Differential equation questions, rate of change

In summary, differential equations are mathematical equations that relate the rate of change of a variable to its current value. They are important for modeling and analyzing real-world phenomena that involve rates of change. The method for solving a differential equation depends on its type and complexity. Initial conditions play a crucial role in determining the specific solution to a differential equation. Finally, differential equations are used to calculate rates of change at any point in time by solving the equation and finding the value of the derivative at that point.
  • #1
AdamNailor
22
0
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 modelAnother 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. ?

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 don't have a clue what to do. I don't know how to separate the r from the dr/dt to find the radius form this relationship.

Thanks in advance. Adam.
 
Last edited:
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  • #2
wait! the latex didn't work, i should have previewed first. I'll just edit it...
 
  • #3
in order to solve for the radius you need to solve the differential equation. try separation of variables.
 
  • #4
AdamNailor said:
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. ?



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 don't have a clue what to do. I don't 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!
 
  • #5
HallsofIvy said:
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.
 
  • #6
AdamNailor said:
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. ?



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 don't have a clue what to do. I don't 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.
 

1. What is a differential equation?

A differential equation is a mathematical equation that relates the rate of change of a variable to its current value. It involves derivatives, which represent the instantaneous rate of change of a function.

2. What is the importance of differential equations?

Differential equations are used to model and analyze real-world phenomena that involve rates of change, such as population growth, chemical reactions, and motion. They are essential in many fields of science and engineering.

3. 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 an integrating factor. In some cases, a differential equation may need to be solved numerically using a computer.

4. What is the role of initial conditions in differential equations?

Initial conditions are values that are known when solving a differential equation. They represent the starting point of a system and are used to determine the specific solution to the equation. Without initial conditions, a differential equation would have an infinite number of solutions.

5. How are differential equations used to calculate rates of change?

Differential equations provide a mathematical framework for calculating rates of change at any point in time. By solving the equation, we can determine the value of the derivative at a specific point, which represents the instantaneous rate of change at that point.

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