A problem on differential equations

In summary, the boy takes approximately 3.6 minutes to finish his 250 gram burger, with a consumption rate of 100-1/900m^2 grams per minute, where m is the mass eaten.
  • #1
chwala
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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##
 
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  • #2
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})##
 
  • #3
consumption rate is ##100-(1/900)m^2##
 
  • #4
chwala said:
consumption rate is ##100-(1/900)m^2##
What about the initial size of the burger - 25 grams or 250 grams?
 
  • #5
the initial burger is 250grams. it is eaten (all of it) until it becomes 0 grams.
 
  • #6
chwala said:

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.
 
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  • #7
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## ?
 
  • #8
chwala said:
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:
chwala said:
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.
 
  • #9
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?
 
  • #10
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.
 
  • #11
ok sam let me have a look at my equations again, greetings from Africa
 
  • #12
chwala said:
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##.
 
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  • #13
thanks now clear , we have
##k=0##
then
## t= 1.5ln 550-1.5ln50##
=##1.5ln(550/50)= 1.5ln11 = 3.59 ## minutes
 

1. What is a differential equation?

A differential equation is a mathematical equation that relates the rate of change of a function to the function itself. It typically involves derivatives, which represent the rate of change, and the function itself.

2. Why are differential equations important?

Differential equations are used to model many physical phenomena and are essential in fields such as physics, engineering, and economics. They allow us to understand and predict the behavior of complex systems.

3. What are the different types of differential equations?

The three main types of differential equations are ordinary differential equations (ODEs), partial differential equations (PDEs), and stochastic differential equations (SDEs). ODEs involve a single independent variable, PDEs involve multiple independent variables, and SDEs include random components.

4. How do you solve a differential equation?

The method for solving a differential equation depends on its type and complexity. Some common techniques include separation of variables, substitution, and using specific formulas for certain types of equations. In some cases, it may not be possible to find an exact solution, and numerical methods must be used.

5. What are applications of differential equations?

Differential equations are used in many real-world applications, such as modeling population growth, predicting weather patterns, designing electrical circuits, and analyzing fluid dynamics. They are also important in the development of technological advancements, such as control systems and computer graphics.

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