Question

Formulating a differential equation

Answer 1

Bit too complicated for me :3
maybe @cwrw238 can help :)

Answer 2

oo...my professor didn't go into detail on those word problems. sorry x.x

Answer 3

I'm not very sure as I'm still learning this but I'm going to give it a try.
Let w(t) be the weight at time(t).
\[\Delta w= [(1600-850-15w(t))\Delta t]/10000\]
\[dw/dt = 3/40 - \frac{ 3 }{ 2000 }w\]

Answer 4

i can't seem to solve it, would you mind giving your solution?

Answer 5

I obtained the same equation and seems good to me. And this appears to be an application of integrating factors on a linear DE. Do you know about integrating factors?

Answer 6

I think it can be solved by separation of variables as well

Answer 7

Oh, you're right. That's also a possibility. :)

Answer 8

okay i'm abit confused because there is no other variable, like it is only 3/40, how to tackle this?

Answer 9

ohh maybe im just confused with the variables itself haha, correct me if im wrong the IF=e^-(3/2000)t is it?

Answer 10

\[\frac{ dw }{ dt }=\frac{ 3 }{40 }-\frac{ 3 }{ 2000 }w\]
rearrange
\[\frac{ dw }{\frac{ 3 }{40}-\frac{ 3 }{2000 }w}=dt\]
then carry on with separation of variables

Answer 11

That is close, but we don't even need the negative sign because when we bring over the part with w:
w' + 3/2000 w = 3/40
u = e^(3/2000 t) which has derivative u' = +3/2000 e^(3/2000 t) = 3/2000 u

Answer 12

oh ya, haha i just roughly guessed :p anw what i was saying i was so used to doing with dy/dx that i confused the variables, anyway, what do u think is the answer? like what does it mean by her weight ultimately approach equillibrium?

Answer 13

I'm thinking we're supposed to look at the asymptotic behavior of the function, as time goes to infinity. A lot of differential equations is mainly those exponential functions, and if you specifically have: y =e^(-at) a>0 , t going to infinity, this eventually goes to 0.

Although we need to use our initial condition of weight being 60 as well after we solve because we will find that the constant from integration is nonzero.

Answer 14

I get \[w(t)=10\exp(-\frac{ 3 }{ 2000 }t)+50\]
so as t approaches infinity the exponential approaches 1.
So, yes, her weight approaches equilibrium

Answer 15

ohh.. that makes it more clearer to me now, okay uhm in a literal sense now, does she loes weight or maintain her weight at 60 kg? haha

Answer 16

*lose

Answer 17

Sorry, exponential approaches zero

Answer 18

Her initial weight is 60 at time t=0.
As t increases, w(t) tends to 50.
So she loses weight until her weight approaches asymptotically to 50.

Answer 19

okay thank you! super clear now! :D