Question

Cal 2: Please help me set up this integral. A metal company has their production increasing at a rate proportional to the product of the number of units N and the time t in years

Answer 1

so a rate of change is proportional to ..... well whats your best attempt?

Answer 2

dp/dn = NT?
which becomes integral (1/n)dp = integral t dt

Answer 3

dn/dt = nt

seperates as

dn/n = t dt

Answer 4

is that the right integral?

Answer 5

integrate both sides

ln(n) = 1/2 t^2 + C ; base e it

n = C e^(t^2/2)

Answer 6

i cant see why it wouldnt be ...

Answer 7

can i give you the other half of the question? and we could try and figure it out?

Answer 8

by all means ..

Answer 9

400 units are presently produced. In 2 years, 1200 units are expected. What will be the production in 4 years?

Answer 10

when t=0, we are presently at 400; e^0 = 1 co C = 400

n(t) = 400 e^(t^2/2)

what is n(2) ?

Answer 11

err, t=4 that is

Answer 12

the final answer is 32 400 units

Answer 13

do we know that according to an answer key?

Answer 14

yes

Answer 15

dn/dt is the rate of the number of items per time frame ... does that define the rate of production to you?

Answer 16

not really, i don't understand, do you get the same answer as in the answer key?

Answer 17

well my 400 e^(t^2/2) did not get the same results no which is why im wondering about the setup

Answer 18

never really went over modeling with differentials

Answer 19

this is a differential equation

Answer 20

its ment to be setup as a differential equation.

Answer 21

basically with the info that is given , we must make the initial formula then make it into a integral so ex the (dn/dt) = nt then put that into a integral

Answer 22

once we solve for N. We must find out the missing variables ex. C and also variable T

Answer 23

once we do that, we figure out the question. In this case how long will it be in 4 years.

Answer 24

it would seem to me that the production defines the number of items made, and that the rate of change of production is the rate at which the number of items is changing

yeah, and we have to model it with the information given.

dn/dt = nt , it seperates

dn/n = t dt , it integrates

ln(n) = t^2/2 + C

and we solve for n

n = Ce^(t^2/2)

the number produced over a given time frame.

Answer 25

yes now we solve for the question.

Answer 26

So for ex. N(0) = 400, N(2) =1200, so what's N(4)=?

Answer 27

thats the concept yes ... and i assume we use the formula from part A to account for it ...

Answer 28

basically we have to find C and also T

Answer 29

the missing variables

Answer 30

t is given, its the number of years

Answer 31

true so we need to find C.

Answer 32

When we let T=0, c=400

Answer 33

if our model is correct

400 = C e^0 ... C = 400

Answer 34

yes

Answer 35

now how do we get the answer..

Answer 36

if our model is accurate

400 e^(2^2/2) = 1200, but its not ... its 2956

which leads me to thing that my thought of dn/dt isnt going to play well and we need another strategy

Answer 37

i guess so.

Answer 38

Well actually isnt it (E^4^2/2)?

Answer 39

the final answer is 32 400 units

Answer 40

lets do this

dp/dt = nt see where it takes us ... consider n a constant

p(t) = n t^2/2 + C , dunno how good it is but at lets see if it fails or not

400 = n + C
1200 = 2n + C

maybe?

Answer 41

i guess

Answer 42

- 400 = -n - C
1200 = 2n + C
-------------
800 = n

C = 400

p(t) = 800 t^2/2 + 400

does p(4) get us the soltion?

Answer 43

6800 ... sigh, well im at a loss

Answer 44

it should

Answer 45

LOL

Answer 46

@ rational any ideas?

Answer 47

tag didnt stick lol @rational , any ideas?

Answer 48

idk man, these questions confuse me

Answer 49

they dont inspire me much either

p = Ce^(t^2/2) + K

400 = Ce^0 + K
1200 = Ce^2 + K
----------------
800 = C(e^2-1)


C = 800/(e^2-1)
K = 400 - 800/(e^2-1)

p(t) = 800 e^(t^2/2)/(e^2-1) + 400 - 800/(e^2-1)

p(t) = 800 (e^(t^2/2)-1)/(e^2-1) + 400

p(4) = 373 532.8

not it ....

Answer 50

my teacher had a ex.

Answer 51

a different question obv.

Answer 52

dp/dt =Kp

Answer 53

which becomes integral 1/p dp = integral k dt

Answer 54

then it becomes p = e^kt + c

Answer 55

how come k does become k^2/2

Answer 56

k is a constant with respect to t

Answer 57

unless its a function of t

Answer 58

so its not the same to the question i have?

Answer 59

dp/dt = kp

dp/p = k dt

ln(p) = kt + C

p = Ce^(kt)

Answer 60

C = 400

1200 = 400e^(2k)

ln(3) = 2k

ln(3)/2 = k

try it

p = 400 e^(t ln(3)/2)

Answer 61

p(4) = 3600 in that case ...

Answer 62

http://www.wolframalpha.com/input/?i=exponential+function+%280%2C400%29%2C%282%2C1200%29%2C%284%2C32400%29

now we have an issue .. there is not best fit exponential function that fits those 3 points exactly