how do i set this up?! a bacteria culture initially contains 100 dells and grows at a rate proportional to its size. after an hour the population has incresed to 420.
a)find an expression for the number of bacteria after t hours.
b)find the number of bacteria after 3 hours
c)find the rate of growth after 3 hours
d)when will the population reach 10,000

Question

anonymous · Answer

i have trouble converting it to numbers. here what i have so far and i working on more. i use the P(t)=P(0)e^kt

anonymous · Answer

so P(0)=100 P(3)=420

anonymous · Answer

i set the equation P(3)=100e^3k=420 i solved and got k=ln42/3

anonymous · Answer

is a [100e^[(ln42/3)(t)]

anonymous · Answer

b) i just plug in 3 for t to the equation in a) ?

anonymous · Answer

for d) i put equation a) equal to 10,000 right?

anonymous · Answer

im stuck on c) though. how do you do that?

anonymous · Answer

I guess that you can derive the expression you got before to obtain dP/dt. And then you sustitute the value of 3 on it.
.

anonymous · Answer

i got it down to 100(42^t/3) i dont how to dervie?

anonymous · Answer

I think your initial conditions are wrong. You wrote P(3) = 420, and on the problem says P(1) = 420

anonymous · Answer

$$420 = 100e^{k(1)}$$

anonymous · Answer

so, $$k = \ln4.2$$

anonymous · Answer

ok crap .yeah your right! good catch

anonymous · Answer

$$P = 100e^{\ln(4.2)t}$$

anonymous · Answer

if you derive you get: $$\frac{d}{dt}100e^{(\ln 4.2)k}= 100(\ln4.2)e^{(\ln 4.2)k}$$

anonymous · Answer

sorry its t instead of k

anonymous · Answer

im trying to get waht you got for the derviative right now. its not coming out for me.

anonymous · Answer

what did you get?

anonymous · Answer

instead of ln4.2 i get 4.2k

anonymous · Answer

remember that $$\frac{d}{dx}e^{ax} = ae^x$$

anonymous · Answer

Sorry i missed an "a"
$$\frac{d}{dx}e^{ax}=ae^{ax}$$

anonymous · Answer

oh. ok. got it now. thanks

anonymous · Answer

You're welcome.