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Solution Manual for Calculus For Biology and Medicine, 4th Edition

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     S OLUTIONS M ANUAL R OGER L IPSETT TO ACCOMPANY C ALCULUS FOR B IOLOGY AND M EDICINE F OURTH E DITION Claudia Neuhauser University of Minnesota Marcus L. Roper University of California—Los Angeles

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    Contents 1 Preview and Review 5 1.1 Precalculus Skills Diagnostic Test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 1.2 Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9 1.3 Elementary Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23 1.4 Graphing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50 Chapter 1 Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 78 2 Discrete-Time Models, Sequences, and Difference Equations 87 2.1 Exponential Growth and Decay . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87 2.2 Sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97 2.3 Modeling with Recursion Relations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113 Chapter 2 Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 126 3 Limits and Continuity 133 3.1 Limits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 133 3.2 Continuity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 147 3.3 Limits at Infinity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 156 3.4 Trigonometric Limits and the Sandwich Theorem . . . . . . . . . . . . . . . . . . . . . . . 159 3.5 Properties of Continuous Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 164 3.6 A Formal Definition of Limits (Optional) . . . . . . . . . . . . . . . . . . . . . . . . . . . 169 Chapter 3 Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 173 4 Differentiation 185 4.1 Formal Definition of the Derivative . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 185 4.2 Properties of the Derivative . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 191 4.3 The Power Rule, the Basic Rules of Differentiation, and the Derivatives of Polynomials . . 201 4.4 The Product and Quotient Rules, and the Derivatives of Rational and Power Functions . 214 4.5 The Chain Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 224 4.6 Implicit Functions and Implicit Differentiation . . . . . . . . . . . . . . . . . . . . . . . . 232 4.7 Higher Derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 239 4.8 Derivatives of Trigonometric Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 242 4.9 Derivatives of Exponential Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 247 4.10 Derivatives of Inverse Functions, Logarithmic Functions, and the Inverse Tangent Function 254 4.11 Linear Approximation and Error Propagation . . . . . . . . . . . . . . . . . . . . . . . . . 262 Chapter 4 Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 266

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    2 CONTENTS 5 Applications of Differentiation 275 5.1 Extrema and the Mean-Value Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 275 5.2 Monotonicity and Concavity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 288 5.3 Extrema and Inflection Points . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 302 5.4 Optimization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 308 5.5 L’Hˆ opital’s Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 320 5.6 Graphing and Asymptotes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 331 5.7 Recurrence Equations: Stability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 351 5.8 Numerical Methods: The Newton-Raphson Method . . . . . . . . . . . . . . . . . . . . . . 362 5.9 Modeling Biological Systems Using Differential Equations . . . . . . . . . . . . . . . . . . 369 5.10 Antiderivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 375 Chapter 5 Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 386 6 Integration 399 6.1 The Definite Integral . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 399 6.2 The Fundamental Theorem of Calculus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 418 6.3 Applications of Integration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 430 Chapter 6 Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 457 7 Integration Techniques and Computational Methods 465 7.1 The Substitution Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 465 7.2 Integration by Parts and Practicing Integration . . . . . . . . . . . . . . . . . . . . . . . . 471 7.3 Rational Functions and Partial Fractions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 486 7.4 Improper Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 509 7.5 Numerical Integration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 522 7.6 The Taylor Approximation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 539 7.7 Tables of Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 546 Chapter 7 Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 552 8 Differential Equations 565 8.1 Solving Differential Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 565 8.2 Equilibria and Their Stability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 587 8.3 Differential Equation Models . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 615 8.4 Integrating Factors and Two-Compartment Models . . . . . . . . . . . . . . . . . . . . . . 634 Chapter 8 Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 648 9 Linear Algebra and Analytic Geometry 655 9.1 Linear Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 655 9.2 Matrices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 667 9.3 Linear Maps, Eigenvectors, and Eigenvalues . . . . . . . . . . . . . . . . . . . . . . . . . . 683 9.4 Demographic Modeling . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 707 9.5 Analytic Geometry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 714 Chapter 9 Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 724 10 Multivariable Calculus 733 10.1 Functions of Two or More Independent Variables . . . . . . . . . . . . . . . . . . . . . . . 733 10.2 Limits and Continuity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 743 10.3 Partial Derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 748 10.4 Tangent Planes, Differentiability, and Linearization . . . . . . . . . . . . . . . . . . . . . . 757 10.5 The Chain Rule and Implicit Differentiation . . . . . . . . . . . . . . . . . . . . . . . . . . 766

