 0, that is A = 0. (b) If t > 0 then t 3 > 0, so A 2 > 0, which is satisfied for any A 0. (c) If t < 0 then t 3 < 0, but A 2 > 0, which is not satisfied for any A. 64. (a) If t = 0 then t 4 = 0, so A 2 = 0, that is A = 0. (b) If t > 0 then t 4 > 0, so A 2 > 0, which is satisfied for any A 0. (c) If t < 0 then t 4 > 0, so A 2 > 0, which is satisfied for any A (a) If t = 0 then t 4 = 0, so A 2 = 0, that is A = 0. (b) If t > 0 then t 4 > 0, but A 2 < 0, which is not satisfied for any A. (c) If t < 0 then t 4 > 0, so A 2 < 0, which is not satisfied for any A. 66. (a) If t = 0 then t 3 = 0, so A 2 = 0, that is A = 0. (b) If t > 0 then t 3 > 0, but A 2 < 0, which is not satisfied for any A. (c) If t < 0 then t 3 < 0, so A 2 < 0, which is satisfied for any A (a) If t = 0 then = A, so A =. (b) If t > 0 then t 3 + >, so A >. (c) If t < 0 then t 3 + <, so A <. 68. The equation At 2 + = 0 is the same as At 2 =. (a) If t = 0 then 0 =, which means that no value of A will work. (b) If t > 0 then t 2 > 0, so we need A < 0 to make At 2 < 0. (c) If t < 0 then t 2 > 0, so we need A < 0 to make At 2 < (a) If t = 0 then At 2 = 0 for any A. (b) If t > 0 then t 2 > 0, so At 2 = 0 only if A = 0. (c) If t < 0 then t 2 > 0, so At 2 = 0 only if A = The equation A 2 t 2 + = 0 is the same as A 2 t 2 =. The left-hand side is never negative, while the right-hand side is always negative. Thus, no values of A can make this true. (a) No A. (b) No A. (c) No A. 7. (a) If x = than A = 4. (b) If x > then A > 4. (c) There is no solution for A < (a) If x = then A = 4. (b) If x > then A > 4. (c) This equation has a solution for all values of A. So no A. 73. (a) If x = then A = 4. (b) If x > then A < 4. (c) This equation has no solution for A > (a) If x = then A = 4. (b) If x > then 0 < A < 4. (c) This equation has no solution for A (a) We see that the graphs cross in the first quadrant. We can find the point of intersection more accurately by zooming in on the graph. We see that the point of intersection occurs at approximately x = 9.6. See Figure 7.4.

19 7.3 SOLUTIONS 233 y 200 x = x Figure 7.4 (b) Using algebra, we have.3x 3 = 20x x 2 = x =.3 = We used only the positive square root since we wanted the point of intersection in the first quadrant. 76. (a) We see that the graphs cross in the first quadrant. We can find the point of intersection more accurately by zooming in on the graph. We see that the point of intersection occurs at approximately x = 7.. See Figure y x = x Figure 7.5 (b) Using algebra, we have 0.2x 5 = 000x 2 x 3 = = 5000 x = (5000) /3 = 7.00.

20 234 Chapter Seven /SOLUTIONS Solutions for Section 7.4 EXERCISES. We have y = kx 5. We substitute y = 744 and x = 2 and solve for k: The formula is 2. We have y = kx 3, with k = The formula is y = kx = k(2 5 ) k = = = y = 23.25x 5. y = 0.35x We have y = kx 2. We substitute y = 000 and x = 5 and solve for k: The formula is y = kx = k(5 2 ) k = = = 40. y = 40x We have y = kx 4. We substitute y = 0.25 and x = 3 and solve for k: The formula is y = kx = k(3 4 ) k = 0.25 = 0.25 = y = 0.25x We have s = k t. We substitute s = 00 and t = 50 and solve for k: s = k t 00 = k 50 k = = The formula is s = 4.42 t. 6. We have A = k/b 3. We substitute A = 20.5 and B = 4 and solve for k: A = k B 3 The function is 20.5 = k ( 4) 3 k = 20.5 ( 4) 3 = 32. A = 32 B 3.

