What happens when time approaches infinity in differential equation $mfracdvdt = mg - kv$.

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We know that the differential equation for velocity is $m , dv/dt = mg - kv$



where $k$ is air drag. What I am wondering is what happens as time approaches infinity. How is terminal velocity expressed in terms of this equation?










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  • Since the equation is "autonomous" you might consider the result if the initial condition is for velocity $v$ to equal terminal velocity. Hint: $dv/dt = 0$.
    – hardmath
    Sep 9 '17 at 23:34














up vote
1
down vote

favorite












We know that the differential equation for velocity is $m , dv/dt = mg - kv$



where $k$ is air drag. What I am wondering is what happens as time approaches infinity. How is terminal velocity expressed in terms of this equation?










share|cite|improve this question























  • Since the equation is "autonomous" you might consider the result if the initial condition is for velocity $v$ to equal terminal velocity. Hint: $dv/dt = 0$.
    – hardmath
    Sep 9 '17 at 23:34












up vote
1
down vote

favorite









up vote
1
down vote

favorite











We know that the differential equation for velocity is $m , dv/dt = mg - kv$



where $k$ is air drag. What I am wondering is what happens as time approaches infinity. How is terminal velocity expressed in terms of this equation?










share|cite|improve this question















We know that the differential equation for velocity is $m , dv/dt = mg - kv$



where $k$ is air drag. What I am wondering is what happens as time approaches infinity. How is terminal velocity expressed in terms of this equation?







calculus differential-equations






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edited Sep 1 at 3:13









Nosrati

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asked Sep 9 '17 at 23:31









John Smith

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  • Since the equation is "autonomous" you might consider the result if the initial condition is for velocity $v$ to equal terminal velocity. Hint: $dv/dt = 0$.
    – hardmath
    Sep 9 '17 at 23:34
















  • Since the equation is "autonomous" you might consider the result if the initial condition is for velocity $v$ to equal terminal velocity. Hint: $dv/dt = 0$.
    – hardmath
    Sep 9 '17 at 23:34















Since the equation is "autonomous" you might consider the result if the initial condition is for velocity $v$ to equal terminal velocity. Hint: $dv/dt = 0$.
– hardmath
Sep 9 '17 at 23:34




Since the equation is "autonomous" you might consider the result if the initial condition is for velocity $v$ to equal terminal velocity. Hint: $dv/dt = 0$.
– hardmath
Sep 9 '17 at 23:34










2 Answers
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When $mg=kv,$ then the right side of your differential equation is $0.$ That means $dv/dt=0$ so the velocity doesn't change. Solving $mg= kv$ for $v$ gives you terminal velocity, and that is the limiting velocity as $ttoinfty.$






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    Let $kgeq0$ then
    begineqnarray*
    mg-kv &=& ma \
    ma+kv &=& mg ~~~~~~;~~~~~~a=v'\
    v'+frackmv &=& g
    endeqnarray*
    which is differential equation and with integration factor $displaystyle I=e^frackmt$ is
    $$e^frackmtv=int e^frackmtg~dt+C=fracmgke^frackmt+C$$
    hence $displaystyle v=fracmgk+Ce^-frackmt$ is velocity equation. If falling be in free thus $v_0=0$ and $displaystyle C=-fracmgk$, so
    $$displaystyle v=fracmgk(1-e^-frackmt)$$
    is velocity equation. It's graph shows the velocity doesn't acceed a value $displaystyle v_T=fracmgk$ because
    $$|v|=Big|fracmgkbig(1-e^-frackmtbig)Big|leqfracmgk=v_T$$
    and this value $v_T$ is Terminal Velocity. Also
    $$displaystylelim_ttoinftyv=lim_ttoinftyfracmgk(1-e^-frackmt)=v_T$$
    this subject is true when an object falls through a fluid, like raindrop.






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      2 Answers
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      2 Answers
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      When $mg=kv,$ then the right side of your differential equation is $0.$ That means $dv/dt=0$ so the velocity doesn't change. Solving $mg= kv$ for $v$ gives you terminal velocity, and that is the limiting velocity as $ttoinfty.$






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        up vote
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        When $mg=kv,$ then the right side of your differential equation is $0.$ That means $dv/dt=0$ so the velocity doesn't change. Solving $mg= kv$ for $v$ gives you terminal velocity, and that is the limiting velocity as $ttoinfty.$






        share|cite|improve this answer






















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          up vote
          1
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          When $mg=kv,$ then the right side of your differential equation is $0.$ That means $dv/dt=0$ so the velocity doesn't change. Solving $mg= kv$ for $v$ gives you terminal velocity, and that is the limiting velocity as $ttoinfty.$






          share|cite|improve this answer












          When $mg=kv,$ then the right side of your differential equation is $0.$ That means $dv/dt=0$ so the velocity doesn't change. Solving $mg= kv$ for $v$ gives you terminal velocity, and that is the limiting velocity as $ttoinfty.$







          share|cite|improve this answer












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          answered Sep 10 '17 at 0:11









