Question

Can someone please help me with this question??!!

An epidemic follows the curve

P = 500 / 1+20,000e^(-0.549t)

; where t is in years. How fast is the epidemic growing after 10 years? (Round your answer to two significant digits.)

Answer 1

well you know that the instantaneous rate of growth is the first derivative of the function

P(t) = 500/ 1+ 20000e^(-0.549t)
P'(t) = -(20000 e^(-0.549t) *(-0.549)/(1+20000e^(-0.549t)^2

using the chain rule and the quotient rule.

= 20000 * 0.549 * e^(-0.549t) / (1+ 20000 e^(-0.549t)^2
= 10980 e^(-0.549)/ (1+ 20000 e^(-0.549t)^2
Plug in 10 for t

= 10980 e^(-5.49)/ (1+ 20000 e^(-5.49))^2

Answer 2

@seitys thanks a lot for your help, i got an answer and would like if you can make sure its right ; i got 0.00649174.. now how would i format that in millions?

Answer 3

My arithmetic is off. I forgot the 500 in P'(t).

I got 3.245870376.

Answer 4

ok so where would the 500 be in the equation?

Answer 5

the derivative should be 500(20000 e^(-0.549t) * (-0.549) / (1+20000 e^(-0.549t)^2

which simplifies to 5.49 x 10^6/(e^0.549t(20000/e^0.549t) + 1)^2

Answer 6

from there i plug in 10 to get the answer?

Answer 7

anywhere you see a t which is e^0.549(10) = e^5.49

Answer 8

because this is a function of dy with respect to dt (change in population with respect to change in time) and your change in time is 10 years.

Answer 9

so if i wanted to find it for 20 years. it would be 5.49 x 10^6/(e^0.549(2)(20000/e^0.549t(2)) + 1)^2. ??

Answer 10

your t is 20 so anywhere you see a t, replace it with 20. e^(0.549)(20) = e^(10.98)

Answer 11

ok im working on it. i will let u know about my answer

Answer 12

@seitys i got 168.1648...

Answer 13

I got 52.035...

Answer 14

The important thing is the calculus here. Just be careful with the arithmetic. I know I always find something on exams when I double check my work.

Answer 15

5.49 x 10^6/(e^0.549t(20000/e^0.549t) + 1)^2

Answer 16

i used this equation to plug in 20, maybe the closing partheses are misplaced?

Answer 17

It seems like you are missing a set of parenthesis around the ^2 and the 20000.

Answer 18

only the ((20000/e^0.549t)+1)^2 is squared. without the extra parenthesis you are squaring the first e^0.549t as well.

Answer 19

im having trouble still.

Answer 20

last question and then i can figure out what i did wrong. what would the equation be when i plug in 30

Answer 21

same equation with 0.549 multiplied by 30 instead of 20. in other words, anywhere you see a t, replace with 30.

Answer 22

i got 112.35

Answer 23

I got .38505...

Answer 24

5.49x10^6/(e^(0.549 (30)) (20000/e^(0.549 (30))+1)^2)

that should be the equation for 30

Answer 25

ohh, geez i keep plugging the equation wrong. thanks for being patient with me sei. :)

Answer 26

np