21 7.4 SOLUTIONS Substituting into the general formula c = kd 2, we have 50 = k(5) 2 or k = 50/25 = 2. So the formula for c is When d = 7, we get c = 2(7) 2 = 98. c = 2d Substituting into the general formula c = k/d 2, we have 50 = k/5 2 or k = 250. So the formula for c is When d = 7, we get c = 250/7 2 = For some constant k, we have S = kh 2. c = 250 d We know that E is proportional to v 3, so E = kv 3, for some constant k.. We have r = k A = ka /2. 2. For some constant k, we have K = kv When x is doubled, we have So y is multiplied by a factor of When x is doubled, we have So y is multiplied by a factor of /8. 5. Dividing both sides by x, we get When x is doubled, we have So y is multiplied by a factor of /2. 6. Multiplying both sides by x 4, we get When x is doubled, we have So y is multiplied by a factor of 6. PROBLEMS 7. (a) We have T = kr 2 D 4. (b) We substitute R = 300D and simplify: new value of y = k(2x) 3 = k 2 3 x 3 = 8(kx 3 ). new value of y = So T is a power function of D. (c) We substitute D = 0.25 R and simplify: So T is a power function of R. k (2x) = k 3 8x = 3 8 k x. 3 y = k x. new value of y = k 2x = 2 k x. y = kx 4. new value of y = k(2x) 4 = k 2 4 x 4 = 6(kx 4 ). T = kr 2 D 4 = k(300d) 2 D 4 = 90,000kD 6. T = kr 2 D 4 = kr 2 (0.25 R) 4 = (0.25) 4 kr 2 R 2 = kR 4.

22 236 Chapter Seven /SOLUTIONS 8. (a) Since the rate R varies directly with the fourth power of the radius r, we have the formula R = kr 4, where k is a constant. (b) Given R = 400 for r = 3, we can determine the constant k. So the formula is (c) Evaluating the formula at r = 5 yields 400 = k(3) = k(8) k = = R = 4.938r (a) T is proportional to the fourth root of B, and so (b) 48 = k (5230) /4 and so k = 48/(5230) /4 = 7.4. (c) Since T = 7.4B /4, for a human with B = 70 we have R = 4.938(5) 4 = cm3 sec. T = k 4 B = kb /4. T = 7.4(70) /4 = 50.3 seconds. It takes about 50 seconds for all the blood in the body to circulate and return to the heart. 20. (a) We have R = kv 2, where k is the constant of proportionality. Thus, R = (20) 2 = 7228 feet =.37 miles. (b) If R is now measured in meters, R = 7228 ft = (7228/3.28) meters = 2204 meters, and V = 20 miles/hour = 20(609 meters)/3600 sec = meters/sec. So the new constant of proportionality is 2204/ = Since N is inversely proportional to the square of L, we have N = k L 2. As L increases, N decreases, so there are more species at small lengths. 22. We need to solve y = kx p for p and k. We know that y = 2 when x =. Since we also have y = k p = k when x =, we have k = 2. To solve for p, use the fact that y = 8 when x = 2 and also y = 2 2 p when x = 2, so 2 2 p = 8, giving 2 p = 4, so p = 2. Thus, y = 2x 2. We must check that all the points in the table satisfy this equation. They do. 23. We need to solve y = kx p for p and k. To solve for p, take the ratio of any two values of y, say the values corresponding to x = 3 and x = 2, namely y = 36 and y = 6: 36 6 = 9 4. Since y = k 3 p when x = 3 and y = k 2 p when x = 2, we have k 3 p ( ) k 2 = 3p 3 p p 2 = = 9 p 2 4. Since ( 3 2 )p = 9 4, we know p = 2. Thus, y = kx2. To solve for k, use any point from the table. Since y = 6 when x = 2, we have k 2 2 = 4k = 6, so k = 4. Thus, y = 4 x 2. We must check that all the points in the table satisfy this equation. They do.

