A problem on differential equations

[chwala]

Homework Statement

A boy is eating a 250 gram burger. When he has eaten a mass m grams, his rate of consumption is given by $100-1/900m^2$ grams per minute. How long does it take him to finish the meal?[/B]

Homework Equations

The Attempt at a Solution

$dm/dt=100-1/900m^2 , dm/dt=(90000-m^2)/900 ⇒dt/dm=900/(90000-m^2), t=∫900/(300+m)(300-m) dm$
let
$u=(300+m), t=∫900/(u(600-u))du= 900/(u(600-u))= (A/u +B/u)$ on solving, $A=1.5$and $B=1.5$ thus
$t=∫(1.5/u+1.5/(600-u))du =1.5lnu+1.5ln(600-u)+k =1.5ln(300+m)+1.5ln(300-m)$
using
$t(250)=0$ we have
$0=1.5ln(300+250)+1.5ln(300-250)+k$ where k=-1.5ln27500$thus$ t=1.5ln(300+m)+1.5ln(300-m)-1.5ln27500
at $m=0$

i assumed that here the burger would have been eaten and no remnants
$t=1.5ln300+1.5ln300-1.5ln27500$
$t=1.78$
the textbook answer on this is $t=3.6 minutes$

 

[SteamKing]

Is this a 25 gram burger or a 250 gram burger?

A 25 gram burger is about 1 bite.

Also, is the consumption rate $100 - (\frac{1}{900}m^2)$ or $(100 - \frac{1}{900m^2})$

 

[chwala]

consumption rate is $100-(1/900)m^2$

 

[SteamKing]

consumption rate is $100-(1/900)m^2$

What about the initial size of the burger - 25 grams or 250 grams?

 

[chwala]

the initial burger is 250grams. it is eaten (all of it) until it becomes 0 grams.

 

[SteamKing]

Homework Statement

A boy is eating a 250 gram burger. When he has eaten a mass m grams, his rate of consumption is given by $100-1/900m^2$ grams per minute. How long does it take him to finish the meal?

Homework Equations

The Attempt at a Solution

$$\frac{dm}{dt}=100-\frac{1}{900}m^2$$
$$\frac{dm}{dt}=\frac{(90000-m^2)}{900}$$
⇒$$\frac{dm}{dt}=\frac{900}{(90000-m^2)}$$
$$t=\int \frac{900}{(300+m)(300-m)} dm$$
let $u=(300+m)$

$$t=\int \frac{900}{u(600-u)}du$$
$$\frac{900}{u(600-u)}= \frac{A}{u} + \frac{B}{600-u}$$
on solving, $A=1.5$ and $B=1.5$

After this point, you have some errors in your integration for the second partial fraction term.

 

[chwala]

I have seen the error:
$t=1.5 ln u - 1.5 ln (600-u) + k$
eventually

$t = 1.5 ln 300 - 1.5 ln 300 - 1.5 ln 11$
$t = -3.6min$
why do we have a negative? is it taking care of the fact that the burger was initially whole and with every bite it kept on decreasing in mass value until the point
$m=0$ ?

 

[Samy_A]

I have seen the error:
$t=1.5 ln u - 1.5 ln (600-u) + k$
eventually

$t = 1.5 ln 300 - 1.5 ln 300 - 1.5 ln 11$
$t = -3.6min$
why do we have a negative? is it taking care of the fact that the burger was initially whole and with every bite it kept on decreasing in mass value until the point
$m=0$ ?

$m$ is the mass eaten:

A boy is eating a 250 gram burger. When he has eaten a mass m grams, his rate of consumption is given by $100-1/900m^2$ grams per minute.

 

[chwala]

am not getting you $dt/dm=1.5/(300+m)-1.5/(300-m)$
and which is equal to
$1.5/u+1.5/(300-(u-300)$ =
$1.5/u-1.5/(600-u)$ =
$900/(u(600-u))$ =
$900/(300+m)(300-m)$ =
$900/(90000-m^2)$ as indicated in my original post. my question still is why do we have $-3.6$minutes?

 

[Samy_A]

Your calculation looks correct to me (I thought there was a sign error, but I was wrong about that).
Your interpretation of $m$ is wrong. $m$ is the mass eaten, so at t=0, m=0, and when the boy finishes his hamburger, m=250 grams.

 

[chwala]

ok sam let me have a look at my equations again, greetings from Africa

 

[Samy_A]

ok sam let me have a look at my equations again, greetings from Africa

To be totally clear, your formula for $t$ is correct, but the error starts when you compute the integration constant $k$.

 

[chwala]

thanks now clear , we have
$k=0$
then
$t= 1.5ln 550-1.5ln50$
=$1.5ln(550/50)= 1.5ln11 = 3.59$ minutes