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    CONTENTS 3 10.6 Directional Derivatives and Gradient Vectors . . . . . . . . . . . . . . . . . . . . . . . . . 770 10.7 Maximization and Minimization of Functions . . . . . . . . . . . . . . . . . . . . . . . . . 777 10.8 Diffusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 805 10.9 Systems of Difference Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 808 Chapter 10 Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 824 11 Systems of Differential Equations 831 11.1 Linear Systems: Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 831 11.2 Linear Systems: Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 863 11.3 Nonlinear Autonomous Systems: Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . 873 11.4 Nonlinear Systems: Lotka-Volterra Model for Interspecific Interactions . . . . . . . . . . . 895 11.5 More Mathematical Models . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 908 Chapter 11 Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 924 12 Probability and Statistics 935 12.1 Counting . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 935 12.2 What is Probability? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 940 12.3 Conditional Probability and Independence . . . . . . . . . . . . . . . . . . . . . . . . . . . 948 12.4 Discrete Random Variables and Discrete Distributions . . . . . . . . . . . . . . . . . . . . 956 12.5 Continuous Distributions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 979 12.6 Limit Theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1000 12.7 Statistical Tools . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1011 Chapter 12 Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1020

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    Chapter 1 Preview and Review 1.1 Precalculus Skills Diagnostic Test 1. (a) y = x + 5 (b) y = 2 x − 1 (c) y = 3 x − 5 (Review 1.2.2) 2. (a) 5 9 (80 − 32) = 240 9 ≈ 26 . 7 ◦ C (b) Inverting the given formula gives x = 9 5 y + 32, so that 9 5 ( − 10) + 32 = − 18 + 32 = 14 ° F (c) Set y = x in the conversion formula and solve for x : x = 5 9 ( x − 32) = 5 9 x − 160 9 ⇒ 4 9 x = − 160 9 ⇒ 4 x = − 160 ⇒ x = − 40 . (Review 1.2.2) 3. A circle with center ( x, y ) = ( − 1 , 5) and with radius 3. (Review 1.2.3) 4. (a) 180 π · π 7 = 180 7 ≈ 25 . 7 ° . (b) x = − π 3 + 2 nπ and x = − 2 π 3 + 2 nπ for any n ∈ Z . (c) Divide both sides of the identity sin 2 θ + cos 2 θ = 1 by cos 2 θ : sin 2 θ + cos 2 θ cos 2 θ = 1 cos 2 θ so tan 2 θ + 1 = sec 2 θ. (d) π 12 , 7 π 12 , 3 π 4 (Review 1.2.4) 5. (a) (i) 2 7 / 3 (ii) 2 8 / 3 (b) Take base 2 logarithms of both sides: x log 2 4 = log 2 1 2 , so 2 x = − 1 and therefore x = − 1 2 .

