What happens when time approaches infinity in differential equation $mfracdvdt = mg - kv$.
Clash Royale CLAN TAG#URR8PPP
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?
calculus differential-equations
add a comment |Â
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?
calculus differential-equations
Since the equation is "autonomous" you might consider the result if the initial condition is for velocity $v$ to equal terminal velocity. Hint: $dv/dt = 0$.
â hardmath
Sep 9 '17 at 23:34
add a comment |Â
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?
calculus differential-equations
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
calculus differential-equations
edited Sep 1 at 3:13
Nosrati
22.1k61747
22.1k61747
asked Sep 9 '17 at 23:31
John Smith
6819
6819
Since the equation is "autonomous" you might consider the result if the initial condition is for velocity $v$ to equal terminal velocity. Hint: $dv/dt = 0$.
â hardmath
Sep 9 '17 at 23:34
add a comment |Â
Since the equation is "autonomous" you might consider the result if the initial condition is for velocity $v$ to equal terminal velocity. Hint: $dv/dt = 0$.
â 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: $dv/dt = 0$.
â 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: $dv/dt = 0$.
â hardmath
Sep 9 '17 at 23:34
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
1
down vote
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 $ttoinfty.$
add a comment |Â
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.
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
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 $ttoinfty.$
add a comment |Â
up vote
1
down vote
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 $ttoinfty.$
add a comment |Â
up vote
1
down vote
up vote
1
down vote
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 $ttoinfty.$
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 $ttoinfty.$
answered Sep 10 '17 at 0:11
Michael Hardy
206k23187466
206k23187466
add a comment |Â
add a comment |Â
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.
add a comment |Â
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.
add a comment |Â
up vote
0
down vote
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.
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.
answered Sep 10 '17 at 1:55
Nosrati
22.1k61747
22.1k61747
add a comment |Â
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2423282%2fwhat-happens-when-time-approaches-infinity-in-differential-equation-m-fracdv%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Since the equation is "autonomous" you might consider the result if the initial condition is for velocity $v$ to equal terminal velocity. Hint: $dv/dt = 0$.
â hardmath
Sep 9 '17 at 23:34