Simple Integral

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  • #1
Radarithm
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Homework Statement


[tex]F=-mg-m\alpha v[/tex][tex]F=m\frac{dv}{dt} =[/tex]
[tex]\int_0^v \frac{dv'}{g+\alpha v'} = -\int_0^t dt'[/tex]
This equals:
[tex]\ln \frac{1+\alpha v}{g} = -\alpha t[/tex]

I don't understand why it does; I keep getting an incorrect answer. Can anyone tell me what I'm doing wrong?

Homework Equations


The Attempt at a Solution



let [tex]u=g+\alpha v'[/tex][tex]du=g+vdv[/tex] so [itex]du-g=vdv[/itex]
[tex]-g+\int_0^v udu = \ln (g+\alpha v)-g[/tex]
This has to do with getting the velocity as a function of time; I don't know if it should be in the physics section or not but this is a problem I'm having with calculus.
 
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Answers and Replies

  • #2
1,540
135

Homework Statement


[tex]F=mg-m\alpha v[/tex][tex]F=m\frac{dv}{dt} =[/tex]
[tex]\int_0^v \frac{dv'}{g+\alpha v'} = -\int_0^t dt'[/tex]
This equals:
[tex]\ln \frac{1+\alpha v}{g} = -\alpha t[/tex]

I don't understand why it does; I keep getting an incorrect answer. Can anyone tell me what I'm doing wrong?

Homework Equations


The Attempt at a Solution



let [tex]u=g+\alpha v'[/tex][tex]du=g+vdv[/tex] so [itex]du-g=vdv[/itex]
[tex]-g+\int_0^v udu = \ln (g+\alpha v)-g[/tex]
This has to do with getting the velocity as a function of time; I don't know if it should be in the physics section or not but this is a problem I'm having with calculus.

Is ##F=mg-m\alpha v## or ##F=mg+m\alpha v## ?
 
  • #3
Radarithm
Gold Member
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Is ##F=mg-m\alpha v## or ##F=mg+m\alpha v## ?

[tex]F=-mg-m\alpha[/tex]
 
  • #4
HallsofIvy
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Homework Helper
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Homework Statement


[tex]F=-mg-m\alpha v[/tex][tex]F=m\frac{dv}{dt} =[/tex]
[tex]\int_0^v \frac{dv'}{g+\alpha v'} = -\int_0^t dt'[/tex]
This equals:
[tex]\ln \frac{1+\alpha v}{g} = -\alpha t[/tex]

I don't understand why it does; I keep getting an incorrect answer. Can anyone tell me what I'm doing wrong?

Homework Equations


The Attempt at a Solution



let [tex]u=g+\alpha v'[/tex][tex]du=g+vdv[/tex] so [itex]du-g=vdv[/itex]
This makes no sense. You chose to use v' as the variable of integration. Why does it become v when you find du? Also g is a constant so its derivative is 0. If [itex]u= g+ \alpha v'[/itex] then [itex]du= \alpha dv'[/itex] so that [itex]dv'= du/\alpha[/itex]
[tex]\int_0^v \frac{dv'}{g+ \alpha v'}= \frac{1}{\alpha}\int_g^{g+\alpha v} \frac{du}{u}[/tex]

[tex]-g+\int_0^v udu = \ln (g+\alpha v)-g[/tex]
This has to do with getting the velocity as a function of time; I don't know if it should be in the physics section or not but this is a problem I'm having with calculus.
 
  • #5
Radarithm
Gold Member
158
2
This makes no sense. You chose to use v' as the variable of integration. Why does it become v when you find du? Also g is a constant so its derivative is 0. If [itex]u= g+ \alpha v'[/itex] then [itex]du= \alpha dv'[/itex] so that [itex]dv'= du/\alpha[/itex]
[tex]\int_0^v \frac{dv'}{g+ \alpha v'}= \frac{1}{\alpha}\int_g^{g+\alpha v} \frac{du}{u}[/tex]

Why do the limits of integration change from [itex]0 \to v(t)[/itex] to [itex]g \to g+\alpha v[/itex] ?
 
  • #6
1,540
135
Why do the limits of integration change from [itex]0 \to v(t)[/itex] to [itex]g \to g+\alpha v[/itex] ?

Because the variable of integration has changed from v' to u .

Use the relation between v' and u (u=g+αv') to find the corresponding new limits .

When v' = 0 → u = g (lower limit)
When v' = v → u = g+αv (upper limit)
 
  • #7
HallsofIvy
Science Advisor
Homework Helper
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Because you changed variables from v' to [itex]u= g+ \alpha v'[/itex]. When v'= 0, [itex]u= g+ \alpha(0)= g[/itex] and when v'= v, [itex]u= g+ \alpha v[/itex].
 
  • #8
Radarithm
Gold Member
158
2
edit: double posted. sorry
 
  • #9
Radarithm
Gold Member
158
2
Because the variable of integration has changed from v' to u .

Use the relation between v' and u (u=g+αv') to find the corresponding new limits .

When v' = 0 → u = g (lower limit)
When v' = v → u = g+αv (upper limit)

Because you changed variables from v' to [itex]u= g+ \alpha v'[/itex]. When v'= 0, [itex]u= g+ \alpha(0)= g[/itex] and when v'= v, [itex]u= g+ \alpha v[/itex].

Thanks for the help guys. I appreciate it.
 

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