Thread: Find the rate at which the population is growing

A population of 3000 bacteria is introduced into a culture and grows in number according to the formula
P(t)= 3000(1 + 3t/100+ t^2),

where t is measured in hours. Find the rate at which the population is growing when t=1

Well, the rate of which it's growing will be told by the derivative of the population equation.

so take the derivative of P(t):
P(t) simplified is: 3000+60t+3000t^2

derivative of P(t)=90+6000t

then plug in 1 for t

90+12000=12090

so when t=1 the population is growing at a rate of 62 bacteria per hour

(plug into calculator to check)

Originally Posted by woohoo

Well, the rate of which it's growing will be told by the derivative of the population equation.

so take the derivative of P(t):
P(t) simplified is: 3000+60t+3000t^2

derivative of P(t)=90+6000t

then plug in 1 for t

90+12000=12090

so when t=1 the population is growing at a rate of 62 bacteria per hour

(plug into calculator to check)

It should be some kinda decimal...

Originally Posted by beachbunny619

It should be some kinda decimal...

humm...

I'm not quite sure then.

Sorry about that!

Anyone else?
Explain for the both of us?

Originally Posted by beachbunny619

A population of 3000 bacteria is introduced into a culture and grows in number according to the formula
P(t)= 3000(1 + 3t/100+ t^2),

where t is measured in hours. Find the rate at which the population is growing when t=1

i'm pretty sure to find the rate when t = 1 you take the derivative of the original function

so P'(t)=3000(3/100 + 2t)
and plug in t = 1 and get 6090 bacteria/hr

i think thats the answer, i was struggling with this today and someone showed me how to do it. so i just learned it...

Originally Posted by beachbunny619

It should be some kinda decimal...

well thats what i thought too but if its the instant growth rate, not the relative growth rate, then its usually a # larger than the answer to the non-derivative function