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Unformatted text preview: MATH 138 Calculus 2, Solutions to Assignment 7 1: The amount A ( t ) of a radioactive substance satisfies the DE A ( t ) = k A ( t ) for some constant k < 0. The substance has a halflife of 10 seconds, which means that A (10) = 1 2 A (0). If A (5) = 100 then find the exact time t at which A ( t ) = 20. Solution: This DE is linear, since we can write it in the form A + kA = 0. An integrating factor is λ = e R k dt = e k t and the general solution is A ( t ) = e k t Z dt = c e kt . Note that A (0) = c , so c is the initial amount. Since the halflife is 10, we have A (10) = 1 2 c = ⇒ c e 10 k = 1 2 c = ⇒ e 10 k = 1 2 = ⇒ e 10 k = 2 = ⇒ 10 k = ln2 = ⇒ k = 1 10 ln2 . and so A ( t ) = c e ( t/ 10) ln 2 = c 2 t/ 10 . Also, we have A (5) = 100 = ⇒ c 2 1 / 2 = 100 = ⇒ c = 100 √ 2 , and so A ( t ) = ( 100 √ 2 ) 2 t/ 10 . Finally, we have A ( t ) = 20 ⇐⇒ ( 100 √ 2 ) 2 t/ 10 = 20 ⇐⇒ 2 t/ 10 = 100 √ 2 20 = 5 √ 2 = √ 50 ⇐⇒ t 10 = log 2 √ 50 = 1 2 log 2 50 ⇐⇒ t = 5log 2 50 . 2: A pot of boiling water is removed from the heat and placed on a table in a room. The temperature T ( t ) of the water at time t satisfies Newton’s Law of Cooling , that is T ( t ) = k ( C T ( t ) ) for some constant k > 0, where C is the room temperature. After 2 minutes, the water has cooled from 100 ◦ to 84 ◦ . After another 2 minutes, it has cooled to 72 ◦ . What is the temperature in the room? Solution: This DE is linear since we can write it in the form T + kT = kC . An integrating factor is λ = e R k dt = e kt , and the general solution is T ( t ) = e kt Z kC e kt = e kt ( Ce kt + a ) = C + ae kt . Since T (0) = 100 we have C +...
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This note was uploaded on 09/18/2008 for the course MATH MAth 138 taught by Professor Marshman during the Spring '08 term at Waterloo.
 Spring '08
 MARSHMAN
 Calculus

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