Question

Growth and Decay assistance

Answer 1

I want to know how to do these problems I dont just want the answer

Answer 2

ok

Answer 3

\[A=A_0 e^{kt } \text{ where } A_0 \text{ represents initial amount } \text{ and } k \text{ represents rate}\\ \text{ and } t \text{ represents time}\]

Answer 4

ok cool, that makes sense, I wish my teacehr gave me that formula

Answer 5

oops I Bimistake left the pre Exponential constant

Answer 6

lololo dont worry bout it @goformit100

Answer 7

thnxs for the assitance guys u guys rock

Answer 8

A_0 is given

Answer 9

Can you tell me what A_0 is?

Answer 10

yah give me a sec

Answer 11

I said A_0 represented the initial amount.
You don't have to do any math.
All you have to do is read the first line.

Answer 12

5000

Answer 13

lol

Answer 14

right so we can go ahead in put that value in for A_0

Answer 15

so we have
\[A=5000e^{kt}\]

Answer 16

we need to find k using the second part of that first line

Answer 17

this will require some math after some substitution

Answer 18

It says we will get 8000 in 10 minutes
do you know where to put these values?

Answer 19

A=5000e^30t

Answer 20

where does 30 come from?

Answer 21

We haven't got to that part of the question.

Answer 22

Try answering my question.

Answer 23

t represents time again
so it should be obvious where 10 goes
A is what we get so it should be obvious where 8000 goes

Answer 24

30 is the time at which the bacteria is left to grow from time 0, and according to the proble it grows exponentially

Answer 25

right now we are trying to find the rate k

Answer 26

so how do we find the constant k?

Answer 27

I was trying to tell you but you are jumping ahead

Answer 28

ok

Answer 29

The general equation to do this is...\[{\rm number~ of ~ bacteria = initial~number~of~bacteria} \times e^{at}\]Here is why it makes sense:

The number of bacteria will increase with increase in time. There will also be an "exponential" increase, which means that you are raising \(t\) to a power of \(e\). So the number of bacteria as a function of time will look something like \(e^t\).

But wait, when we plug in \(t = 0\) to the above equation, we get \(1\). It's not necessary that we start with one bacterium. So we multiply that quantity by the initial number of bacteria. Thus, our function looks like this: \(n_{\rm initial}\times e^{t}\).

But wait again! It's not really necessary that for a given time \(t\), we'll just raise it to the power of \(e\). We can raise it to the power of \(e\) and then square the whole thing which would still be an exponential function by definition! For that, we multiply \(t\) by another number to ensure that we're not missing anything else out of the image. This number is also the referred to as "the rate of exponential growth".

Finally, our function boils down to, in proper notation,\[n(t) = n_0 e^{at}\]Here, your function is...\[n(t) = 5000e^{at}\]Now, if you plug in \(t = 10 \), you should be getting the right-hand-side to be equal to the left-hand-side. At time = 10 minutes, the number of bacteria is \(8000\). The right hand side is \(5000e^{10a}\). Equate both of them and find the value of \(a\). When you find the value of \(a\), you will know what your function is. You can play around with that function to determine the number of bacteria at other values of \(t\), which includes 30 minutes.

Answer 30

wow too much work!!

Answer 31

We have \[A=A_0e^{kt}\]
It was clear what the initial was which was 5000.
\[A=5000e^{kt}\]
Now we still have the unknown k.
But omg guess what we are also given 8000 in 10 minutes.
If we substitute these values in we will have only one unknown which would be k.
k is a constant number, it doesn't change. So plugging in these values will be what we need in order to solve for k.