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    6 Preview and Review (c) Since 10 000 = 10 4 , we have log 10 10 000 = 4. (d) log 10 3 x + log 10 5 x = log 10 (3 x · 5 x ) = log 10 (15 x 2 ). (e) ln ( x 2 ) + ln x = 2 ln x + ln x = 3 ln x . Solving 3 ln x = 2 gives ln x = 2 3 , so that x = e 2 / 3 . (f ) If ln x = 3, then x = e 3 , so that log 10 x = log 10 ( e 3 ) = 3 log 10 e . (Review 1.2.5) 6. (a) Using the quadratic formula, the roots are − 1 ± √ 1 2 − 4 · 1 · 1 2 · 1 = − 1 2 ± i √ 3 2 . (b) (1 + i )(2 − i ) = 1 · 2 + 1 · ( − i ) + i · 2 − i 2 = 3 + i . (c) z + z = ( a + bi ) + ( a − bi ) = 2 a ∈ R (Review 1.2.6) 7. (a) (i) f ( x ) ∈ [0 , 1] (ii) f ( x ) ∈ [0 , ∞ ) (b) (i) f (3) = √ 3 (ii) ( f ◦ g )( x ) = f ( ( x + 1) 2 ) = √ ( x + 1) 2 = | x + 1 | . (iii) ( g ◦ f )(4) = g ( √ 4) = g (2) = (2 + 1) 2 = 9. (c) Since f ( x ) = | x | ≥ 0, then ( f ◦ f )( x ) = | f ( x ) | = f ( x ) (Review 1.3.1) 8. (a) For large values of x , (iii) grows much faster due to the presence of the x 3 term. For example, when x = 6, p 1 ( x ) = 6, p 2 ( x ) = 18, and p 3 ( x ) = 72. (b) (i) The highest power of r that appears is 2, so this is a degree 2 polynomial in r . (ii) Since the velocity should not be negative we must have 1 − r 2 a 2 ≥ 0, so that r ≤ a . Since r is a distance, it must be nonnegative. Putting these together gives 0 ≤ r ≤ a for the domain. (iii) Since 1 − r 2 a 2 ≤ 1 for the allowable values of r , the range is 0 ≤ u ( r ) ≤ u 0 . (iv) The maximum value occurs when r = 0; that value is u 0 . u ( r ) = 1 2 u 0 when 1 − r 2 a 2 = 1 2 , which happens when r = a √ 2 . (Review 1.3.2) 9. (a) For the three values of c , we have r (1) = 1 1 + 10 ≈ 0 . 091 , r (2) = 2 2 + 10 ≈ 0 . 167 , r (3) = 3 3 + 10 ≈ 0 . 231 . The rate of metabolization is largest when c = 3. (b) Since r (5) = 5 5 + 10 = 1 3 , r (10) = 10 10 + 10 = 1 2 , the rate of metabolization increases by less than double.

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    1.1 Precalculus Skills Diagnostic Test 7 (c) From part (b), r (5) = 1 3 ; to get r ( c ) = 2 3 we need r ( c ) = c c + 10 = 2 3 , so that 3 c = 2( c + 10). Solving gives c = 20. (d) No. r (10) = 1 2 . Doubling the rate of metabolization would increase it to 1. But since the denominator of r ( c ) is always strictly greater than the numerator, its value can never be 1. (Review 1.3.3) 10. The relationship is N = kA 1 / 5 . (a) We want to find A so that kA 1 / 5 = 1 2 kA 1 / 5 0 . Cancelling the k ’s gives A 1 / 5 = 1 2 A 1 / 5 0 . Now raise both sides to the fifth power, giving A = A 0 32 . The new habitat area is A 0 32 . (b) We want to find A so that kA 1 / 5 = 1 3 kA 1 / 5 0 . Cancelling the k ’s gives A 1 / 5 = 1 3 A 1 / 5 0 . Now raise both sides to the fifth power, giving A = A 0 243 . The new habitat area is A 0 243 . (c) We want to find A so that kA 1 / 5 = 2 kA 1 / 5 0 . Cancelling the k ’s gives A 1 / 5 = 2 A 1 / 5 0 . Now raise both sides to the fifth power, giving A = 32 A 0 . The new habitat area is 32 A 0 . (Review 1.3.4) 11. (a) Since 1000 = N (0) = N 0 e r · 0 = N 0 we get N 0 = 1000. Since 2000 = N (2) = 1000 e r · 2 , we have 2 = e 2 r . Take natural logarithms, giving 2 r = ln 2 so that r = ln 2 2 . The formula is therefore N ( t ) = 1000 e t ln 2 / 2 . (b) Solving N ( t ) = 3000 gives e t ln 2 / 2 = 3, so that t ln 2 2 = ln 3, and then t = 2 ln 3 ln 2 ≈ 3 . 17. (c) Solving N ( t ) = 4000 gives e t ln 2 / 2 = 4, so that t ln 2 2 = ln 4, and then t = 2 ln 4 ln 2 = 4. (Review 1.3.5) 12. (a) f − 1 ( x ) = √ x − 1 (b) f − 1 ( x ) = − 1 + e x/ 2 (c) f − 1 ( x ) = x 1 / 5 (Review 1.3.6) 13. (a) (i) ln x + ln( x 2 + 1) = ln( x · ( x 2 + 1)). (ii) log ( x 1 / 3 ) − log ( ( x + 1) 1 / 3 ) = 1 3 log x − 1 3 log( x + 1) = 1 3 (log x − log( x + 1)) = 1 3 log x x +1 . (iii) 2 + log 2 x = log 2 4 + log 2 x = log 2 (4 x ). (b) (i) ln 7 ln 2 (ii) ln 6 ln 10 (iii) ln 2 ln x (Review 1.3.7) 14. (a) (i) The amplitude is 2, and the period is 2 π . (ii) The amplitude is 2, and the period is 2 π 3 . (iii) The amplitude is 3, and the period is 2 π π/ 2 = 4. (b) (i) The range is all of R . The maximum domain is all of R except for the points (2 n +1) 2 π where n is an integer (since tangent is undefined at odd multiples of π / 2 ).

