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Population growth

  • Thread starter ProBasket
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  • #1
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A bacteria culture starts with 300 bacteria and grows at a rate proportional to its size. After 3 hours, there are 9000 bacteria

A.) Find an expression for the number of bacteria after t hours. for this part, i got [tex]300*e^{1.133732*t}[/tex]

B.) Find the number of bacteria after 4 hours. well using the expression from above and subbing in 4, i get 27965.04104

C.) Find the growth rate after 4 hours. this just means that i need to solve for k right?
p(t) = 300e^{kt}
p(3) = 300e^{3k} = 9000
solved for k and got 1.133% right?

D.) After how many hours will the population reach 30000
well setting the equation from part A equal to 30000 and solved for t and got 4.319991


i know that at least one of these are wrong, but i cant figure out which one. can someone tell me what im doing wrong?
 

Answers and Replies

  • #2
375
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I think Part C is wrong. Isn't the growth rate = dp/dt = kp? (k is constant, by the way, and is equal to 1.133732). So use your answer from part B to get the answer to part C. (Simple as multiplying by k).
 
  • #3
140
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awesome, thanks for the help
 
  • #4
xanthym
Science Advisor
410
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ProBasket said:
A bacteria culture starts with 300 bacteria and grows at a rate proportional to its size. After 3 hours, there are 9000 bacteria

A.) Find an expression for the number of bacteria after t hours. for this part, i got [tex]300*e^{1.133732*t}[/tex]

B.) Find the number of bacteria after 4 hours. well using the expression from above and subbing in 4, i get 27965.04104

C.) Find the growth rate after 4 hours. this just means that i need to solve for k right?
p(t) = 300e^{kt}
p(3) = 300e^{3k} = 9000
solved for k and got 1.133% right?

D.) After how many hours will the population reach 30000
well setting the equation from part A equal to 30000 and solved for t and got 4.319991


i know that at least one of these are wrong, but i cant figure out which one. can someone tell me what im doing wrong?
From problem statement:
{(dB/dt) = k*B} ⇒ B(t) = B0*exp(k*t)
{B(t=0) = 300} ⇒ B0 = 300
{B(t=3) = 9000} ⇒ 9000 = 300*exp{k*(3)} ⇒ k = (1/3)*Loge{9000/300} = (1.1337325)

ITEM #A:
B(t) = 300*exp{(1.1337325)*t}

ITEM #B:
B(4) = 300*exp{(1.1337325)*(4)} = (27,965 bacteria)

ITEM #C:
(dB/dt)t=4 = k*B(4) = (1.1337325)*(27965) = (31,705 bacteria/hr)

ITEM #D:
(30000) = 300*exp{(1.1337325)*t} ⇒ t = (1.1337325)(-1)*Loge{30000/300}
t = (4.06195 hr)


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