Integrating the Function: $\frac{dv}{dt}=g-\frac{c}{m}v$

In summary: Hi everyoneI am looking at integrating this function,\frac{dv}{dt}=g-\frac{c}{m}vSo far i have rearranged the function to get,(\int1/g - \int1/(cv/m))* dv= \intdt ThanksJBIn summary, the conversation discusses solving a differential equation using integration. The equation in question is \frac{dv}{dt}=g-\frac{c}{m}v, and there is a mistake in the rearrangement that was attempted. The correct method involves separation and a substitution. There is also a mention of initial conditions.
  • #1
bortonj88
16
0
Hi everyone

I am looking at integrating this function,

[tex]\frac{dv}{dt}[/tex]=g-[tex]\frac{c}{m}[/tex]v

So far i have rearranged the function to get,

([tex]\int[/tex]1/g - [tex]\int[/tex]1/(cv/m))* dv= [tex]\int[/tex]dt

Thanks

JB
 
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  • #2
Are you saying:
[tex]\frac{dv}{dt}=g-\frac{c_V}{m}[/tex]
and this rearranges to
[tex]\int\frac{1}{g}dv -\int\frac{1}{\frac{c_V}{m}}dv=\int dt[/tex]
If so, there is a mistake because you can't split apart a denominator like that.
 
  • #3
bortonj88 said:
Hi everyone

I am looking at integrating this function,

[tex]\frac{dv}{dt}[/tex]=g-[tex]\frac{c}{m}[/tex]v

So far i have rearranged the function to get,

([tex]\int[/tex]1/g - [tex]\int[/tex]1/(cv/m))* dv= [tex]\int[/tex]dt

Thanks

JB
There are at least a couple of ways of solving this differential equation, one of which is separation.
[tex]\frac{dv}{dt} = g -\frac{c}{m}v[/tex]

[tex]\Rightarrow \frac{dv}{g -\frac{c}{m}v} = dt[/tex]

[tex]\Rightarrow \frac{dv}{gm -cv} = \frac{1}{m}dt[/tex]

Tip: Use only one pair of tex tags per line rather than sprinkling them in the line.
 
  • #4
yeah, sorry, i made a mistake with that.

It was meant to read;

[tex]\int[/tex](1/(g-(c/m)v)) dv = [tex]\int[/tex]dt

Thanks
JB
 
  • #5
[tex]
\Rightarrow \frac{dv}{gm -cv} = \frac{1}{m}dt
[/tex]

Yeah i had this, i then integrated to get
[tex]
\Rightarrow ln{gm -cv} = \frac{1}{m}t
[/tex]
 
  • #6
sorry, meant to read

[tex]

\Rightarrow ln({gm -cv}) = \frac{1}{m}t

[/tex]
 
  • #7
bortonj88 said:
[tex]
\Rightarrow \frac{dv}{gm -cv} = \frac{1}{m}dt
[/tex]

Yeah i had this, i then integrated to get
[tex]
\Rightarrow ln{gm -cv} = \frac{1}{m}t
[/tex]
What was the substitution you used on the left side? Also, where is the constant of integration on the right side?

[tex]ln(gm -cv) = \frac{1}{m}t[/tex]
 
  • #8
thats the problem, i couldn't figure out how to do the integral on the left hand side, so I used quickmath.

If you have a suggestion please let me know, as this isn't homework, my housemate gave it to me to try and solve and now its really bugging me!

As for the constant, inital conditions are v=0 when t=0, and therefore the constant is not used, or so i thought
 
  • #9
The substitution is u = gm - cv, so du = -cdv. You need a factor of -c on the left, which you can get by multiplying the numerator by -c and also multiplying by -1/c.

Here's the left side.

[tex]\int \frac{dv}{gm -cv} = \frac{-1}{c}\int \frac{-cdv}{gm -cv} [/tex]
[tex] = \frac{-1}{c}\int \frac{du}{u} [/tex]

Can you finish it. Don't forget to undo the substitution.

Regarding the constant, it's safer to put it in, and then if it's zero you can remove it.
 

FAQ: Integrating the Function: $\frac{dv}{dt}=g-\frac{c}{m}v$

What is the meaning of "Integrating the Function"?

Integrating the function in this context refers to finding the mathematical expression for velocity (v) as a function of time (t) by solving the differential equation $\frac{dv}{dt}=g-\frac{c}{m}v$. This allows us to understand how the velocity of an object changes over time under the influence of gravity (g) and air resistance (c/m).

How does the acceleration due to gravity affect the velocity of an object?

The acceleration due to gravity (g) is a constant force that pulls objects towards the center of the Earth. In this equation, it is represented as a positive term, meaning it causes the velocity of an object to increase over time. This is why objects dropped from a height accelerate towards the ground.

What does the term "c/m" represent in the equation?

The term "c/m" represents the ratio of air resistance (c) to the mass (m) of the object. This value is typically small, meaning the effect of air resistance on the velocity of an object is relatively small. However, it can still have a significant impact on the motion of objects with large surface areas, like parachutes or feathers.

What is the significance of solving this equation for real-world applications?

This equation is useful for predicting the motion of objects in the presence of both gravity and air resistance. It can be used to understand the behavior of objects such as falling bodies, projectiles, or even airplanes. Solving this equation allows us to make informed decisions and design more efficient and effective systems.

Are there any limitations to this equation?

While this equation is a good approximation for many real-world scenarios, it does have some limitations. For example, it assumes a constant value for air resistance, which may not be the case in all situations. It also does not take into account other factors that may affect the motion of an object, such as wind or friction. Additionally, it only applies to objects moving in a vertical direction and cannot be used for objects moving horizontally.

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