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    8 Preview and Review (ii) The range is [ − 1 , 1], since − 1 ≤ cos x ≤ 1 for all x . The domain is R . (c) Compress cos x by a factor of 2 in the x -direction and stretch it by a factor of 3 in the y -direction to get 3 cos 2 x . (Review 1.3.8) 15. - 6 - 4 - 2 2 x - 0.5 0.5 y ( a ) - 4 - 2 2 4 x 0.5 1. 1.5 y ( b ) - 4 - 2 2 4 x - 0.5 0.5 y ( c ) - 4 - 2 2 4 x - 0.5 0.5 y ( d ) - 8 - 4 4 8 x - 0.5 0.5 y ( e ) (Review 1.4.1) 16. (a) (i) 2 kg (ii) 30 kg (iii) 100 kg (b) Adult weight Puppy weight = 20 kg 0 . 4 kg = 50 (c) The point representing the cat will lie one tick below the Jack Russell Terrier. (Review 1.4.2) 17. (a) Since both axes are log scales, this is a power relationship; since the slope of the line is 0 . 41, the relationship is D = k A 0 . 41 . (b) Since only the vertical axis is a log scale, this is an exponential relationship; since the slope of the line is − 1, the relationship is N = k 10 − m .

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    1.2 Preliminaries 9 (c) Since only the vertical axis is a log scale, this is an exponential relationship; since the slope of the line is 0 . 86, the relationship is N = k 10 0 . 86 t ≈ k 7 . 24 t . (d) Since only the vertical axis is a log scale, this is an exponential relationship; since the slope of the line is − 0 . 09, the relationship is N = k 10 − 0 . 09 t ≈ k 0 . 813 t . (Review 1.4.3) 18. (a) This is the graph of patient 1 shifted two hours to the right, so it is graph 4. (b) This graph should start from zero at 8AM just as for patient 1, but it should have a longer tail since it takes longer to enter the bloodstream. This is graph 3. (c) This graph should have the same shape as the graph of patient A, which is graph 4, but is half as high. This is graph 2. (Review 1.4.4) 1.2 Preliminaries  1.2.1 1. (a) Walking 4 units to the right and to the left from − 1 we get the numbers 3 and − 5, respectively. (b) | x − ( − 1) | = 4, so x + 1 = ± 4, yielding x = 3 or x = − 5. 2. With three numbers there will be three pairwise distances: |− 5 − 2 | = |− 7 | = 7 , | 2 − 7 | = |− 5 | = 5 and |− 5 − 7 | = |− 12 | = 12. 3. (a) 2 x + 4 = ± 6, so either 2 x = 2 giving the solution x = 1 or 2 x = − 10 and the other solution is x = − 5. (b) x − 3 = ± 2, from which we see that x = 5 or x = 1. (c) 2 x − 3 = ± 5, so 2 x = 8 giving x = 4 or 2 x = − 2 and x = − 1. (d) 1 − 5 x = ± 6, so that 5 x = 1 ± 6. Then 5 x = − 5, so x = − 1, or 5 x = 7, so x = 7 5 . 4. (a) The equation implies ± (2 x + 4) = ± (5 x − 2), which reduces to two equations 2 x + 4 = 5 x − 2 and 2 x + 4 = − (5 x − 2), so the solutions are: x = 2 and x = − 2 7 . (b) As in part (a) we get 1 + 2 u = ± (5 − u ), and solving both gives 1 + 2 u = 5 − u with solution u = 4 3 , and 1 + 2 u = − (5 − u ) = u − 5 with solution u = − 6. (c) As in part (a) we get the equations 4 + t 2 = ± ( 3 2 t − 2) that solves to t = 6 and t = − 1. (d) As in part (a) 2 s − 6 = ± (3 − s ); solving both equations gives s = 3 and s = 3. So s = 3 is the only solution. 5. (a) We can rewrite the absolute value as two inequalities − 4 ≤ 5 x − 2 ≤ 4 ⇒ − 4 + 2 ≤ 5 x ≤ 4 + 2 ⇒ − 2 ≤ 5 x ≤ 6 ⇒ − 2 5 ≤ x ≤ 6 5 . (b) Rewriting as a pair of inequalities gives 3 − 4 x < − 8 or 3 − 4 x > 8. Solving the first we get: 4 x > 11, or x > 11 4 , and solving the second we get 4 x < − 5, or x < − 5 4 . So x < − 5 4 or x > 11 4 . (c) Rewriting as a pair of inequalities gives 7 x + 4 ≥ 3, which solves as x ≥ − 1 7 , and 7 x + 4 ≤ − 3 which gives 7 x ≤ − 7 or x ≤ − 1. So x ≤ − 1 or x ≥ − 1 7 .

