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exp growth_decay

# Solution use the carbon-14 rate constant k = ln2 5700

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Unformatted text preview: Solution: Use the carbon-14 rate constant k =- ln2 5700 to find t when y ( t ) = . 3 y . y ( t ) = y e kt . 3 y = y e kt . 3 = e kt ln0 . 3 = ln e kt = kt t = ln0 . 3 k =- 5700ln0 . 3 ln2 ≈ 9901 years Example 3 Newton’s Law of Cooling. Find time t given T ,T S and temperature T at time t . A hot bowl of soup cools from 70 ◦ C to 60 ◦ C after 10 minutes inside a re- frigerator that measures- 5 ◦ C . How many more minutes will it take for the soup to reach 40 ◦ C ? Solution: First find rate constant k given T S =- 5 , T = 70 , and T (10) = 60 . T- T S = ( T- T S ) e kt T (10)- (- 5) = (70- (- 5)) e k (10) 60 + 5 = 75 e 10 k 65 = 75 e 10 k 65 75 = e 10 k ln 65 75 = ln e 10 k = 10 k k = ln(65 / 75) 10 ≈ - . 0143 . Then find time t when T = 40 and T = 60 . T- T S = ( T- T S ) e kt 40 + 5 = (60- (- 5)) e kt 45 65 = e kt ln 45 65 = kt t = ln(45 / 65) k = 10ln(45 / 65) ln(65 / 75) ≈ 25 . 7 min Example 4 Continuously Compounded Interest. Find time t given interest rate r and an amount relative to A . How long will it take an investment to increase in value by 50% if the interest rate is 4% compounded continuously? Solution: We wish to find t when A ( t ) = 1 . 5 A given interest rate r = 0 . 04 . A ( t ) = A e rt 1 . 5 A = A e . 04 t 1 . 5 = e . 04 t ln1 . 5 = 0 . 04 t t = ln1 . 5 . 04 ≈ 10 . 14 years S. Chang CU-Boulder Dept of Applied Mathematics...
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Solution Use the carbon-14 rate constant k = ln2 5700 to...

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