Question

A population of bacteria is changing at the rate of dP/dt = 2000/(1+0.2t), where t is the time in days. The initial population when t=0 is 1000. Determine the population after 10 days.

Answer 1

1000*log(1+0.2t), then put 10 into the new function.

Answer 2

I get the answer 1439
dP=2000/(1+0.2t) dt
integrate the right hand term from 0 to 10
left hand term, integrate from 1000 to P
you can get the answer

Answer 3

Note that we can integrate this to receive a non-linear expression.

A hint:
\[
\int\frac{dP}{dt}dt=2000\int\frac{1}{1+.2t}dt
\]Which turns to:
\[
\int dP=2000\int\frac{1}{1+.2t}dt
\]Using u-substitution for \(t\), you'll receive the function, which you can use to evaluate what you're given.

Answer 4

@LolWolf -- I never really understood the whole u-substitution thing; it just makes things more confusing, lol. Though it probably is supposed to do the opposite.

Answer 5

use Woframalpha .com. your question is a integral question.

Answer 6

All right, I'll explain what u-sub is, it's actually pretty intuitive, and kind of nice.

Note the chain rule in differentiation:
\[
\frac{d}{dx}f(g(x))=f'(g(x))g'(x)
\]Now, integrate both sides of the expression, where we receive:
\[
f(g(x))=\int f'(g(x))g'(x)\;dx
\]But we also know, by definition:
\[
f(g(x))=\int f'(g(x))\;dg
\]So
\[
f(g(x))=\int f'(g(x))\frac{dg}{dx}\;dx=\int f'(g(x))\;dg
\]So this actually turns into:
\[
\int f'(g(x))\frac{dg}{dx}\;dx=\int f'(g(x))\;dg=f(g(x))
\]Right? So, u-substitution is doing exactly that, we're changing the variable of integration based on another function.

Answer 7

Oh, all right. So, let me see if I follow this:\[\int\limits_{}^{}\frac{ 1 }{ 1+.2t }dt\]f(t) = \[1*(1+.2t)^-1\]
So, does that mean u would be the demonimator of the fraction? Sorry, bear with me if I'm wrong, trying to u nderstand.

Answer 8

Yep, that works. So, we now have:
\[
u=1+.2t
\]So then:
\[
du=\;?
\]

Answer 9

du = \[0+0.2 = 0.2\]

Answer 10

Don't forget your dt. du=0.2dt But yes. you are correct.

Answer 11

So, back to your original integral. You need to solve for dt. So, dt=du/0.2 =5du. So, your integral changes to...

Answer 12

\[\int\limits_{}^{}\frac{ 5du }{ u}\]

Answer 13

Oh, right since dt was solved for; this is still making little sense, but I'm trying to understand, so. Would you have to plug the values back in for du and u?

Answer 14

Well, the nice thing about using u-sub, is that this is now a very easy function to integrate. It integrates into 5* ln |u|. Now you can go back and replace what u is equal to.

Answer 15

OF course don't forget your constant of integration so that you can evaluate your initial condition.

Answer 16

so \[\int\limits_{}^{}\frac{ 5(0.2dt) }{ 5*\ln \left| 1+0.2t \right|}\]

Answer 17

Oh, I wouldn't need dt in the numerator. Would just be 5(0.2), right?

Answer 18

Well, not quite, the integral(anti-derivative) has been found, so we would remove the integral sign, as well as dropping the entire numerator. Your antiderivat of the entire u-sub is going to be 5* ln |1+0.2t| + C.

Answer 19

Now, you have a 2000 sitting out in front of that integral, so you will need to multiply the result by 2000, giving your 10000*ln |1+0.2t| + C. Now, you must find C when t=0, then you will have your function for any time t.

Answer 20

So, P(t)= 10000*ln|1+0.2t| + C, and you know P(0)=1000, so plug that information into your function and you should be able to determine C.

Answer 21

Questions, confusion?

Answer 22

Sorry, still working it out over here.

Answer 23

ok.

Answer 24

I got -9000.

Answer 25

Oh, wait, wrong... lol.

Answer 26

C=1000

Answer 27

good

Answer 28

Now, combine that with the first part and you have ur function for any time, t. So, all that is left is to find P(10).

Answer 29

So, P(t) = \[10000*\ln \left| 1+0.2t \right|+1000\]
P(10) = 10000*ln|1+0.2(10)|+1000

Answer 30

perfect! what is that value?

Answer 31

So P = 110861?
:D

Answer 32

hmmm...one too many decimals in that answer. Try again.

Answer 33

11986.1

Answer 34

There we go!

Answer 35

Wow! Thank you so much. I appreciate all this help with math problems that I have little idea how to do. I would give you 1 million medals if I could.

Answer 36

lol...nah, it's my job. For some reason I like to do it at work and at home.

Answer 37

Good luck with the rest of your work. I will be on for a bit longer if you get stuck on another one.