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    10 Preview and Review (d) Rewriting gives − 7 < 3 + 2 x < 7 ⇒ − 7 − 3 < 2 x < 7 − 3 ⇒ − 10 < 2 x < 4 ⇒ − 5 < x < 2 . 6. (a) − 6 < 2 x + 3 < 6; subtracting 3 gives − 9 < 2 x < 3, so that − 9 2 < x < 3 2 . (b) Rewriting as a pair of inequalities gives 3 − 4 x ≥ 2 or 3 − 4 x ≤ − 2, so that x ≤ 1 4 or x ≥ 5 4 . (c) − 1 ≤ x + 5 ≤ 1, which is the same as − 6 ≤ x ≤ − 4. (d) Since no absolute value can be negative, there are no values of x that will satisfy the equation.  1.2.2 7. Use the point slope formula y − y 0 = m ( x − x 0 ). In this case we get y − 2 = − 2( x − 3) ⇒ y − 2 = − 2 x + 6 ⇒ 2 x + y − 8 = 0 . 8. Use the point slope formula y − y 0 = m ( x − x 0 ). In this case we get y − ( − 1) = 1 4 ( x − 2) ⇒ y + 1 = 1 4 x − 1 2 ⇒ − x + 4 y + 6 = 0 . 9. Use the point slope formula y − y 0 = m ( x − x 0 ). In this case we get y − ( − 2) = − 3( x − 0) ⇒ y + 2 = − 3 x ⇒ 3 x + y + 2 = 0 . 10. Use the point slope formula y − y 0 = m ( x − x 0 ). In this case we get y − 5 = 1 2 ( x − ( − 3)) ⇒ 2 y − 10 = x + 3 ⇒ − x + 2 y − 13 = 0 . 11. First compute the slope using the two given points: m = y 2 − y 1 x 2 − x 1 = 4 − ( − 3) 1 − ( − 2) = 7 3 . Now use the point slope formula y − y 0 = m ( x − x 0 ) with ( x 0 , y 0 ) = (1 , 4) (we could equally well use the other point, ( − 2 , − 3)): y − 4 = 7 3 ( x − 1) ⇒ 3( y − 4) = 7( x − 1) ⇒ 3 y − 12 = 7 x − 7 ⇒ − 7 x + 3 y − 5 = 0 . 12. First compute the slope using the two given points: m = y 2 − y 1 x 2 − x 1 = − 4 − 4 1 − ( − 1) = − 4 . Now use the point slope formula y − y 0 = m ( x − x 0 ) with ( x 0 , y 0 ) = ( − 1 , 4) (we could equally well use the other point, (1 , − 4)): y − 4 = − 4( x − ( − 1)) ⇒ y − 4 = − 4 x − 4 ⇒ 4 x + y = 0 .

