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anonymous
 one year ago
Consider an organism growing according to S(t)=S(0)e^at. Suppose a=0.001/s, and S(0)= 1 mm. At time 1000 s, S(t)=2.71828 mm. How close must t be to 1000s to guarantee a size within 0.1 mm of 2.71828mm?
anonymous
 one year ago
Consider an organism growing according to S(t)=S(0)e^at. Suppose a=0.001/s, and S(0)= 1 mm. At time 1000 s, S(t)=2.71828 mm. How close must t be to 1000s to guarantee a size within 0.1 mm of 2.71828mm?

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DrummerGirl3
 one year ago
Best ResponseYou've already chosen the best response.1S(t)=S0eat You're looking to find the values of t such that S(t)=2.71828±0.1 S0eat=S(t) eat=S(t)S0 Take natural log of both sides atlne=lnS(t)S0 lne=1, divide both sides by a and evaluate t=1alnS(t)S0=10.001s−1ln2.71828mm−0.1mm1mm=ln(2.61828)∗1000s= For the second value, use 2.81828 instead of 2.61828.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I do not understand why you multiplied S(0) by S(t) in that first step. Shouldn't you divide it, not multiply?

campbell_st
 one year ago
Best ResponseYou've already chosen the best response.1well using the information your initial model for the growth is \[S(t) = 1 \times e^{0.001 \times t}\] you are also told when t = 1000 the organism is 2.71828 mm long. so you are being asked to find the time when the organism is 2.61828 long and also find t when the organism is 2.81828 mm long. these 2 values represent 0.1 mm above and below the length when t = 1000. so you need to solve 2 equations for t Equation 1 \[2.61828 = 1 \times e^{0.001t}\] equation 2 \[2.81828 = 1 \times e^{0.001t}\] so to solve the equations take the natural log of both sides then divide than answers by 0.001 hope it helps

campbell_st
 one year ago
Best ResponseYou've already chosen the best response.1the 2 equations can be simplified to \[2.61828 = e^{0.001t}\] and \[2.81828 = e^{0.001t}\] so take the natural log of both sides and then solve for t
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