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Unformatted text preview: Answer Guide 5 Math 408C: Unique Numbers 56975, 56980, and 56985 Tuesday, September 29, 2009 Homework problems Section 3.7 16. (a) Using V ( r ) = 4 3 &r 3 , we compute the average rates of change V (8) & V (5) 8 & 5 = 4 9 & (512 & 125) = 172 & ¡ m 3 / ¡ m, V (6) & V (5) 6 & 5 = 4 3 & (216 & 125) = 364 3 & ¡ m 3 / ¡ m, V (5 : 1) & V (5) 5 : 1 & 5 = 40 3 & & (5 : 1) 3 & 125 ¡ ¡ 102 : 01 & ¡ m 3 / ¡ m. (b) Because V ( r ) = 4 &r 2 , we have V (5) = 100 & ¡ m 3 / ¡ m. (c) The change & V in volume of the round cell is approximately equal to the change & r in radius times the surface area of the inner sphere. This means that & V ¡ & r (4 &r 2 ) . In other words, & V & r ¡ 4 &r 2 = S ( r ) = V ( r ) : 32. Note: P ( t ) = r (1 & P ( t ) =P c ) P ( t ) & ¢P ( t ) is an example of a &di/erential equation¡. These are used extensively in science and engineering to model natural processes. This equation describes how a population changes in time. Solving the di/erential equation means ¢nding a function P ( t ) with a given initial value P (0) . You will study di/erential equations if you go on to take M427K. (a) The population is stable if and only if dP dt = 0 . (b) Observe that 0 = dP dt = r ¢ 1 & P P c £ P & ¢P 1 has the solution P = 0 and the solution P = P c & 1 & & r ¡ : Plugging in P c = 10 ; 000 , r = 0 : 05 , and & = 0 : 04 , we get P = 2000 . Notice that this is smaller than the equilibrium population P c = 10 ; 000 one would have if there were no &shing. (c) Plugging in r = & , we get P = 0 . So this rate of &shing drives the population to extinction. Section 3.8 12. Let t denote the time in minutes. Let x ( t ) denote the diameter, in centimeters. And let S ( x ( t )) denote the surface area in cm 2 . We are given that & 1 cm 2 /s = dS dt = dS dx dx dt : Because S ( x ) = 4 ¡ ¢ x 2 £ 2 = ¡x 2 , we know that dS dx = 2 ¡x . Therefore, we have dx dt = & 1 dS dx = & 1 2 ¡x ; which says that the rate of decrease of diameter is 1 2 &x . Plugging in x = 10 , we get 1 20 & ....
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