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    1.2 Preliminaries 11 13. First compute the slope using the two given points: m = y 2 − y 1 x 2 − x 1 = 1 − 3 2 − 0 = − 1 . Now use the point slope formula y − y 0 = m ( x − x 0 ) with ( x 0 , y 0 ) = (0 , 3) (we could equally well use the other point, (2 , 1)): y − 3 = − 1( x − 0) ⇒ y − 3 = − x ⇒ x + y − 3 = 0 . 14. First compute the slope using the two given points: m = y 2 − y 1 x 2 − x 1 = 5 − ( − 1) 4 − 1 = 6 3 = 2 . Now use the point slope formula y − y 0 = m ( x − x 0 ) with ( x 0 , y 0 ) = (1 , − 1) (we could equally well use the other point, (4 , 5)): y − ( − 1) = 2( x − 1) ⇒ y + 1 = 2 x − 2 ⇒ 2 x − y − 3 = 0 . 15. Horizontal lines are always of the form y = k ; since this line goes through ( 4 , 1 4 ) , its equation is y = 1 4 . The standard form is 4 y − 1 = 0. 16. Horizontal lines are always of the form y = k ; since this line goes through (0 , − 1), its equation is y = − 1. The standard form is y + 1 = 0. 17. Vertical lines are always of the form x = k ; since this line goes through ( − 2 , 0), its equation is x = − 2. The standard form is x + 2 = 0. 18. Vertical lines are always of the form x = k ; since this line goes through (2 , − 3), its equation is x = 2. The standard form is x − 2 = 0. 19. Use the slope-intercept form y = mx + b . We are given m = 3 and b = 2, so the equation is y = 3 x + 2, which in standard form is 3 x − y + 2 = 0. 20. Use the slope-intercept form y = mx + b . We are given m = − 1 and b = 5, so the equation is y = − x + 5, which in standard form is x + y − 5 = 0. 21. Use the slope-intercept form y = mx + b . We are given m = 1 2 and b = 2, so the equation is y = 1 2 x + 2, or 2 y = x + 4, which in standard form is x − 2 y + 4 = 0. 22. Use the slope-intercept form y = mx + b . We are given m = − 1 3 and b = 1 3 , so the equation is y = − 1 3 x + 1 3 , which in standard form is x + 3 y − 1 = 0. 23. Use the point slope formula y − y 0 = m ( x − x 0 ) where ( x 0 , y 0 ) = (1 , 0) is the given point. In this case we get y − 0 = − 2( x − 1) ⇒ y = − 2 x + 2 ⇒ 2 x + y − 2 = 0 . 24. Use the point slope formula y − y 0 = m ( x − x 0 ) where ( x 0 , y 0 ) = ( − 2 , 0) is the given point. In this case we get y − 0 = 1( x − ( − 2)) ⇒ y = x + 2 ⇒ x − y + 2 = 0 . 25. Use the point slope formula y − y 0 = m ( x − x 0 ) where ( x 0 , y 0 ) = ( − 1 2 , 0 ) is the given point. In this case we get y − 0 = − 1 2 ( x − ( − 1 2 )) ⇒ y = − 1 2 x − 1 4 ⇒ 2 x + 4 y + 1 = 0 .