23 24. Solve for y by taking the ratio of (say) the values of y when x = and when x = 2: 6 = 6. We know y = k 2 p when x = 2 and y = k p when x =. Thus, k 2 p k p = 2p p = ( 2 ) p = SOLUTIONS 237 Thus p = 4. To solve for k, note that y = k 2 4 = 6k when x = 2. Thus, 6k = 6. Thus, k =. This gives y = x 4. We must check that all the points in the table satisfy this equation. They do. 25. Solve for y by taking the ratio of, say, the values of y when x = 2 and x = 8/5 /5 = 8. We know y = k 2 p when x = 2 and y = k p when x =. Thus, k 2 p k p = 2p = 8. Then p = 3. To solve for k, note that y = k 2 p = k 2 3 = 8k when x = 2. Thus we have 8k = 8. Then k = /5, 5 which gives y = (/5)x 3. We must check that all the points in the table satisfy this equation. They do. 26. (a) We have (b) We have 4 f(2r) f(r) = 3 π(2r)3 4 3 πr3 4 3 π 4 3 = π 8r r 3 3 = 8. 4 f(r) f( r) = 3 πr π( 2 r)3 = = = 8. 8 (c) Both expressions equal 8 which means that in both cases, the larger sphere has eight times the volume of the smaller. The first result tells us that the volume of a sphere of radius 2r is eight times the volume of a sphere of radius r. The second result tells us that the volume of a sphere of radius r is eight times the volume of a sphere of radius r. 2 In both cases, the result shows that a sphere whose radius is twice that of a smaller sphere has eight times its volume. We see that, algebraically, it does not matter if we say the larger sphere has a radius twice the size of the smaller, or instead the smaller sphere has a radius half the size of the larger. 27. (a) We have ( ) r g 0 = g(r) 4 π 3 4 π 3 ( r km 0 = km km = 00 kmr 2 r 3 8 r3 ) 2 r r 2 = 00.

24 238 Chapter Seven /SOLUTIONS (b) We have g(r) g(0r) = kmr 2 km(0r) 2 = km km r r 2 = 00 = 00. (c) Both expressions equal 00 which means that in both cases, the force on the closer object is 00 times as great as the force on the more distant object. The first result tells us that the force on an object at a distance of r/0 is 00 times the force at a distance of r. The second result tells us that the force on an object at a distance of r is 00 times the force at a distance of 0r. In both cases, the result shows that the force on an object is 00 times the force an an object of identical mass that is ten times farther away. This is telling us that, algebraically, it does not matter if we say the more distant object is at a distance ten times as great, or instead the closer object is at a distance one-tenth as great. 28. When the side is x, the area is x 2. (a) When the side is 2x, So area is multiplied by a factor of 4. (b) When the side is 3x, Area = (2x) 2 = 2 2 x 2 = 4x 2. Area = (3x) 2 = 3 2 x 2 = 9x 2. So area is multiplied by a factor of 9. (c) When the side is x, 2 ( ) 2 ( ) 2 Area = 2 x = x 2 = 2 4 x2. So area is multiplied by a factor of /4. (d) When the side is 0.x, Area = (0.x) 2 = (0.) 2 x 2 = 0.0x 2. So area is multiplied by a factor of When the side is x, the volume is x 3. (a) When the side is 2x, Volume = (2x) 3 = 2 3 x 3 = 8x 3. So volume is multiplied by a factor of 8. (b) When the side is 3x, Volume = (3x) 3 = 3 3 x 3 = 27x 3. So volume is multiplied by a factor of 27. (c) When the side is x, 2 ( ) 3 ( ) 3 Volume = 2 x = x 3 = 2 8 x3. So volume is multiplied by a factor of /8. (d) When the side is 0.x, Volume = (0.x) 3 = (0.) 3 x 3 = 0.00x 3. So volume is multiplied by a factor of The area of a circle of radius r is πr 2. If r is halved, then we get a new circle of radius (/2)r, which has area ( ) 2 ( ) 2 π 2 r = π r 2 = 2 4 πr2. So the area is multiplied by /4.

