Mixing
Volume of a leaky tank
$begingroup$
A tank of water is initially leaking at a rate of $1$ litre per minute. The rate it continues to leak at is proportional to the volume in the tank. After $7$ minutes, the tank is half full. What is the capacity of the tank?
I set $dV/dt=-kV$ where $V$ is volume and $k$ is constant.
Plugging in the initial conditions of $dV/dt(t=0)=-1$ tells me $k=V_{t=0}^{-1}$ and so I have $dV/dt=-V_{t=0}^{-1} V$. Now we integrate to get $ln{V}=-V_{t=0}^{-1} t$ or $V=-e^{V^{-1}_{t=0}t}$.
Using $V(t=7)=frac{1}{2} V_{t=0}^{-1}$, we find $frac{1}{2} V_{t=0}^{-1} = -e^{7V_{t=0}^{-1}}$. Does this have any real solutions?
ordinary-differential-equations
edited Mar 14 '18 at 14:59
Rodrigo de Azevedo
12.9k41856
asked Feb 26 '16 at 11:57
user11128user11128
1335
$endgroup$
add a comment |
$begingroup$
A tank of water is initially leaking at a rate of $1$ litre per minute. The rate it continues to leak at is proportional to the volume in the tank. After $7$ minutes, the tank is half full. What is the capacity of the tank?
I set $dV/dt=-kV$ where $V$ is volume and $k$ is constant.
Plugging in the initial conditions of $dV/dt(t=0)=-1$ tells me $k=V_{t=0}^{-1}$ and so I have $dV/dt=-V_{t=0}^{-1} V$. Now we integrate to get $ln{V}=-V_{t=0}^{-1} t$ or $V=-e^{V^{-1}_{t=0}t}$.
Using $V(t=7)=frac{1}{2} V_{t=0}^{-1}$, we find $frac{1}{2} V_{t=0}^{-1} = -e^{7V_{t=0}^{-1}}$. Does this have any real solutions?
ordinary-differential-equations
edited Mar 14 '18 at 14:59
Rodrigo de Azevedo
12.9k41856
asked Feb 26 '16 at 11:57
user11128user11128
1335
$endgroup$
add a comment |
$begingroup$
A tank of water is initially leaking at a rate of $1$ litre per minute. The rate it continues to leak at is proportional to the volume in the tank. After $7$ minutes, the tank is half full. What is the capacity of the tank?
I set $dV/dt=-kV$ where $V$ is volume and $k$ is constant.
Plugging in the initial conditions of $dV/dt(t=0)=-1$ tells me $k=V_{t=0}^{-1}$ and so I have $dV/dt=-V_{t=0}^{-1} V$. Now we integrate to get $ln{V}=-V_{t=0}^{-1} t$ or $V=-e^{V^{-1}_{t=0}t}$.
Using $V(t=7)=frac{1}{2} V_{t=0}^{-1}$, we find $frac{1}{2} V_{t=0}^{-1} = -e^{7V_{t=0}^{-1}}$. Does this have any real solutions?
ordinary-differential-equations
edited Mar 14 '18 at 14:59
Rodrigo de Azevedo
12.9k41856
asked Feb 26 '16 at 11:57
user11128user11128
1335
$endgroup$
A tank of water is initially leaking at a rate of $1$ litre per minute. The rate it continues to leak at is proportional to the volume in the tank. After $7$ minutes, the tank is half full. What is the capacity of the tank?
I set $dV/dt=-kV$ where $V$ is volume and $k$ is constant.
Plugging in the initial conditions of $dV/dt(t=0)=-1$ tells me $k=V_{t=0}^{-1}$ and so I have $dV/dt=-V_{t=0}^{-1} V$. Now we integrate to get $ln{V}=-V_{t=0}^{-1} t$ or $V=-e^{V^{-1}_{t=0}t}$.
