Question

A stable population P of mosquitoes (measured in thousands), is observed to suddenly undergo a period of acceleration, given by

http://i48.tinypic.com/10z4olv.jpg

Answer 1

where did the log come from ?

Answer 2

antiderivative of 1/x

Answer 3

i don't see a \(\frac{1}{x}\)

Answer 4

take the anti derivative get \(2t-2\sqrt{t-4}\)

Answer 5

frankly i have no idea what the constant would be in this case, but lets assume for the moment it is 0
then take the anti derivative again

Answer 6

i get \(t^2-\frac{4}{3}(t+4)^{\frac{3}{2}}+C\)

Answer 7

since the initial population was 1000 and since the function is in units of 1000 we can put \(f(0)=1=-\frac{4}{3}(4)^{\frac{3}{2}}+C\) and solve for \(C\)

Answer 8

i get \(C=1+\frac{32}{3}\) whatever that is

Answer 9

Nothing I try in relation to any of that is right

Answer 10

How long are you going to assume that the constant of the first anti derivative is 0?

Answer 11

Would "a stable population" imply that \(p'(0)=0\)?

Answer 12

i guess so

Answer 13

otherwise you cannot really do the problem

Answer 14

I was assuming it was some positive constant..

Answer 15

dang but you're right

Answer 16

okay so I'm going to see how that changes things

Answer 17

Umm, so we have \(p''(t) = 2-\frac{1}{\sqrt{t+4}}\)? For some reason my anti derivative is \(p'(t)=2t+2\sqrt{t+4}+C\).

Answer 18

yes that's what I just got

Answer 19

So if \(p'(0)=0\) then: \[0=2(0)+2\sqrt{(0)+4}+C\]

Answer 20

And I'm getting \(C = -4\)

Answer 21

Then \(p'(t) = 2t+2\sqrt{t+4}-4\) and we have to find the anti-derivative of that.

Answer 22

I'm getting \[ \large
p''(t) = t^2+4(t+4)^{3/2}-4t+C
\]

Answer 23

@byerskm2 What are you getting?

Answer 24

I got the same thing. I'm going to try it in the hw box

Answer 25

Still need to solve for that last \(C\)

Answer 26

But we don't have a value for it

Answer 27

We know \(p''(0)=1000\) so:\[\large
1000 = (0)^2+2((0)+4)^{3/2}-4(0)+C
\]

Answer 28

c=1000

Answer 29

Nope.

Answer 30

no, wait

Answer 31

984

Answer 32

Wait, I messed up a bit

Answer 33

is it 4/3 in front of the ()

Answer 34

it's not \(p''(t)\) it's actually just \(P(t)\).

Answer 35

And the 2 should have been a 4.

Answer 36

\[
\large
1000 = (0)^2+4((0)+4)^{3/2}-4(0)+C
\]

Answer 37

hmm 968

Answer 38

Okay so the end result:

Answer 39

\[
\large
P(t) = t^2+4(t+4)^{3/2}-4t+968
\]

Answer 40

But they might want a square root, I dunno. like \[
\large
p''(t) = t^2+4\sqrt{(t+4)^3}-4t+C
\]

Answer 41

yeah they didn't like th 3/2

Answer 42

Okay niether one of those worked this is getting ridiculous

Answer 43

*neither

Answer 44

Hmmm, hold on, they said the population is in thousands...

Answer 45

We didn't convert perhaps.

Answer 46

\[
\large
1= (0)^2+4((0)+4)^{3/2}-4(0)+C
\]

Answer 47

So maybe \(C=-31\)

Answer 48

\[\large
P(t) = t^2+4(t+4)^{3/2}-4t-31\]

Answer 49

It should be growing though so C should be positive

Answer 50

Why would \(C\) have to be positive? It's just a constant of integration. That \(t+4\) ensure the population is above \(0\) when \(t=0\).

Answer 51

I feel like we messed up somewhere near the beginning or something. this shouldn't be that hard

Answer 52

Well did you try it with the -31?

Answer 53

i did

Answer 54

:(

Answer 55

Okay, I'll try again. I think you didn't actually get the anti derivative because I probably made a mistake.

Answer 56

I'm starting over

Answer 57

Okay, so I did make a mistake...

Answer 58

\[
P'(t) = 2t-2\sqrt{t+4}+4
\]

Answer 59

I doubt we both made the exact same mistake...

Answer 60

I've got 4/15 this time wtf

Answer 61

oh youre talking about p'

Answer 62

ok I got 2/3

Answer 63

2/3?

Answer 64

sorry 2t+ 2/3(t+4)^3/2 +C

Answer 65

Nope... how are you getting that?

Answer 66

when I check with the derivative so that I get 1 out front

Answer 67

Okay so for \[
\int 2-\frac{1}{\sqrt{t+4}} dt
\]I first get this:\[
2t-\int\frac{1}{\sqrt{t+4}} dt
\]I do u substitution: \(u=t+4\) and \(du=dt\). \[
\int \frac{1}{\sqrt{u}}du = \int u^{-1/2}du = \frac{u^{1/2}}{1/2} = 2\sqrt{u} = 2\sqrt{t+4}
\]So at the very end we get \[
P'(t) = 2t-2\sqrt{t+4}+C
\]

Answer 68

\[0=2(0)-2\sqrt{(0)+4}+C=-2\sqrt{4}+C=-4+C\implies C=4
\]So\[
P'(t)=2t-2\sqrt{t+4}+4
\]

Answer 69

Then we find \(P(t)\).

Answer 70

@byerskm2 get it?

Answer 71

yeah I get the concept but it's still wrong

Answer 72

this is what we got last time

Answer 73

I think I'll just try this again tomorrow. Thanks for all of your work :) it helped to see the integration by parts :)

Answer 74

That was u substitution, not integration by parts.

Answer 75

oh damn that's what i meant

Answer 76

Lol it's been a very long day...