Falling skydiver differential equation

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Homework Help Overview

The problem involves a differential equation governing the velocity of a skydiver falling under the influence of gravity and drag. The equation is presented in the form m dv/dt = mg - kv^2, where the variables and constants include the mass of the skydiver, gravitational acceleration, and a drag coefficient.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss rearranging the differential equation and express confusion about the applicability of integrating factors due to the presence of v squared. There are attempts to explore different forms of the equation and the implications of treating certain variables as constants.

Discussion Status

Some participants have provided guidance on separating the equation and have questioned the appropriateness of using integrating factors. There is an ongoing exploration of the correct method to approach the problem, with some participants seeking clarification on the roles of the variables involved.

Contextual Notes

Participants note that k, m, and g are constants, and there is a discussion about the nature of the independent and dependent variables in the context of motion and differential equations.

darryw
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Homework Statement


The velocity v(t) of a skydiver falling to the ground is governed by

m dv/dt = mg - kv^2

where m is the skydiver's mass, g is the acceleration due to gravity, k > 0 is the drag
coefficient, and v(t) >= 0.

Solve this equation for v(t) with the initial condition v(0) = 0.


Homework Equations



I have been doing other diff equations but when its a word problem it confuses me.. So, i treat m, g and k as constants.
I rearranged equation into this:

dv/dt + (k/m)v^2 = g

so now i can use integrating factor..

mu(x) = e^integ (k/m) ...but I am going to stop here already, in case I am already on wrong track.. The fact that v is squared changes things, no? (cant use integrating factor?) If this is ok though, i will proceed.. Thanks a lot for any help/tips

The Attempt at a Solution




 
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darryw said:

Homework Equations



I have been doing other diff equations but when its a word problem it confuses me.. So, i treat m, g and k as constants.
I rearranged equation into this:

dv/dt + (k/m)v^2 = g

so now i can use integrating factor..

mu(x) = e^integ (k/m) ...but I am going to stop here already, in case I am already on wrong track.. The fact that v is squared changes things, no? (cant use integrating factor?) If this is ok though, i will proceed.. Thanks a lot for any help/tips

Instead of putting it in that form, put it in a different form

\frac{dv}{dt} = g- \frac{k}{m}v^2

\frac{1}{g- \frac{k}{m}v^2} \frac{dv}{dt} = 1

Split into partial fractions.
 
I don't understand the reason for putting it in that form?
 
darryw said:
I don't understand the reason for putting it in that form?

You had it in the form of

dv/dt + (k/m)v2 = g


this will not give you an integrating factor of e∫(k/m)dt


had the v2, just been 'v', then it would work fine.
 
ok thankjs, .so in the form you have it i can get an integ factor? I mean, this is the method i need to use to solve right? (integrating factor) (as opposed to separable equation)
 
darryw said:
ok thankjs, .so in the form you have it i can get an integ factor? I mean, this is the method i need to use to solve right? (integrating factor) (as opposed to separable equation)
Forget integrating factors for this problem. The integrating factor business applies to problems in this form:
y' + P(x)y = Q(x)
What you have doesn't fit in this form.

rock.freak667 has shown you how to separate the equation and has given you advice on how to continue.
 
I need to confirm something before I proceed solving: in this equation, t in independent variable, v is dependent variable. k,m, and g are all constants. Is that correct?

Also, if i compare this to the previous equations I've been solving, t is like x (independent), and v is like y (dependent). and also t will always be independent because its time in these motion differential equations right? thanks
 
Yes.
 

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