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    12 Preview and Review 26. Use the point slope formula y − y 0 = m ( x − x 0 ) where ( x 0 , y 0 ) = ( − 1 2 , 0 ) is the given point. In this case we get y − 0 = 1 5 ( x − ( − 1 2 )) ⇒ y = 1 5 x + 1 10 ⇒ 2 x − 10 y + 1 = 0 . 27. Since the line we want is parallel to x + 2 y − 4 = 0, it must have the same slope. Now, x + 2 y − 4 = 0 implies that 2 y = − x + 4, or y = − 1 2 x + 2, so the line has slope m = − 1 2 . Now use the point slope formula y − y 0 = m ( x − x 0 ) where ( x 0 , y 0 ) = (2 , − 3) is the given point. In this case we get y − ( − 3) = − 1 2 ( x − 2) ⇒ y + 3 = − 1 2 x + 1 ⇒ 2 y + 6 = − x + 2 ⇒ x + 2 y + 4 = 0 . 28. Since the line we want is parallel to x − 2 y + 4 = 0, it must have the same slope. Now, x − 2 y + 4 = 0 implies that 2 y = x + 4, or y = 1 2 x + 2, so the line has slope m = 1 2 . Now use the point slope formula y − y 0 = m ( x − x 0 ) where ( x 0 , y 0 ) = (1 , 2) is the given point. In this case we get y − 2 = 1 2 ( x − 1) ⇒ y − 2 = 1 2 x − 1 2 ⇒ x − 2 y + 3 = 0 . 29. The line passing through (0 , 2) and (3 , 0) has slope m = 0 − 2 3 − 0 = − 2 3 , so the line we want has the same slope. Now use the point slope formula y − y 0 = m ( x − x 0 ) where ( x 0 , y 0 ) = ( − 1 , − 1) is the given point. In this case we get y − ( − 1) = − 2 3 ( x − ( − 1)) ⇒ y + 1 = − 2 3 x − 2 3 ⇒ 2 x + 3 y + 5 = 0 . 30. The line passing through (0 , − 4) and (2 , 1) has slope m = 1 − ( − 4) 2 − 0 = 5 2 , so the line we want has the same slope. Now use the point slope formula y − y 0 = m ( x − x 0 ) where ( x 0 , y 0 ) = (2 , − 1) is the given point. In this case we get y − ( − 1) = 5 2 ( x − 2) ⇒ y + 1 = 5 2 x − 5 ⇒ 5 x − 2 y − 12 = 0 . 31. The line 2 y − 5 x + 7 = 0 can be written as 2 y = 5 x − 7, or y = 5 2 x − 7 2 , so it has slope 5 2 . Since the line we want is perpendicular to it, it must have slope m ⊥ = − 1 m = − 2 5 . It passes through (1 , 4), so using the point slope formula y − y 0 = m ( x − x 0 ) we get y − 4 = − 2 5 ( x − 1) ⇒ 5 y − 20 = − 2 x + 2 ⇒ 2 x + 5 y − 22 = 0 . 32. The line x − 2 y + 3 = 0 can be written as 2 y = x + 3, or y = 1 2 x + 3 2 , so it has slope m 0 = 1 2 . Since the line we want is perpendicular to it, it must have slope m = − 2 (since for perpendicular lines we must have m · m 0 = − 1). It passes through (1 , − 1), so using the point slope formula y − y 0 = m ( x − x 0 ) we get y − ( − 1) = − 2( x − 1) ⇒ y + 1 = − 2 x + 2 ⇒ 2 x + y − 1 = 0 .

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    1.2 Preliminaries 13 33. The given line, passing through ( − 2 , 1) and (1 , − 2), has slope m = y 2 − y 1 x 2 − x 1 = ( − 2) − 1 1 − ( − 2) = − 3 3 = − 1 . Therefore the slope of the perpendicular line is m ⊥ = − 1 m = 1. It passes through (5 , − 1), so using the point slope formula y − y 0 = m ( x − x 0 ) we get y − ( − 1) = 1( x − 5) ⇒ y + 1 = x − 5 ⇒ x − y − 6 = 0 . 34. The given line, passing through ( − 2 , 0) and (1 , 1), has slope m = y 2 − y 1 x 2 − x 1 = 1 − 0 1 − ( − 2) = 1 3 . Therefore the slope of the perpendicular line is m ⊥ = − 1 m = − 3. It passes through (4 , − 1), so using the point slope formula y − y 0 = m ( x − x 0 ) we get y − ( − 1) = − 3( x − 4) ⇒ y + 1 = − 3 x + 12 ⇒ 3 x + y − 11 = 0 . 35. The line we want is horizontal and passes through (1 , 3). Since all horizontal lines are of the form y = k , here we must have y = 3, or y − 3 = 0. 36. The line we want is horizontal and passes through (1 , 5). Since all horizontal lines are of the form y = k , here we must have y = 5, or y − 5 = 0. 37. The line we want is vertical and passes through ( − 2 , 3). Since all vertical lines are of the form x = k , here we must have x = − 2, or x + 2 = 0. 38. The line we want is vertical and passes through (3 , 1). Since all vertical lines are of the form x = k , here we must have x = 3, or x − 3 = 0. 39. A line perpendicular to a horizontal line is a vertical line. Since all vertical lines are of the form x = k , and the line passes through (1 , − 3), we must have x = 1, or x − 1 = 0. 40. A line perpendicular to a horizontal line is a vertical line. Since all vertical lines are of the form x = k , and the line passes through (1 , 3), we must have x = 1, or x − 1 = 0. 41. A line perpendicular to a vertical line is a horizontal line. Since all horizontal lines are of the form y = k , and the line passes through (7 , 3), we must have y = 3, or y − 3 = 0. 42. A line perpendicular to a vertical line is a horizontal line. Since all horizontal lines are of the form y = k , and the line passes through ( − 2 , 5), we must have y = 5, or y − 5 = 0. 43. (a) We convert each of these into centimeters, remembering that the formula y = 30 . 5 x requires that x be measured in feet: (i) y = 30 . 5 · 6 ft = 183 cm (ii) y = 30 . 5 · 4 12 ft ≈ 30 . 5 · 0 . 1667 ft ≈ 10 . 2 cm (iii) 1 ft , 7 in = 1 + 7 12 ft ≈ 1 . 583 ft, so y ≈ 30 . 5 · 1 . 583 ft ≈ 48 . 3 cm. (iv) y = 30 . 5 · 20 . 5 12 ft ≈ 30 . 5 · 1 . 708 ft ≈ 52 . 1 cm (b) Divide both sides of y = 30 . 5 x by 30 . 5, giving y 30 . 5 = x , or x = y 30 . 5 . Since x is measured in feet and y in centimeters, this gives a way to convert from centimeters to feet.

