Question

A bacteria culture grows with constant relative growth rate. After 2 hours there are 400 bacteria and after 8 hours the count is 50,000. Find the rate of growth after 7 hours. P'(7) = ?

Answer 1

equation follows as P(7) = 80e^(r(7))

Answer 2

i need to take the derivative of P(7)

Answer 3

can u give what is initial equation of P

Answer 4

P(0) = 80

Answer 5

thatz it ??

Answer 6

how did u get e^(r(7))

Answer 7

from the growth rate formula of C*exp^(rt)

Answer 8

rt = rate ??

Answer 9

p(0)=80

Answer 10

r = rate t = time

Answer 11

P(t)=80*e^(r*t)

Answer 12

yes

Answer 13

we need to find r first

Answer 14

after 2 hour
400=80*e^2r ........(1)

Answer 15

.8047

Answer 16

r=0.8047 u got ??

Answer 17

yes

Answer 18

so we need to find P'(7)
first find P(t)

Answer 19

oh i mean P ' (t)

Answer 20

okay

Answer 21

natural log?

Answer 22

no no just differentiate

Answer 23

P(t)=80*e^(0.8047*t)

Answer 24

\[P'(t)=\frac{ d P(t) }{ dt }\]

Answer 25

P'(t)=0.8047*80*e^(0.8047*t)

Answer 26

now plug in t=7

Answer 27

what rule did u use to differentiate?

Answer 28

\[P'(7)=0.8047*80*e^{0.8047*7}\]

Answer 29

u know the differentiation of e^2x ??

Answer 30

it doesn't change right?

Answer 31

yeah but 2 will come froward

Answer 32

=2*e^2x

Answer 33

right ??

Answer 34

i believe so haha, ill try out the answer i get when plugging in t and see if its right

Answer 35

similarly 0.8047 sill come forward

Answer 36

it worked! thank you man

Answer 37

welcome :)

Answer 38

\[P'(7)={125\over2}\cdot \ln \left( 5 \right)\cdot \sqrt5\]

Answer 39

how???

Answer 40

80*0.8047=64.3
125/2=62.5

Answer 41

sorry ... 80 times that

Answer 42

\[P'(7)={5000}\cdot \ln \left( 5 \right)\cdot \sqrt5 \]