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?

Question

drummergirl3 · Answer

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 · Answer

got it??

anonymous · Answer

I do not understand why you multiplied S(0) by S(t) in that first step.  Shouldn't you divide it, not multiply?

campbell_st · Answer

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 · Answer

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