25 7.4 SOLUTIONS If the cube has side length s, then one of its faces has area s 2. Since there are six faces, the entire cube has surface area 6s 2. If s is increased by 0%, then the side length is multiplied by., so the new side length is (.)s, and the new surface area is 6(. s) 2 = 6(.) 2 s 2 = (.) 2 (6s 2 ) =.2 6s 2. So the surface area is increased by 2%. 32. If the cube has side length s, then it has volume s 3. If s is increased by 0%, then the side length is multiplied by., so the new side length is (.)s, and the volume is So the volume is increased by 33.%. New volume = (. s) 3 = (.) 3 s 3 =.33 s 3 =.33 Old volume. 33. (a) For both birds and mammals the life span increases with body size. Thus the larger mammal will have the longer life span. Because the graph of the bird life span is above that of the mammal, the bird has the greater life span, for a fixed body size. (b) A body size of 0 would mean no life span! From this and the graph we can construct the table and the corresponding graph. This is very close to a line with slope 57/25 = Thus, on average a bird will live over twice as long as a mammal of the same size. Table 7. Body size Mammal life span Bird life span bird lifespan mammal lifespan Figure 7.6 ( (c) (i) From L M =.8W 0.20 LM ) /0.20 we find W = which, when substituted in L B = 28.3W 0.9, gives.8 ( ) 0.9/0.20 LM L B = This is the function that is plotted in part (b). Also, 0.9/0.20 = 0.95 is very.8 close to, so the graph looks like the line L B = 28.3 LM = 2.398LM..8 ( ) (ii) To have L M = L B we would need.8w 0.20 = 28.3W 0.9, or.8w 0.0 = 28.3, so W = = kg. (The largest animal is the blue whale which is about 0 5 kg. The earth weighs about kg.) If there were an bird or mammal of this size, its life span would be.8 ( ) 0.20 = years. This is unrealistic. 34. If z is proportional to a power of y we have z = k y n, where k is a constant. If y is proportional to a power of x we have y = k 2x m, where k 2 is a constant. So z = k y n = k (k 2x m ) n = (k k n 2 )x nm, so z is proportional to a power of x. 35. If z is proportional to a power of x we have z = k x n, where k is a constant. If y is proportional to the same power of x we have y = k 2x n, where k 2 is a constant. So z + y = k x n + k 2x n = (k + k 2)x n, so z + y is proportional to a power x.

