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Help w/ partial fractions

  • #1
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Homework Statement


Consider an object that is coasting horizontally subject to a drag force f = -bv = cv^2. Write down Newton's second law...

The Attempt at a Solution



So I did all of the steps leading up to this:

[tex]m∫\frac{dv}{bv+cv^2}=-t dt[/tex]

Using partial fractions I get [tex]\frac{1}{bv+cv} = \frac{c_1}{v}+\frac{c_2}{b+cv}[/tex]
[tex]1 = c_1(b+cv)+c_2v[/tex]

Usually in partial fractions you have something like [tex]c_1(x-a)[/tex] where a is a real number. However there is no real number there so I am confused on what to do.
 

Answers and Replies

  • #2
gabbagabbahey
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Homework Statement


Consider an object that is coasting horizontally subject to a drag force f = -bv = cv^2.
I assume you mean [itex]f=-bv-cv^2[/itex]?

[tex]1 = c_1(b+cv)+c_2v[/tex]

Usually in partial fractions you have something like [tex]c_1(x-a)[/tex] where a is a real number. However there is no real number there so I am confused on what to do.
Is [itex]b[/itex] not a real number? "confused:

The equation must hold for all values of [itex]v[/itex], so pick a couple of convenient values and solve for [itex]c_1[/itex] and [itex]c_2[/itex]...
 
  • #3
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I assume you mean [itex]f=-bv-cv^2[/itex]?



Is [itex]b[/itex] not a real number? "confused:

The equation must hold for all values of [itex]v[/itex], so pick a couple of convenient values and solve for [itex]c_1[/itex] and [itex]c_2[/itex]...
No [tex]b = βD[/tex]

Where β is a constant and D is the diameter of the object.

I don't understand the second part of your statement.
 
  • #4
gabbagabbahey
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No [tex]b = βD[/tex]

Where β is a constant and D is the diameter of the object.
Maybe you should clarify what you mean by "real number". The usual definition is given here.

I don't understand the second part of your statement.
Well, for example, when [itex]v=0[/itex] your equation tells me that [itex]1= c_1(b+(0))+c(0)=c_1 b[/itex], so [itex]c_1=\frac{1}{b}[/itex]. Pick one more value of [itex]v[/itex] in order to solve for [itex]c_2[/itex].
 
  • #5
SammyS
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Homework Statement


Consider an object that is coasting horizontally subject to a drag force f = -bv = cv^2. Write down Newton's second law...

The Attempt at a Solution



So I did all of the steps leading up to this:

[tex]m∫\frac{dv}{bv+cv^2}=-t dt[/tex]

Using partial fractions I get [tex]\frac{1}{bv+cv} = \frac{c_1}{v}+\frac{c_2}{b+cv}[/tex]
[tex]1 = c_1(b+cv)+c_2v[/tex]

Usually in partial fractions you have something like [tex]c_1(x-a)[/tex] where a is a real number. However there is no real number there so I am confused on what to do.
Write [tex]1 = c_1(b+cv)+c_2v[/tex] as [tex]1 = c_1\,b+(c_1\,c+c_2)v\ .[/tex]

This must be true for all v, so equate coefficients of powers of v on the left side (only a constant term) with coefficients of powers of v on the right side.
 
  • #6
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Maybe you should clarify what you mean by "real number". The usual definition is given here.



Well, for example, when [itex]v=0[/itex] your equation tells me that [itex]1= c_1(b+(0))+c(0)=c_1 b[/itex], so [itex]c_1=\frac{1}{b}[/itex]. Pick one more value of [itex]v[/itex] in order to solve for [itex]c_2[/itex].
I mean exactly what I said by the real number.

Yes I have solved for c_1. But could not solve for c_2, up until I thought about substituting c_1 = 1/b into the equation that solved for c_2.

I let v = 1 and I obtain the following, [tex]c_2= 1 - \frac{c}{b}[/tex]

Is this correct?
Write [tex]1 = c_1(b+cv)+c_2v[/tex] as [tex]1 = c_1\,b+(c_1\,c+c_2)v\ .[/tex]

This must be true for all v, so equate coefficients of powers of v on the left side (only a constant term) with coefficients of powers of v on the right side.
Your wording confuses me here.
 
  • #7
vela
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If your problem was that you were confused because the equation didn't have actual numbers in it instead of variables representing those numbers, you just have to get over it. If not, I don't think anyone here understands what you meant when you said "there is no real number there."
 
  • #8
SammyS
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... [tex]1 = c_1\,b+(c_1\,c+c_2)v\ .[/tex]
This must be true for all v, so equate coefficients of powers of v on the left side (only a constant term) with coefficients of powers of v on the right side.
Your wording confuses me here.
Well, it was difficult to word that correctly, but I think I did.

What is the constant term (coefficient of v0) on the left hand side of the equation?

What is the constant term (coefficient of v0) on the right hand side of the equation?

Equate these.

***************
What is the coefficient of v (that's v1) on the left hand side of the equation? (It's zero. Right?)

What is the coefficient of v (that's v1) on the right hand side of the equation?

Equate these.
 
  • #9
gabbagabbahey
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I mean exactly what I said by the real number.
Again, it is not clear (to me anyway) what you mean. Is your concern that you aren't given numerical values for [itex]b[/itex] and [itex]c[/itex]? If so, you need to recall what was no doubt taught to you (or should have been taught to you) in high school algebra: we often use variables in equations to represent different quantities so that we can find solutions to problems of a certain type, rather than having to solve the same type of problem every time we have different values of those quantities. Your solution to this problem will be of the same form for any Real (as in not complex or imaginary) values of the coefficients [itex]b[/itex] and [itex]c[/itex], so rather than picking a single set of numerical values for those coefficients, and solving the problem just for that set of values, it makes more sense to solve it with the coefficients represented as variables [itex]b[/itex] and [itex]c[/itex].

Yes I have solved for c_1. But could not solve for c_2, up until I thought about substituting c_1 = 1/b into the equation that solved for c_2.

I let v = 1 and I obtain the following, [tex]c_2= 1 - \frac{c}{b}[/tex]

Is this correct?
That's close, but not quite correct. You should double check your algebra (your method sounds correct). It might be easier if you look at [itex]v=-\frac{b}{c}[/itex] instead, but you should get the same result no matter which value of [itex]v[/itex] you choose
 

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