anonymous
  • anonymous
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?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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DrummerGirl3
  • DrummerGirl3
S(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.
DrummerGirl3
  • DrummerGirl3
got it??
anonymous
  • anonymous
I do not understand why you multiplied S(0) by S(t) in that first step. Shouldn't you divide it, not multiply?

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campbell_st
  • campbell_st
well 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
  • campbell_st
the 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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