Using $V(t=7)=frac{1}{2} V_{t=0}^{-1}$, we find $frac{1}{2} V_{t=0}^{-1} = -e^{7V_{t=0}^{-1}}$. Does this have any real solutions?
ordinary-differential-equations
ordinary-differential-equations
edited Mar 14 '18 at 14:59
Rodrigo de Azevedo
12.9k41856
asked Feb 26 '16 at 11:57
user11128user11128
1335
edited Mar 14 '18 at 14:59
Rodrigo de Azevedo
12.9k41856
asked Feb 26 '16 at 11:57
user11128user11128
1335
edited Mar 14 '18 at 14:59
Rodrigo de Azevedo
12.9k41856
edited Mar 14 '18 at 14:59
Rodrigo de Azevedo
12.9k41856
edited Mar 14 '18 at 14:59
Rodrigo de Azevedo
12.9k41856
12.9k41856
asked Feb 26 '16 at 11:57
user11128user11128
1335
asked Feb 26 '16 at 11:57
user11128user11128
1335
asked Feb 26 '16 at 11:57
user11128user11128
1335
1335
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
You forgot the integration constant when solving the DE. The general solution to $frac{dV}{dt}=-kV$ is $$V(t)=e^{-kt+c}=e^ce^{-kt}$$
where the constant $e^c=V_0$. You then have the equation $$frac{1}{2}V_0=V_0 e^{-7/V_0}$$
Let me know if the path from here isn't clear, and I'll elaborate.
answered Feb 26 '16 at 12:06
Bobson DugnuttBobson Dugnutt
8,53831939
$endgroup$
add a comment |
$begingroup$
There should be a constant of integration when you integrate. Moreover, when you say $ln V=-ct$, you get $V=e^{-ct}$, not $V=-e^{ct}$.
Let the capacity be $V_0$.
$$frac{dV}{dt}=-kV$$
$$[ln V]_{V_0}^{V}=-k[t]_{0}^{t}$$
$$lnleft(frac{V}{V_0}right)=-kt$$
Moreover, at $t=0$, $frac{dV}{dt}=-1$ and $V=V_0$.
Thus,
$$-1=-kV_0$$
$$k=frac{1}{V_0}$$
which gives,
$$lnleft(frac{V}{V_0}right)=-frac{1}{V_0}t$$
At $t=0$, $V=frac{V_0}{2}$
Thus,
$$lnleft(frac{frac{V_0}2}{V_0}right)=-frac{1}{V_0}7$$
$$lnleft(frac{1}{2}right)=-frac{1}{V_0}7$$
$$V_0=frac{7}{ln 2}$$
with appropriate units.
Regarding the other part of the question, it can be seen by drawing a graph that $frac{x}{2}=-e^{7x}$ has a solution in the fourth quadrant. Clearly, if $x$ denoted the volume of the tank, then there would be no solution.
answered Feb 26 '16 at 12:14
GoodDeedsGoodDeeds
10.3k31335
$endgroup$
$begingroup$
Hi thanks for this. Actually this was a mistake on my part when I re-did it today. I did it originally in a test a few days ago and got the same answer you have above and it was marked incorrect which got me puzzled. Oh well...
$endgroup$
– user11128
Feb 26 '16 at 14:23
$begingroup$
@user11128 There is one missing point - I assumed that the tank is initially full, but the question does not mention it anywhere.
$endgroup$
– GoodDeeds
Feb 26 '16 at 15:20
$begingroup$
Yes I think we are to assume this. Don't know what the problem was then.....alas
$endgroup$
– user11128
Feb 26 '16 at 15:51
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You forgot the integration constant when solving the DE. The general solution to $frac{dV}{dt}=-kV$ is $$V(t)=e^{-kt+c}=e^ce^{-kt}$$
where the constant $e^c=V_0$. You then have the equation $$frac{1}{2}V_0=V_0 e^{-7/V_0}$$
Let me know if the path from here isn't clear, and I'll elaborate.
answered Feb 26 '16 at 12:06
Bobson DugnuttBobson Dugnutt
8,53831939
$endgroup$
add a comment |
$begingroup$
You forgot the integration constant when solving the DE. The general solution to $frac{dV}{dt}=-kV$ is $$V(t)=e^{-kt+c}=e^ce^{-kt}$$
where the constant $e^c=V_0$. You then have the equation $$frac{1}{2}V_0=V_0 e^{-7/V_0}$$
Let me know if the path from here isn't clear, and I'll elaborate.
answered Feb 26 '16 at 12:06
Bobson DugnuttBobson Dugnutt
8,53831939
$endgroup$
add a comment |
$begingroup$
You forgot the integration constant when solving the DE. The general solution to $frac{dV}{dt}=-kV$ is $$V(t)=e^{-kt+c}=e^ce^{-kt}$$
where the constant $e^c=V_0$. You then have the equation $$frac{1}{2}V_0=V_0 e^{-7/V_0}$$
Let me know if the path from here isn't clear, and I'll elaborate.
