setting up diff eqns: oil leaking from barrel; bush fires

[Daniiel]

Hey i've been studying over my extension math revision and i just cant get my head around these two questions. I'm sure i've been setting up the differential equations wrong but im not sure whats right.

i have been for q1, dr/dt = y^(1/2). fliping to dt/dr then intergrating just some assitance with setting up the begining equation will be enough to get me going its just the start of these i get confused about The test is tomorow and tuesday so im kind of freaking out. heres the two questions, thanks in advance.

1. Oil is leaking out of a barrel loaded in a cargo ship. The rate at which the oil level is dropping seems proportional to the square root of the level of oil in the barrel at that time.
Write a differential equation for y(t), the level of oil (in cms) and time (hours).
Find a general solution for the equation.
If the barrel was filled up to 144cm when it started leaking and the level dropped to 80cm in four hours, how long will it take for the barrel to empty?

2. An investigation is under way in the department of forests, land and conservation to improve our understanding of the rate at which bush fires spread. A bush fire has broken out and a spotter plane has already taken some photographs. The first photograph shows a fire area of half a hectare. Comparison with a photograph taken six hours later shows that the fire seems to be spreading uniformly in all directions and has burned an additional 1700 square metres.
Use differential equations and find two possible models for the spread of the fire.

 

[arthur ohlsten]

Re: setting up diff eq

dy/dt=ky^1/2
dy= ky^1/2 dt
y^(-1/2) dy = kdt integrate
y^1/2 / [1/2] =kt +c
2 y^1/2 = kt +c

when t=0 y=144 substitute
2[sqrt[144]] =k0+c
c=24

2y^1/2 = kt +24

when t=4 y=80 substitute
2 sqrt80 =k[4] +24
2[4sqrt5]-24 =4k
k= 2sqrt5 -6

2y^1/2 =[2sqrt5 -6] t +24 answer
y^1/2 =sqrt5 -3] t +12
when y=0 what is t ?
set y=0
0=[sqrt5 -3]t +12
t= 12 / [3-sqrt5] answer

Please check for errors

Arthur

 

[arthur ohlsten]

Re: setting up diff eq

the area A that has been burned is:
A=pi r^2
where r is the radius of th burnt area

the rate of expansion is dr/dt. Buy dr/dt is a constant k
dA/dt = 2 pi r k
dA = 2pi r k dt
A=pi R^2 k t +c


at t=0 A(0) = 1/2 hectare
A(0)=pi r^2 k[0]+c
c=A(0)

A= pi r^2 k t +A(0)
at t=6 A(6) = 1700 metres sq. + A(0) solve for r(6) and substitute

A(6)= pi [r(6)^2] 6 k +A(0) solve for k

k is rate of the fire is advancing

Arthur

 

[Daniiel]

Re: setting up diff eq

thanks loads for your help
i think thats right
but would the dy/dt=ky^1/2 be dy/dt= - ky^1/2 because its decreasing?

i agree with it all but again im pretty hopeless at finding errors
what about these two lines
y^1/2 / [1/2] =kt +c
2 y^1/2 = kt +c
is that simplified right?

 

[arthur ohlsten]

Re: setting up diff eq

K is the constant of proportionality
assume K = 13 as a hypothetical answer

if you were to replace K in the original with -K you would obtain -k=-13 as a answer
The y61/2 is for the square root of the height of oil, and K is the constant of proportionlity. The proper value for K will be obtained regardless if you write the equation dy/dt =K y^1/2 or dy/dt = -K y^1/2

To check to see if I am right solve both possible equations and see if the answers differ

Arthur

 

[arthur ohlsten]

Re: setting up diff eq

yes it is simplified correctly

example
10 / [1/2]= 10[2]

to divide by a fraction you invert the fraction and multiply

y^1/2 /[ 1/2] = [2/1]y^1/2

Arthur

 

[Daniiel]

Re: setting up diff eq

thankyou very much aurther

 

[DrToddMatthews]

Re: setting up diff eq

thanks loads for your help

Help? Give me a break. He did the entire problem for you.

Are there fees at this site or can anybody get their homework done for free?

 

[Deleted member 4993]

Re: setting up diff eq

Are there fees at this site or can anybody get their homework done for free?

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