Question

PLEASE HELP ANYONE!!!
The number of bacteria p in a pond after t hours is p(t). The more bacteria there are, the more rapidly the number increases. This is expressed by the equation p'(t)=6p(t). Find the most general exponential function of the form A times e^Kt which satisfies the equation. (A and K are constants. Hint: substitute this function into the equation and see what you get).

p(t)=

If there were 1000 bacteria at t=0, how many are there 3 hours later?
p(3)=

Answer 1

Hmm so this is a separable differential equation.
This will work out maybe a little different than you're used to. :o

Answer 2

\[\Large \frac{dp}{dt}\quad=\quad 6p\]

Answer 3

Let's ignore the hint for now and just do it the legit way XD

Answer 4

So we want to start by `separating` the variables.
We want to get the `p's and dp` on one side, and the `t's and dt` on the other.

So we'll "multiply" the dt to the other side.
And divide both sides by p.\[\Large \frac{dp}{p}=6dt\]

Answer 5

See how we have dp and p on the same side now? :O
Making a little bit of sense maybe? :x

Answer 6

So our next step is to integrate.\[\Large \int\limits\frac{dp}{p}=\int\limits6dt\]

Answer 7

Where you at sam? XD wake up!!

Answer 8

im here lol

Answer 9

are you still there?

Answer 10

So what do we get when we integrate the p side? :D

Answer 11

\[\Large \int\limits \frac{1}{p}dp\quad=\quad?\]

Answer 12

ummm.. 1

Answer 13

No we can't apply the power rule to a term with a -1 exponent.

Can you think of a special function that gives us 1/x when we take it's derivative?

Answer 14

its not 1

Answer 15

im lost honestly

Answer 16

\[\Large (\ln x)' \quad=\quad \frac{1}{x}\]Right? +_+

Answer 17

yes yes i see

Answer 18

So for the left side,\[\Large \color{royalblue}{\int\limits\limits\frac{dp}{p}}=\int\limits\limits6dt\]We'll get,\[\Large \color{royalblue}{\ln p+C}=\int\limits\limits\limits6dt\]

Answer 19

We add a +C when we do indefinite integrals :D

Answer 20

How bout the integral on the right side?\[\Large 6\int\limits \;dt \quad=\quad ?\]

Answer 21

ummm...

Answer 22

im sorry my brain isnt as sharp as it should be
since its like 11:16 here

Answer 23

XD fair

Answer 24

\[\Large \ln p+C=6\color{royalblue}{\int\limits(1)dt}\]When we integrate 1, with respect to t, it becomes t.
You can always take the derivative to check.
Derivative of t is 1, yes? :x
\[\Large \ln p+C=6\color{royalblue}{(t+c)}\]

Answer 25

yes

Answer 26

Distribute the 6,\[\Large \ln p+C=6t+c\](Absorb the 6 into the c, cuz whatev, it doesn't matter, c is arbitrary).

Answer 27

Subtract C from each side,\[\Large \ln p\quad=\quad 6t+\color{orangered}{c-C}\]Again, since the constants are arbitrary (could take on any value), we can combine them into a new constant.\[\Large \ln p\quad=\quad 6t+\color{orangered}{d}\]

Answer 28

ok

Answer 29

We want to solve for p.
So from here we'll have to do a sneaky little math trick.
We'll `exponentiate` each side.
Rewrite each side as an exponent with base e.\[\LARGE e^{\ln p}\quad=\quad e^{6t+d}\]

Answer 30

Since the `exponential function` and the `log function` are inverses of one another, they "undo" each other.
So we simply get p on the left side.\[\LARGE p=e^{6t+d}\]

Answer 31

From here we want to apply a rule of exponents:\[\Large \color{teal}{e^{a+b}\quad=\quad e^a\cdot e^b}\]

Answer 32

So applying this rule gives us:\[\LARGE p\quad=\quad e^{6t}\cdot e^d\]

Answer 33

Again, we're going to do something annoying :)
e^d is simple a weird looking constant, we'll label it as something a little more manageable\[\Large p\quad=\quad A\cdot e^{6t}\].

Answer 34

Soooo I think that's the form they were looking for +_+
That takes care of the first part.

Answer 35

let me see

Answer 36

\[\Large p(t)\quad=\quad A\cdot e^{6t}\]

Answer 37

is their possibly another way webwork can accept this answer like the issue earlier?

Answer 38

Oh hmm I remember that...

A*exp(6t)

yah try that.

Answer 39

correct!

Answer 40

Next part is kinda tricky...
we have to use the first piece of information to find A.

Then we plug in our A value and find p(t) at t=3.

Answer 41

ok

Answer 42

1000 bacteria at t=0.\[\Large p(0)\quad=\quad 1000\qquad\to\qquad Ae^{6(0)}\quad=\quad1000\]

Answer 43

So what do we get for A?

Answer 44

hold on

Answer 45

1000^18

Answer 46

\[\Large Ae^{0}=1000\]
\[\Large e^0\quad=\quad?\]

Answer 47

1

Answer 48

\[\Large Ae^{0}=1000\]\[\Large A(1)=1000\]\[\Large A=1000\]

Answer 49

my bad

Answer 50

\[\Large p(t)\quad=\quad A\cdot e^{6t}\]\[\Large p(t)\quad=\quad1000\cdot e^{6t}\]

Answer 51

ok

Answer 52

\[\Large p(3)\quad=\quad1000\cdot e^{6(3)}\]

Answer 53

1000*e^18

Answer 54

yes.
Your webworks might want you to put it in as a decimal value.
Or maybe like this: 1000*exp(18)

i'm not sure which.

Answer 55

it works thanks!!!!!!!!