solns8 - Math 1A UCB Spring 2010 A Ogus Solutions1 for...

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Math 1A — UCB, Spring 2010 — A. Ogus Solutions 1 for Problem Set 8 § 3.7 #18. If a tank holds 5000 gallons of water which drains from the bottom of the tank in 40 minutes, the volume remaining in the tank after t minutes is V = 5000(1 - t 40 ) 2 . Solution. The rate at which the water is draining is V 0 ( t ) = 2 · 5000(1 - t 40 )( - 1 40 ) = - 250(1 - t 40 ). (a) V 0 (5) = - 250(1 - 1 8 ) = - 7 8 250. (b) V 0 (10) = - 250(1 - 1 4 ) = - 3 4 250. (c) V 0 (20) = - 250(1 - 1 2 ) = - 1 2 250 (d) V 0 (40) = - 250(1 - 40 40 ) = 0 The water is coming out the fastest, when | V 0 ( t ) | is largest, which is when t 40 is smallest. Thus we conclude that the water is coming out fastest at the beginning. The more time goes by the slower the water comes out, until it stops at time 40. § 3.7 #28. The frequency of vibrations of a vibrating violin string is given by f = 1 2 L q T ρ . Solution. (a) df dL = - 1 2 L - 2 q T ρ . df dT = 1 2 L 1 2 q T ρ 1 ρ df = - 1 2 L 1 2 q T ρ T ρ 2 (b) The derivative df dL is negative, hence the graph f ( L ) is decreasing. If we decrease the length of the string, the frequency and hence the pitch is higher. The derivative df dT is positive, since L, ρ are positive. Hence the graph f ( T ) is increasing. Thus, if the tension is increased, then also the pitch increases. The derivative df is negative, hence the graph f ( ρ ) is decreasing. Thus is the linear density is increased, the pitch decreases. § 3.7 #31. If p ( x ) is the total value of the production when there are x workers in a plant, then the average productivity of the workforce at the plant is A ( x ) = p ( x ) x . Solution. A 0 ( x ) = xp 0 ( x ) - p ( x ) x 2 . A 0 ( x ) > 0 iff xp 0 ( x ) - p ( x ) x 2 > 0 iff xp 0 ( x ) - p ( x ) > 0 iff p 0 ( x ) > p ( x ) x . § 3.8 # 3(a). A bacteria culture initially contains 100 cells and grows at a rate proportional to its size. After an hour the population has increased to 420. Find an expression for the number of bacteria after t hours. Solution. The problem tells us that if P is the population, then dP/dt = kP , where k is some fixed number. This is a differential equation with solution P ( t ) = P 0 e kt , where P 0 = 100, the initial population. Since we know 420 = P (1) = 100 e k (1) , we see that k = ln(420 / 100), and so our answer is P ( t ) = 100 e ln(420 / 100) t = 100(4 . 2) t .
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