Solutions Homework 2

Solutions Homework 2 - MATH 106 CALCULUS I FOR BIO& SOC...

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Unformatted text preview: MATH 106 CALCULUS I FOR BIO. & SOC. SCI. FALL 2011 INSTRUCTOR: JOS ´ E MANUEL G ´ OMEZ Solutions Homework 2. (1) (5 Points) A certain strain of bacteria that reproduces asexually triples its size every 45 days. If after 180 days there are 1620 bacteria, how many bacteria were there originally? Solution: Let B ( t ) denote the number of bacteria after t days. Bacteria reproducing asexually is an example of exponential growth and thus we have B ( t ) = B a t , where B is the initial number of bacteria. We would like to determine B . We know that the number of bacteria triples every 45 days. Therefore B (45) = 3 B . From here we get 3 B = B a 45 . Thus a 45 = 3; that is, a = (3) 1 / 45 . On the other hand, since there are 1620 bacteria after 180 days, we have B (180) = 1620. Thus 1620 = B (180) = B a 180 = B (3) 180 / 45 = B 3 4 . This shows that B = 1620 3 4 = 1620 81 = 20 . We conclude that there were 20 bacteria originally. (2) (5 Points) A picture supposedly painted by Vermeer (1632-1675) contains 99.5% of its carbon–14. It is known that carbon-14 has a half life of 5730 years. From this information, can you decide whether or not the picture is a fake? Please explain yourinformation, can you decide whether or not the picture is a fake?...
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This note was uploaded on 01/20/2012 for the course CALC AS.110.106 taught by Professor Josegomez during the Fall '11 term at Johns Hopkins.

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Solutions Homework 2 - MATH 106 CALCULUS I FOR BIO& SOC...

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