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Differential equation questions, rate of change

by AdamNailor
Tags: differential, equation, rate
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AdamNailor
#1
May11-08, 12:31 PM
P: 22
Hello and greetings everyone.

1. A quantity of oil is dropped into water. When the oil hits the water it spreads out as a circle. The radius of the circle is r cm after t seconds and when t = 3 the radius of the circle is increasing at the rate of 0.5 centimetres per second

One observer believes that the radius increases at a rate which is proportional to [tex]\frac{1}{t+1}[/tex]

i) Write down the a differential equation for this situation, using k as a constant of proportionality.

ii) Show that k = 2

iii) Calculate the radius of the circle after 10 seconds according to this model


Another observer belives that the rate of increase of the the radius of the circle is proportional to [tex]\frac{1}{(t+1)(t+2)}[/tex]


iv) Write down a new differential equation for this new situation. Using the same initial conditions as before, find the the new value for the constant

v) Hence solve the differential equation

vi) Calculate the radius of the circle after 10 seconds according to this model.




2. ?



3. The attempt at a solution

I'm pretty sure I can do the first two, i am getting stuck on part iii) however. I may beable to do the rest by myself after this is sorted out, but thought i would post the full Question just incase.

i) [tex]\frac{dr}{dt}=\frac{k}{t+1}[/tex]

ii) 0.5 = k/(3+1) so, k = 2

iii) [tex]\frac{dr}{dt}=\frac{2}{t+1}[/tex]

so.... [tex]\frac{dr}{dt}=\frac{2}{11}[/tex] ??

Then i really dont have a clue what to do. I dont know how to separate the r from the dr/dt to find the radius form this relationship.

Thanks in advance. Adam.
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AdamNailor
#2
May11-08, 12:32 PM
P: 22
wait! the latex didn't work, i should have previewed first. I'll just edit it......
exk
#3
May11-08, 02:03 PM
P: 119
in order to solve for the radius you need to solve the differential equation. try separation of variables.

HallsofIvy
#4
May11-08, 10:09 PM
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Differential equation questions, rate of change

Quote Quote by AdamNailor View Post
Hello and greetings everyone.

1. A quantity of oil is dropped into water. When the oil hits the water it spreads out as a circle. The radius of the circle is r cm after t seconds and when t = 3 the radius of the circle is increasing at the rate of 0.5 centimetres per second

One observer believes that the radius increases at a rate which is proportional to [tex]\frac{1}{t+1}[/tex]

i) Write down the a differential equation for this situation, using k as a constant of proportionality.

ii) Show that k = 2

iii) Calculate the radius of the circle after 10 seconds according to this model


Another observer belives that the rate of increase of the the radius of the circle is proportional to [tex]\frac{1}{(t+1)(t+2)}[/tex]


iv) Write down a new differential equation for this new situation. Using the same initial conditions as before, find the the new value for the constant

v) Hence solve the differential equation

vi) Calculate the radius of the circle after 10 seconds according to this model.




2. ?



3. The attempt at a solution

I'm pretty sure I can do the first two, i am getting stuck on part iii) however. I may beable to do the rest by myself after this is sorted out, but thought i would post the full Question just incase.

i) [tex]\frac{dr}{dt}=\frac{k}{t+1}[/tex]

ii) 0.5 = k/(3+1) so, k = 2
Good!

iii) [tex]\frac{dr}{dt}=\frac{2}{t+1}[/tex]

so.... [tex]\frac{dr}{dt}=\frac{2}{11}[/tex] ??p=
No, that's dr/dt when t= 10. As the problem says, you need to solve the differential equation.

Then i really dont have a clue what to do. I dont know how to separate the r from the dr/dt to find the radius form this relationship.

Thanks in advance. Adam.
If dr/dt= 2/(t+1) then dr= (2/(t+1))dt. It's that easy!
AdamNailor
#5
May12-08, 10:03 AM
P: 22
Quote Quote by HallsofIvy View Post
Good!


No, that's dr/dt when t= 10. As the problem says, you need to solve the differential equation.


If dr/dt= 2/(t+1) then dr= (2/(t+1))dt. It's that easy!

Thanks. I had a mental block remembering that to solve differential equations you have to integrate (kind of fundamental, i know!)

so...

dr= (2/(t+1))dt => r = 2 ln (t+1) + c

t=0, r=0 => c=0

When t = 10, r = 2ln11 = (approx) 4.796

That agrees with the answer in the book. cheers.
flss
#6
May30-10, 10:17 AM
P: 1
[QUOTE=AdamNailor;1726074]Hello and greetings everyone.

1. A quantity of oil is dropped into water. When the oil hits the water it spreads out as a circle. The radius of the circle is r cm after t seconds and when t = 3 the radius of the circle is increasing at the rate of 0.5 centimetres per second

One observer believes that the radius increases at a rate which is proportional to [tex]\frac{1}{t+1}[/tex]

i) Write down the a differential equation for this situation, using k as a constant of proportionality.

ii) Show that k = 2

iii) Calculate the radius of the circle after 10 seconds according to this model


Another observer belives that the rate of increase of the the radius of the circle is proportional to [tex]\frac{1}{(t+1)(t+2)}[/tex]


iv) Write down a new differential equation for this new situation. Using the same initial conditions as before, find the the new value for the constant

v) Hence solve the differential equation

vi) Calculate the radius of the circle after 10 seconds according to this model.




2. ?



3. The attempt at a solution

I'm pretty sure I can do the first two, i am getting stuck on part iii) however. I may beable to do the rest by myself after this is sorted out, but thought i would post the full Question just incase.

i) [tex]\frac{dr}{dt}=\frac{k}{t+1}[/tex]

ii) 0.5 = k/(3+1) so, k = 2

iii) [tex]\frac{dr}{dt}=\frac{2}{t+1}[/tex]

so.... [tex]\frac{dr}{dt}=\frac{2}{11}[/tex] ??

Then i really dont have a clue what to do. I dont know how to separate the r from the dr/dt to find the radius form this relationship.

Thanks in advance. Adam.[/QUOTE

How about placing all terms in r with dr and all terms in t with dt and integrate both sides? Don't forget the constant and substitute given values to calculate the relation.

Could you E-Mail me the question at fayadwali@hotmail.com, and I'll send you the entire result. I can't understand the terms such as ...text/ etc.


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