          Michael Hardy

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              Let $kgeq0$ then
              begineqnarray*
              mg-kv &=& ma \
              ma+kv &=& mg ~~~~~~;~~~~~~a=v'\
              v'+frackmv &=& g
              endeqnarray*
              which is differential equation and with integration factor $displaystyle I=e^frackmt$ is
              $$e^frackmtv=int e^frackmtg~dt+C=fracmgke^frackmt+C$$
              hence $displaystyle v=fracmgk+Ce^-frackmt$ is velocity equation. If falling be in free thus $v_0=0$ and $displaystyle C=-fracmgk$, so
              $$displaystyle v=fracmgk(1-e^-frackmt)$$
              is velocity equation. It's graph shows the velocity doesn't acceed a value $displaystyle v_T=fracmgk$ because
              $$|v|=Big|fracmgkbig(1-e^-frackmtbig)Big|leqfracmgk=v_T$$
              and this value $v_T$ is Terminal Velocity. Also
              $$displaystylelim_ttoinftyv=lim_ttoinftyfracmgk(1-e^-frackmt)=v_T$$
              this subject is true when an object falls through a fluid, like raindrop.






              share|cite|improve this answer
























                up vote
                0
                down vote













                Let $kgeq0$ then
                begineqnarray*
                mg-kv &=& ma \
                ma+kv &=& mg ~~~~~~;~~~~~~a=v'\
                v'+frackmv &=& g
                endeqnarray*
                which is differential equation and with integration factor $displaystyle I=e^frackmt$ is
                $$e^frackmtv=int e^frackmtg~dt+C=fracmgke^frackmt+C$$
                hence $displaystyle v=fracmgk+Ce^-frackmt$ is velocity equation. If falling be in free thus $v_0=0$ and $displaystyle C=-fracmgk$, so
                $$displaystyle v=fracmgk(1-e^-frackmt)$$
                is velocity equation. It's graph shows the velocity doesn't acceed a value $displaystyle v_T=fracmgk$ because
                $$|v|=Big|fracmgkbig(1-e^-frackmtbig)Big|leqfracmgk=v_T$$
                and this value $v_T$ is Terminal Velocity. Also
                $$displaystylelim_ttoinftyv=lim_ttoinftyfracmgk(1-e^-frackmt)=v_T$$
                this subject is true when an object falls through a fluid, like raindrop.






                share|cite|improve this answer






















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                  up vote
                  0
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                  Let $kgeq0$ then
                  begineqnarray*
                  mg-kv &=& ma \
                  ma+kv &=& mg ~~~~~~;~~~~~~a=v'\
                  v'+frackmv &=& g
                  endeqnarray*
                  which is differential equation and with integration factor $displaystyle I=e^frackmt$ is
                  $$e^frackmtv=int e^frackmtg~dt+C=fracmgke^frackmt+C$$
                  hence $displaystyle v=fracmgk+Ce^-frackmt$ is velocity equation. If falling be in free thus $v_0=0$ and $displaystyle C=-fracmgk$, so
                  $$displaystyle v=fracmgk(1-e^-frackmt)$$
                  is velocity equation. It's graph shows the velocity doesn't acceed a value $displaystyle v_T=fracmgk$ because
                  $$|v|=Big|fracmgkbig(1-e^-frackmtbig)Big|leqfracmgk=v_T$$
                  and this value $v_T$ is Terminal Velocity. Also
                  $$displaystylelim_ttoinftyv=lim_ttoinftyfracmgk(1-e^-frackmt)=v_T$$
                  this subject is true when an object falls through a fluid, like raindrop.






                  share|cite|improve this answer












                  Let $kgeq0$ then
                  begineqnarray*
                  mg-kv &=& ma \
                  ma+kv &=& mg ~~~~~~;~~~~~~a=v'\
                  v'+frackmv &=& g
                  endeqnarray*
                  which is differential equation and with integration factor $displaystyle I=e^frackmt$ is
                  $$e^frackmtv=int e^frackmtg~dt+C=fracmgke^frackmt+C$$
                  hence $displaystyle v=fracmgk+Ce^-frackmt$ is velocity equation. If falling be in free thus $v_0=0$ and $displaystyle C=-fracmgk$, so
                  $$displaystyle v=fracmgk(1-e^-frackmt)$$
                  is velocity equation. It's graph shows the velocity doesn't acceed a value $displaystyle v_T=fracmgk$ because
                  $$|v|=Big|fracmgkbig(1-e^-frackmtbig)Big|leqfracmgk=v_T$$
                  and this value $v_T$ is Terminal Velocity. Also
                  $$displaystylelim_ttoinftyv=lim_ttoinftyfracmgk(1-e^-frackmt)=v_T$$
                  this subject is true when an object falls through a fluid, like raindrop.







                  share|cite|improve this answer












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                  share|cite|improve this answer










                  answered Sep 10 '17 at 1:55









                  Nosrati

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