# Math Help - Two differential equation questions

1. ## Two differential equation questions

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.

2. Originally Posted by Daniiel
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?
since we are proportional to the square root of the level of oil in the barrel at that time. we have

$y' = - k \sqrt{y}$ .........we have "-" because it is decreasing. i suppose you can solve that differential equation?

the initial data is $y(0) = 144$. furthermore, we have $y(4) = 80$. you can use this to find $k$ and the constant of integration through simultaneous equations.

finally, you want to find $t$ so that $y(t) = 0$

3. 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]

thats what i had before, but the original ky^1/2 should be -?
and for the second question i did
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

are they right?

4. Originally Posted by Daniiel
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]

thats what i had before, but the original ky^1/2 should be -?
good job.

the value you got for k was negative, so it takes care of that minus sign i had put there.

and for the second question i did
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

are they right?
i said rubbish here

5. if the fire is moving uniformly in all directions wouldnt that make it a circle?
and if A = 5000 r^2 = sqroot[5000/pi] = 39.9

6. Originally Posted by Daniiel
if the fire is moving uniformly in all directions wouldnt that make it a circle?
and if A = 5000 r^2 = sqroot[5000/pi] = 39.9
yes, but you need A = 5000 at the beginning.

$A' = k \pi r^2$ with $A(0) = 5000$ since it is the "area" of the fire that was 1/2 a hectare, and $A(6) = 6700$ since it was 1700 "additional" square meters

now i am going to bed. obviously i am too tired to be answering questions. i've made so many mistakes in this one thread already

7. thankyou for all your help, i really appreciate it you have been a great help

ive go tthis much done for the second question, and the first question completed =D

would anyone know if thats done correctly? and another equation i could get from that?

8. Originally Posted by Daniiel
thankyou for all your help, i really appreciate it you have been a great help

ive go tthis much done for the second question, and the first question completed =D

would anyone know if thats done correctly? and another equation i could get from that?
you know, i think my first equation would work as well. if you check what we did on this problem, we are using $\pi r^2$ as a constant anyway. we don't integrate with respect to $r$ or anything. we simply solve for $r$ when we need to. thus, $A' = k$ should solve as a good model as well. you will get a different constant of integration, but the numbers should work out the same. you can try it and compare them

i addressed how you should modify your first attempt in post #6