answered Feb 26 '16 at 12:06
Bobson DugnuttBobson Dugnutt
8,53831939
$endgroup$
You forgot the integration constant when solving the DE. The general solution to $frac{dV}{dt}=-kV$ is $$V(t)=e^{-kt+c}=e^ce^{-kt}$$
where the constant $e^c=V_0$. You then have the equation $$frac{1}{2}V_0=V_0 e^{-7/V_0}$$
Let me know if the path from here isn't clear, and I'll elaborate.
answered Feb 26 '16 at 12:06
Bobson DugnuttBobson Dugnutt
8,53831939
answered Feb 26 '16 at 12:06
Bobson DugnuttBobson Dugnutt
8,53831939
answered Feb 26 '16 at 12:06
Bobson DugnuttBobson Dugnutt
8,53831939
answered Feb 26 '16 at 12:06
Bobson DugnuttBobson Dugnutt
8,53831939
8,53831939
add a comment |
add a comment |
$begingroup$
There should be a constant of integration when you integrate. Moreover, when you say $ln V=-ct$, you get $V=e^{-ct}$, not $V=-e^{ct}$.
Let the capacity be $V_0$.
$$frac{dV}{dt}=-kV$$
$$[ln V]_{V_0}^{V}=-k[t]_{0}^{t}$$
$$lnleft(frac{V}{V_0}right)=-kt$$
Moreover, at $t=0$, $frac{dV}{dt}=-1$ and $V=V_0$.
Thus,
$$-1=-kV_0$$
$$k=frac{1}{V_0}$$
which gives,
$$lnleft(frac{V}{V_0}right)=-frac{1}{V_0}t$$
At $t=0$, $V=frac{V_0}{2}$
Thus,
$$lnleft(frac{frac{V_0}2}{V_0}right)=-frac{1}{V_0}7$$
$$lnleft(frac{1}{2}right)=-frac{1}{V_0}7$$
$$V_0=frac{7}{ln 2}$$
with appropriate units.
Regarding the other part of the question, it can be seen by drawing a graph that $frac{x}{2}=-e^{7x}$ has a solution in the fourth quadrant. Clearly, if $x$ denoted the volume of the tank, then there would be no solution.
answered Feb 26 '16 at 12:14
GoodDeedsGoodDeeds
10.3k31335
$endgroup$
$begingroup$
Hi thanks for this. Actually this was a mistake on my part when I re-did it today. I did it originally in a test a few days ago and got the same answer you have above and it was marked incorrect which got me puzzled. Oh well...
$endgroup$
– user11128
Feb 26 '16 at 14:23
$begingroup$
@user11128 There is one missing point - I assumed that the tank is initially full, but the question does not mention it anywhere.
$endgroup$
– GoodDeeds
Feb 26 '16 at 15:20
$begingroup$
Yes I think we are to assume this. Don't know what the problem was then.....alas
$endgroup$
– user11128
Feb 26 '16 at 15:51
add a comment |
$begingroup$
There should be a constant of integration when you integrate. Moreover, when you say $ln V=-ct$, you get $V=e^{-ct}$, not $V=-e^{ct}$.
Let the capacity be $V_0$.
$$frac{dV}{dt}=-kV$$
$$[ln V]_{V_0}^{V}=-k[t]_{0}^{t}$$
$$lnleft(frac{V}{V_0}right)=-kt$$
Moreover, at $t=0$, $frac{dV}{dt}=-1$ and $V=V_0$.
Thus,
$$-1=-kV_0$$
$$k=frac{1}{V_0}$$
which gives,
$$lnleft(frac{V}{V_0}right)=-frac{1}{V_0}t$$
At $t=0$, $V=frac{V_0}{2}$
Thus,
$$lnleft(frac{frac{V_0}2}{V_0}right)=-frac{1}{V_0}7$$
$$lnleft(frac{1}{2}right)=-frac{1}{V_0}7$$
$$V_0=frac{7}{ln 2}$$
with appropriate units.
Regarding the other part of the question, it can be seen by drawing a graph that $frac{x}{2}=-e^{7x}$ has a solution in the fourth quadrant. Clearly, if $x$ denoted the volume of the tank, then there would be no solution.
answered Feb 26 '16 at 12:14
GoodDeedsGoodDeeds
10.3k31335
$endgroup$
$begingroup$
Hi thanks for this. Actually this was a mistake on my part when I re-did it today. I did it originally in a test a few days ago and got the same answer you have above and it was marked incorrect which got me puzzled. Oh well...