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    14 Preview and Review (c) (i) x = 195 30 . 5 ≈ 6 . 39 ft (ii) x = 12 30 . 5 ≈ 0 . 39 ft (iii) x = 48 30 . 5 ≈ 1 . 57 ft 44. (a) Since 1 pound equals 2 . 20 kg, let y = 2 . 20 x , where x is kilograms and y is pounds. (b) (i) y = 2 . 20 x , so 63 = 2 . 20 x , or x ≈ 28 . 6 kg. (ii) y = 2 . 20 x , so 5 = 2 . 20 x , or x ≈ 2 . 27 kg. (iii) y = 2 . 20 · 2 . 5 = 5 . 5 pounds. (iv) y = 2 . 20 · 76 ≈ 167 pounds. 45. Distance = rate · time; here time = 15 min = 1 4 h = 0 . 25 h, and the distance is 10 mi. The constant of proportionality is miles per hour or “mph”. Using the given values gives 10 mi = speed · 0 . 25 h ⇒ speed = 10 0 . 25 = 40 mph . The speed (constant of proportionality) is 40 mph. 46. Let y be the number of seeds produced, and x be the biomass. Since y is proportional to x , we have y = mx where m is the constant of proportionality. Then 13 = m · 213, so that m = 13 213 ≈ 0 . 061. 47. 1 ft = 0 . 305 m, so 3 . 279 ft = 1 m. Then 1 m 2 = (1 m)(1 m) = (3 . 279 ft)(3 . 279 ft) ≈ 10 . 75 ft 2 . 48. 1 ha = 10 000 m 2 , and 1 acre = 4046 . 86 m 2 . Therefore the area ratio of a hectare to an acre (that is, the number of acres in a hectare) is 1 ha 1 acre = 10 000 m 2 4046 . 86 m 2 ≈ 2 . 471 acre . 49. (a) Since the relationship is linear, if we let y be the number of liters and x the number of ounces, we have x = my where m = 33 . 81. So the relationship is x = 33 . 81 y , or y ≈ 0 . 0296 x . (b) With x = 12 ounces, we get y = 0 . 0296 · 12 ≈ 0 . 355 l. 50. (a) Since the relationship is linear, if we let y be the number of kilometers and x the number of miles, we have y km = mx mi where m = 1 . 609. So the relationship is y = 1 . 609 x . (b) We have 434 = 1 . 609 x , so that x = 434 1 . 609 ≈ 270 mi. 51. (a) Since one cup of flour weighs 120 g, it follows that 2 . 5 cups of flour weighs 2 . 5 · 120 = 300 g. (b) Since one cup of flour weighs 120 g, one gram of flour is 1 120 cups. Thus 225 g of flour is 225 · 1 120 = 1 . 875 cups. (c) Let y be the number of cups and x the weight in grams. Then from part (b), y = 1 120 x . 52. (a) Let x be the temperature in ◦ C and let y be the temperature in ° F. Then we have the points (0 , 32) and (100 , 212). First find the slope between these two pts: m = 212 − 32 100 − 0 = 180 100 = 18 10 = 9 5 The y -intercept is 32. A linear equation that relates Celsius and Fahrenheit then is y = 9 5 x +32, or F = 9 5 C + 32

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