26 240 Chapter Seven /SOLUTIONS 36. If z is proportional to a power of x we have z = k x n, where k is a constant. If y is proportional to a power of x we have y = k 2x m, where k 2 is a constant. So zy = k x n k 2x m = (k k 2)x n+m, so zy is proportional to a power of x. 37. If z is proportional to a power of x we have z = k x n, where k is a constant. If y is proportional to a different power of x we have y = k 2x m, where k 2 is a constant and n m. So z + y = k x n + k 2x m. Since n m these two terms cannot be combined to be proportional to a power of x, so z + y is not proportional to a power of x. Solutions for Chapter 7 Review EXERCISES. We have y = 3x 2 ; k = 3, p = We have y = 5x /2 ; k = 5, p = /2. 3. We have y = 3 8 x ; k = 3 8, p =. 4. Not a power function. 5. We have y = 5 2 x /2 ; k = 5 2, p = /2. 6. We have y = 9x 0 ; k = 9, p = We have y = 0.2x 2 ; k = 0.2, p = Not a power function. 9. We have y = 5 3 x 3 = 25x 3 ; k = 25, p = We have y = 8x ; k = 8, p =.. We have y = (/5)x; k = /5, p =. 2. Not a power function because of the Rewriting as 7w 7 w we see that w is the base, the exponent is and /7 is the coefficient. 4. When we raise ( 2t) to the third power we obtain 8t 3. Therefore the base is t, the exponent is 3 and and the coefficient is We have 25v 5 /25v 3 = 5v 2, so the base is v, the exponent is 2 and the coefficient is The base is r, the exponent is 3, and the coefficient is 4 3 π. 7. We know that (4x 3 )(3x 2 ) = 2x. Therefore, the base is x, the exponent is, and the coefficient is The base is z, the exponent is 4, and the coefficient is. 9. We have 3 x 2 = 3x 2. The base is x, the exponent is 2, and the coefficient is We have 8 2/x 6 = 8x6 2 = 4x6. The base is x, the exponent is 6, and the coefficient is We have 3( 4r) 2 = 3(6r 2 ) = 48r 2. The base is r, the exponent is 2, and the coefficient is We have 0t = 5 4t = 5 4 t 5. The base is t, the exponent is 5, and the coefficient is / We have (πa)(πa) = π 2 a 2. The base is a, the exponent is 2, and the coefficient is π We have πa + πa = 2πa. The base is a, the exponent is, and the coefficient is 2π. 25. We multiply both sides by x 3, then solve for x 3 and then raise both sides to the /3 rd power: 50 x 3 = = 2.8x 3

27 x 3 = ( ) 50 /3 x = = SOLUTIONS to Review Problems for Chapter Seven We have: 5 x 2 = 8 x 3 5x 3 = 8x 2 x 3 x 2 = 8 5 x = We have: 2 x = 3 2 = 3 x x = 2 3 = 4 x = 4 2 = We have: x 3 = 5 4 = 5 4 x = x 3 x 3 = (0.8) 2 x = 3 + (0.8) 2 = We have: This gives two solutions: and 00 (x 2) 2 = 4 00 = 4(x 2) 2 (x 2) 2 = 25 x 2 = ± 25 = ±5. x 2 = 5 so x = 7, x 2 = 5 so x = 3.

28 242 Chapter Seven /SOLUTIONS 30. We have: A Bx n = C A = BCx n x n = A BC ( ) A /n x =. BC 3. (a) We have: (b) We have: 2π L = R C 2 2π L = R C 2 RC 2 L = L = 2π ( RC 2 2π L = R2 C 4 4π 2. ) 2 2π L = R C 2 2π L = R C 2 C 2 = 2π L R 2π L C = ± R. 32. Since we take an odd root to solve the equation, the equation has only one solution. 33. Since we take an even root to solve the equation, the equation has two solutions. 34. Since we square both sides to solve the equation, the equation has only one solution. 35. Since no real number when raised to an even power gives a negative number, this equation has no solutions. 36. Since we raise both sides of the equation to a power of 5, this equation has only one solution. 37. Since we take an odd root to solve the equation, this equation has only one solution. PROBLEMS 38. (a) We have E = klh 2. (b) Since E = 4 million foot-pounds = 4,000,000 when L = 600 and h = 30, we have 4,000,000 = k 600 (30 2 ) k = 4,000, (30 2 ) = We have E = 7.407Lh 2.

