Question

A culture of bacteria is growing at a rate of 3e ^{.2t} per hour, with t in hours and 0<=t<=20. How many new bacteria will be in the culture after the first five hours?

Answer 1

I have\[ Q'(t) =3e ^{.2t}\]
\[Q(t) = 15e ^{.2t} + c\]

Answer 2

is that correct thinking?

Answer 3

if you're given the growth rate, then use the integral to find the amount at time t

i do not believe you integrate correctly

Answer 4

integrate from 0 to t

where t =5

Answer 5

ohhh yeah that would make sense

Answer 6

hmm ok so that comes out with the correct answer of 25, but when I try it for the next part b) How many new bacteria are introduced in the sixth through the fourteenth hours? I come up with the answer 196 when it should be 205

Answer 7

integrate from 6 to 14

either you integrated incorrectly or the book is wrong

usually its a calculation error so double check that

Answer 8

what happens to the + c on the integration? Does it mean anything?

Answer 9

\[Q(0) = 0 = 15e ^{.2t} + c\] so shouldn't c = -15 and be used when calculating?

Answer 10

i said your integration was incorrect
where in teh world did you get 15

Answer 11

3/.2 = 15?

Answer 12

nvm i misread the .2 as a 2
my mistake

Answer 13

unless Q(0)=0 was given to you, or any other values, you cant really find c

Answer 14

well I Q(t) represents the new bacteria so I assumed that Q(0) would = 0 right?

Answer 15

well no, new bacteria means the c would cancel out, so essentially c=0

Answer 16

well then at Q(0) we would already have 15 bacteria, that doesn't make sense

Answer 17

new bacteria, meaning, over a period of time

Answer 18

you are not calculating for a single point in time

Answer 19

after hte first 5 hours means
\[15e^{.2*5}+c-(15e^{.2*0}+c)\]

i think

Answer 20

ok that makes sense, I guess I'm just a little shaky on my understanding of the constant of integration is all.

Answer 21

so for the second part we have \[15e ^{.2 * 14} - (15e ^{.2 * 6}) = 246.67 - 49.80 = 196.87?\]