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Diff EQs 6.6 Common Problems: Exponential Growth and Decay Compound Interest Radiation and half-life Newton’s law of cooling Other fun topics Suppose we are interested in a quantity that increases at a rate proportional to the amount present. If we also know the amount at a certain time, we can find y as a function of time by solving the differential equation: dy ky dx dy ky dx y ekteC dy kdx y Ce kt y When t = o, y = dy y kdx y0. Solving for C, C = y0 ln y kt C y y0 e kt y ektC Initial Amount The rate of change of a population of rabbits is proportional to the number of rabbits present at any given time. If 10 rabbits are present initially, and 195 rabbits are present in 6 months, how many rabbits will there be in 2 years? dr kr r (0) 10 r (24) ? dt r (6) 195 r Ce kt r 10ekt r 10e.4951t 195 10e k ( 6) r (24 ) 10 e.4951( 24) 19.5 e k ( 6) r (24) 1,446,974 ln 19.5 6k k .4951 Suppose you deposit $800 in an account that pays 6.3% annual interest. How much will you have 8 years later if interest is: A. compounded continuously B. compounded quarterly kt r dA .063 A A 800e .063(8) A P1 dt k dA A $1324.26 .063dt ( 4 )(8) A .063 A 8001 dA 4 A .063dt ln A .063t C A $1319.07 A e.063t C A Ce.063t The rate at which the population of a group of organisms grows is directly proportional to the number of organisms in the population. If the population at time zero is 3,500 and the population after one year is 5,250, what will the population be after 3 years? dy 3 ky y 3500 e t ln 2 dt 3 y y0 e kt y (3) 3500 e ( 3) ln 2 5250 3500ek (1) y(3) 11,812.5 3 ln k 2 Half-Life After t years, half of the original amount y y0 e kt is left 1 y0 y0 e kt 2 1 e kt 2 1 ln kt 2 ln 1 ln 2 k t t is the half-life. ln 2 k K does not t depend on the initial amount! Example ½ life of carbon-14 is 5700 years. Find the age of a sample in which 10% of the original radioactivity is depleted. ln 2 ln 2 90% is k still t 5700 ln 2 present t .9 y 0 y 0 e 5700 ln 2 t .9 e 5700 ln 2 ln .9 t 5700 866 t Newton’s Law of Cooling k T Ts dT dt Rate of change is Proportional to Difference between temp of object’s temp of surroundings and temp wrt time of object When a murder is committed, the body, originally at 37C, cools according to Newton’s Law of Cooling. Suppose on the day of a very difficult calc quiz, Mr. Hopkins’ body is found at 2pm with a body temperature of 35C in a room with a constant temperature of 20C. 2 hours later the body temperature is 30C. Being the bright students you are, you should be able to tell Mr. Neuhard when the heinous crime was committed. k T Ts dT T 20 15e kt 37 20 15e( .203)t dt 30 20 15ek ( 2) dT 17 15e( .203)t T Ts kdt 10 15e k ( 2) 17 2 ln .203t ln 2k ln T Ts kt C 3 15 T Ts Ce kt k .203 .62 t T 20 Ce T 20 15e( .203)t min .62 60 37 min kt hr 35 20 Ce k ( 0) The crime was 15 C committed at 1:23 pm

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posted: | 2/23/2012 |

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