29 To find the units, we solve the equation E = klh 2 for k to obtain k = E Lh 2. The units of E are foot-pounds which are feet pounds, so we have Units of k = The units of k are pounds per square foot. (c) We substitute h = (/4)L = 0.25L and simplify: SOLUTIONS to Review Problems for Chapter Seven 243 Units of E Units of L (Units of h) 2 = foot-pounds feet feet 2 feet pounds = feet 3 = pounds feet 2. E = 7.407Lh 2 = 7.407L(0.25L) 2 = 7.407(0.25) 2 L L 2 = 0.463L 3. (d) We substitute L = 5h and simplify: E = 7.407(5h)h 2 = h We know R is proportional to the fourth power of r, with constant of proportionality 4.94 so we have (a) We substitute R = 500 and solve for r: The radius of the pipe is about 3.72 cm. (b) We have R = 4.94r R = r4 r = R = 4.94r = 4.94r = r4 ( ) 500 /4 r = = 3.72 cm ( 4.94 R ) /4 = ( ) /4 R /4 = 0.67R / We have r = 0.67R /4. (c) Since r =Constant R /4, we see that r is proportional to the /4 th power of R. 40. We have (a) We substitute L = 50 and E = 40,000 and solve for h: E = 7.4Lh 2. 40,000 = 7.4(50)h 2 40, (50) = h2 40,000 h = = ft. 7.4(50)

30 244 Chapter Seven /SOLUTIONS (b) We substitute L = 20 and solve for h: E = 7.4(20)h 2 h 2 = 7.4(20) E h = 7.4(20) E = 7.4(20) E = E /2. We have h = E /2. The coefficient is and the power is /2. 4. We have for some constant k. We solve for R: P = k R 3, P R 3 = k The quantity R is inversely proportional to the cube root of P. 42. (a) We have T = kr 2 D 4 for some constant k. (b) We solve for R: We have R = CT /2 D 2 so n = /2 and m = 2. (c) We solve for D: R 3 = k P ( ) k /3 R = = k/3 Constant =. P P /3 P /3 T = kr 2 D 4 R 2 = T kd 4 T R = kd = T /2 4 k /2 D = 2 (/k/2 )T /2 D 2. T = kr 2 D 4 D 4 = T kr ( 2 ) T /4 D = = T /4 kr 2 k /4 R = /2 (/k/4 )T /4 R /2. We have D = CT /4 R /2 so n = /4 and m = / (i), because for a positive power of x to be greater than, x must be greater than. 44. (i), because for a positive power of x to be greater than, x must be greater than. 45. (iii), because for a positive power of x to be less than, x must be less than. 46. (iii), because for a positive power of x to be less than, x must be less than. 47. (i), because for a positive power of x to be greater than, x must be greater than. 48. (iii), because for a positive power of x to be less than, x must be less than. 49. (i), because for a positive power of x to be greater than, x must be greater than. 50. (ii), because for a non-zero power of x to equal, x must equal. 5. (iii), because for a negative power of x to be greater than, x must be less than. 52. (i), because for a negative power of x to be less than, x must be greater than. 53. (i), because for a negative power of x to be less than, x must be greater than.

31 SOLUTIONS to Review Problems for Chapter Seven (iv), because no even power is negative. 55. (iii), because for a negative power of x to be greater than, x must be less than. 56. (ii), because for a non-zero power of x to equal, x must equal. 57. Taking the square root of both sides of the equation, we get 58. Dividing both sides of the equation by 4, we get 2x = ± 6 = ±4 x = ±2. x 2 = 4 x = ± 4 = ± Dividing both sides of the equation by 6, we get Multiplying both sides by 4, we get x 2 4 =. x 2 = 4 x = ± 4 = ± If we take the cube root of both sides, we cannot solve the equation because 3 x cannot be further simplified. Step (b) is the appropriate step to solve the equation. After subtracting 8 from both sides, we can take the cube root of both sides of the equation to get the solution. 6. Equations (c) and (d) have the same solutions as the given equation. Equation (c) is obtained from the given equation by taking the square root of both sides. Equation (d) is obtained from the given equation by dividing both sides by 9. These operations do not change the equality of the two sides and so the transformed equations have the same solutions as the original given equation. Equation (a) does not have the same solutions as the given equation because the negative root is not included. There is no way to obtain equation (b) from the given equation by carrying out the same operation on both sides of the equation. 62. (a) See Figure 7.7. (b) If s =, then the relation between d and w is given by w = kd. The graph of this equation looks like a straight line through the origin, but the points on the plot do not lie on a straight line through the origin. So it does not support the hypothesis. (c) If s = 2 or 3, then we have w = kd 2 or w = kd 3. Both of these look like graphs that go through the origin and then curve upward. The points on the data plot also look like they lie on a curve that goes through the origin and curves upward, so it is possible that s = 2 or s = 3. (d) Table 7.2 (e) Diameter, d (cm) Weight, w (kg) w/d Overall, the ratios are increasing. If s = 2 were the correct value, the ratios should be all approximations of the constant k, so they should all stay roughly the same. Table 7.3 Diameter, d (cm) Weight, w (kg) w/d

