Chapter 10

# Chapter 10 - 10.1 SOLUTIONS 429 CHAPTER TEN Solutions for...

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Unformatted text preview: 10.1 SOLUTIONS 429 CHAPTER TEN Solutions for Section 10.1 1. The rate of change of P is proportional to P so we have dP dt = kP, for some constant k . Since the population P is increasing, the derivative dP/dt must be positive. Therefore, k is positive. 2. (a) (III) An island can only sustain the population up to a certain size. The population will grow until it reaches this limiting value. (b) (V) The ingot will get hot and then cool off, so the temperature will increase and then decrease. (c) (I) The speed of the car is constant, and then decreases linearly when the breaks are applied uniformly. (d) (II) Carbon-14 decays exponentially. (e) (IV) Tree pollen is seasonal, and therefore cyclical. 3. The rate at which the balance is changing is 5% times the current balance, so we have Rate of change of B = 0 . 05 · Current balance so we have dB dt = 0 . 05 B. 4. The rate of change of Q is proportional to Q so we have dQ dt = kQ, for some constant k . Since the radioactive substance is decaying, the quantity present, Q , is decreasing. The derivative dQ/dt must be negative, so the constant of proportionality k is negative. 5. (a) = (III), (b) = (IV), (c) = (I), (d) = (II). 6. (a) The amount of caffeine, A , is decreasing at a rate of 17% times A , so we have dA dt =- . 17 A. The negative sign indicates that the amount of caffeine is decreasing at a rate of 17% times A . Notice that the initial amount of caffeine, 100 mg, is not used in the differential equation. The differential equation tells us only how things are changing. (b) At the start of the first hour, we have A = 100 . Substituting this into the differential equation, we have dA dt =- . 17 A =- . 17(100) =- 17 mg/hour . We estimate that the amount of caffeine decreases by about (17 mg/hr ) · (1 hr ) = 17 mg during the first hour. This is only an estimate, however, since the derivative dA/dt will not stay constant at- 17 throughout the entire first hour. 7. The amount of alcohol, A , is decreasing at a constant rate of 1 ounce per hour, so we have dA dt =- 1 . The negative sign indicates that the amount of alcohol is decreasing. 8. The balance in the account, B , is increasing at a rate of 4% times B and is decreasing at a rate of 2000 dollars per year. We have Rate of change of B = Rate in- Rate out. dB dt = 0 . 04 B- 2000 . Notice that the initial amount of \$ 25 , 000 in the account is not used in the differential equation. The differential equation tells us only how things are changing. 430 Chapter Ten /SOLUTIONS 9. The amount of morphine, M , is increasing at a rate of 2 . 5 mg/hour and is decreasing at a rate of . 347 times M . We have Rate of change of M = Rate in- Rate out. dM dt = 2 . 5- . 347 M....
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## This note was uploaded on 04/21/2008 for the course MATH 227 taught by Professor ? during the Winter '08 term at University of Cincinnati.

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Chapter 10 - 10.1 SOLUTIONS 429 CHAPTER TEN Solutions for...

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