Question

Could someone help me with an exponential growth word problem (it's an attachment)

Answer 1

1st task is to find the value of k....


so you are looking at

\[P(t) = \frac{400 \times 20000}{400 + (20000 - 400)e^{-kt}}\]

so to find k you need

so you know if the initial population A = 400 and doubles after 1 year

P(t) = 800 when t = 1

substitute and solve for k

when you find k, you will be able to solve for t = 3

Answer 2

800=400×20000/400+(20000−400)e^(−k1)

Answer 3

that seems good... now its the mechanical process of solving for k

Answer 4

Hmmm... that seem to be where I am getting stuck. I'll attempt it, then show you what I get.

Answer 5

well easiest thing to do is swap 800 and the denominator so you have

\[400 + (20000 - 400)e^{_k} = \frac{8000000}{800} \]


then work on from there

Answer 6

20000e^k=10000

Answer 7

not quite...

\[19600e^{-k} = 10000 - 400\]

solve the right side then divide by 19600

Answer 8

you will get a value for e^(-k)

Answer 9

0.4897

Answer 10

great so next take the log of both sides

-k = ln(0.4897)

Answer 11

then divide both sides of the equation by -1

Answer 12

-1=-0.713962--->1= 0.713962

Answer 13

great so you know k...

the last part of the equation is asking for the rate of change, at t = 3

so that means you need the 1st derivative, with repsect to t, and then substitute t = 3

Answer 14

First derivative of t=1

Answer 15

nope find the derivative of

\[P(t) = \frac{8000000}{400 + 19600e^{-0.71366t}}\]

I'd probably use the quotient rule

Answer 16

then substitute t = 3 to find the rate of change

Answer 17

For the derivative, I got (699387. e^(0.71366 t))/(49.+e^(0.71366 t))^2

Answer 18

ok... so then just substitute t = 3 and evaluate for the rate of change

Answer 19

1799.20 THANK YOU!!!