32 246 Chapter Seven /SOLUTIONS This time the ratios are decreasing. In part (d) they were increasing overall. Somewhere in between we should be able to find a value where they are roughly constant. So the correct value of s is somewhere between s = 2 and s = 3. (f) We calculate the ratios with s = 2.5. Table 7.4 Diameter, d (cm) Weight, w (kg) w/d There seems to be still a slight downward trend in the ratios, so we try again with 2.4. Table 7.5 Diameter, d (cm) Weight, w (kg) w/d Here the ratios show no clear upward or downward trend, so we estimate s = 2.4 and we take the average of the ratios as our estimate for k, which is about Thus the power function w = 0.086d 2.4 is a good fit for the data, as shown in Figure 7.8. w (kg) w (kg) d (cm) d (cm) Figure 7.7 Figure (a) Kepler s Law states that the square of the period, P, is proportional to the cube of the distance, d. Thus, we have P 2 = kd 3. Solving for P gives P = kd 3 = kd 3/2 = k d 3/2, where we let k = k. For the earth, P = 365 and d = 93,000,000. Thus, 365 = k (93,000,000) 3/2,

33 so This gives k = 365 (93,000,000) 3/2. SOLUTIONS to Review Problems for Chapter Seven 247 ( ) 365 P = (93,000,000) d 3/2 3/2 d 3/2 d3/2 = 365 = 365. (93,000,000) 3/2 93,000,000 (b) For Mars, d = 42,000,000, so we have P = 365 ( ) 3/2 42,000,000 93,000,000 which gives P = 689 earth days. 64. (a) Since the graph shows values of V that get larger as the values of n get larger, the exponent β must be positive. Since it bends towards the horizontal axis, we have β <. So 0 < β <. (b) The value of V corresponding to n =,000,000 is about V = 0,000. Since n 0.5 =,000, = 000, we have V = = 0n 0.5, so K = 0. Solutions for Solving Drill. We raise both sides to the /3 power: x 3 = 0 x = 0 /3 = We first divide by 5 and then take the square root. Remember that to find all solutions, when we take even roots, we must consider both the positive and the negative roots. We have 5x 2 = 32 x 2 = 6.4 x = ± 6.4 = ± This equation is linear in x. We have 4x + = 8 4x = 7 x = We isolate the t 5, then raise both sides to the /5 power: 2t 5 = 74 t 5 = 37 t = 37 /5 =

34 248 Chapter Seven /SOLUTIONS 5. This equation is linear in p. We have 5p 2 = 3p + 8 2p = 20 p = This equation is linear in t. We have.3t = 6.2.3t = 4.7 t = We isolate the square root, then square both sides. We have 25 t = 8 t = 0.32 t = = We isolate the w 3, then raise both sides to the /3 power. We have 8w = 30 8w 3 = 25 w 3 = 3.25 w = 3.25 /3 = We isolate the p 4, then raise both sides to the /4 power. Since we are taking an even root, we must include both the positive and negative values. We have.2p 4 = 60 p 4 = 50 p = ±50 /4 = ± This is a linear equation. We have 5.2x 7. = 3.9x x = 32.9 x = We isolate the x 2, then take the square root of both sides. Since we are taking an even root, we must include both the positive and negative values. We have 2.3x = x 2 =.3 x 2 = 4.93 x = ± 4.93 = ±2.27.