Bacteria growth rate question

Question

This stupid calculus problems hate my guts and the growth rate one is giving me trouble here is the question

 

A bacteria culture grows at a rate proportional to the number 'y' present at any given time 't' . The bacteria count 'y' was 100 after 3 hours and 280 after 7 hours

 

A) what is the rate of growth?

B)what was the initial amount of bacteria?

C) How much of the bacteria was present after 12 hours?

Answer 1

This is an example of a separable problem:

 

y(t) = population at time t

 

dy/dt = rate of change of population

 

It is given that the rate of change is proportion to the amount of bacteria present, so

 

dy/dt = kt where k is a constant

 

Separate the x and t variables;

 

dy/y = k dt

 

Integrate both sides and get:

 

ln(y) = kt + c where c = another constant

 

Take e^() on both sides and get:

 

y(t) = e^(kt) * e^c

 

Let a = e^c 

 

Then

 

y(t) = a * e^(kt)

 

Now use the two given values to solve for the constants and k and a:

 

y(3) =100

y(7) = 280

 

100 = a * e^(3k)

 

280 = a * e^(7k)

 

Take ratio of bottom equation to top equation to eliminate a and get k:

 

280/100 = e^[(7-3)k] = e^(4k)

 

2.8 = e^(4k)

 

k = ln(2.8) / 4 = 0.2574

 

So far we have y(t) = a * e^(0.2574 t)

 

Now use either of the given values to find a:

 

y(3) = 100

 

100 = a * e^[(0.2574)(3)] = 2.1645 a

 

a = 100 / 2.1645 = 46.2

 

We could have also used y(7) = 280

 

280 = a * e^[(0.2574)(7)] = 6.0605 a, which also gives a = 46.2

 

Put it all together and get:

 

y(t) = 46.2 * e^(0.2574 t) with t in hours

 

Finally, we can answer the 3 questions:

 

(A) Rate of growth:

 

dy/dt = 46.2 * 0.2574 * e^(0.2574 t) = 11.8919 e^(0.2574 t)

 

(B) Initial amount of bacteria:

 

y(0) = 46.2 * e^(0) = 46.2 (which would be rounded off to 46, since there are no fractional bacteria!)

 

(C) Amount after 12 hours:

 

y(12) = 46.2 * e^(0.2574 * 12) = 1014.1 (round off to 1014) 

 

 

 

Answer 2

Hang in there Kris...  ;-)

 

Do you know from your calculus that something that grows at a rate that is proportional

to the number present has a solution that is an exponential function?

 

If so, then you can start with the general form of an exponential equation:

 

y = a b^t       (other variations are  y = a (1+r)^t   or   y = a e^rt  )

 

The two data points they gave us mean that:

100 = a b³

280 = a b⁷

If we divide the second equation by the first equation, then on each side we have:

280/100 = (a b⁷)/(a b³)

 2.8 =  b⁴

This gets rid of the a and now we have that b = 4th-root of 2.8

Using a calculator:  b ~ 1.294

 

A) So the amount is multiplied by 1.294 every hour,

   or we'd say that the rate of growth is 29.4% per hour (this is the "r" in the (1+r) version).

 

Now we know that the equation is    y = a 1.294^t

If we can find "a" then we'll know everything...

Since y should be 100 when t=3 we have:

      100 = a 1.294³

     100 = 2.167 a

so  a = 100/2.167 = 46.15

and our equation is now fully known:

     y = 46.15 1.294^t

and we can find the number at any time:

 

B) At  t=0 we get:  y = 46.15 1.294⁰ = 46.15  .

 

C) At t = 12 we get:   y = 46.15 * 1.294¹² = 46.15 * 22.04 =  1017.1 .