$endgroup$
– user11128
Feb 26 '16 at 14:23
$begingroup$
@user11128 There is one missing point - I assumed that the tank is initially full, but the question does not mention it anywhere.
$endgroup$
– GoodDeeds
Feb 26 '16 at 15:20
$begingroup$
Yes I think we are to assume this. Don't know what the problem was then.....alas
$endgroup$
– user11128
Feb 26 '16 at 15:51
add a comment |
$begingroup$
There should be a constant of integration when you integrate. Moreover, when you say $ln V=-ct$, you get $V=e^{-ct}$, not $V=-e^{ct}$.
Let the capacity be $V_0$.
$$frac{dV}{dt}=-kV$$
$$[ln V]_{V_0}^{V}=-k[t]_{0}^{t}$$
$$lnleft(frac{V}{V_0}right)=-kt$$
Moreover, at $t=0$, $frac{dV}{dt}=-1$ and $V=V_0$.
Thus,
$$-1=-kV_0$$
$$k=frac{1}{V_0}$$
which gives,
$$lnleft(frac{V}{V_0}right)=-frac{1}{V_0}t$$
At $t=0$, $V=frac{V_0}{2}$
Thus,
$$lnleft(frac{frac{V_0}2}{V_0}right)=-frac{1}{V_0}7$$
$$lnleft(frac{1}{2}right)=-frac{1}{V_0}7$$
$$V_0=frac{7}{ln 2}$$
with appropriate units.
Regarding the other part of the question, it can be seen by drawing a graph that $frac{x}{2}=-e^{7x}$ has a solution in the fourth quadrant. Clearly, if $x$ denoted the volume of the tank, then there would be no solution.
answered Feb 26 '16 at 12:14
GoodDeedsGoodDeeds
10.3k31335
$endgroup$
There should be a constant of integration when you integrate. Moreover, when you say $ln V=-ct$, you get $V=e^{-ct}$, not $V=-e^{ct}$.
Let the capacity be $V_0$.
$$frac{dV}{dt}=-kV$$
$$[ln V]_{V_0}^{V}=-k[t]_{0}^{t}$$
$$lnleft(frac{V}{V_0}right)=-kt$$
Moreover, at $t=0$, $frac{dV}{dt}=-1$ and $V=V_0$.
Thus,
$$-1=-kV_0$$
$$k=frac{1}{V_0}$$
which gives,
$$lnleft(frac{V}{V_0}right)=-frac{1}{V_0}t$$
At $t=0$, $V=frac{V_0}{2}$
Thus,
$$lnleft(frac{frac{V_0}2}{V_0}right)=-frac{1}{V_0}7$$
$$lnleft(frac{1}{2}right)=-frac{1}{V_0}7$$
$$V_0=frac{7}{ln 2}$$
with appropriate units.
Regarding the other part of the question, it can be seen by drawing a graph that $frac{x}{2}=-e^{7x}$ has a solution in the fourth quadrant. Clearly, if $x$ denoted the volume of the tank, then there would be no solution.
answered Feb 26 '16 at 12:14
GoodDeedsGoodDeeds
10.3k31335
answered Feb 26 '16 at 12:14
GoodDeedsGoodDeeds
10.3k31335
answered Feb 26 '16 at 12:14
GoodDeedsGoodDeeds
10.3k31335
answered Feb 26 '16 at 12:14
GoodDeedsGoodDeeds
10.3k31335
10.3k31335
$begingroup$
Hi thanks for this. Actually this was a mistake on my part when I re-did it today. I did it originally in a test a few days ago and got the same answer you have above and it was marked incorrect which got me puzzled. Oh well...
$endgroup$
– user11128
Feb 26 '16 at 14:23
$begingroup$
@user11128 There is one missing point - I assumed that the tank is initially full, but the question does not mention it anywhere.
$endgroup$
– GoodDeeds
Feb 26 '16 at 15:20
$begingroup$
Yes I think we are to assume this. Don't know what the problem was then.....alas
$endgroup$
– user11128
Feb 26 '16 at 15:51
add a comment |
$begingroup$
Hi thanks for this. Actually this was a mistake on my part when I re-did it today. I did it originally in a test a few days ago and got the same answer you have above and it was marked incorrect which got me puzzled. Oh well...
$endgroup$
– user11128
Feb 26 '16 at 14:23
$begingroup$
@user11128 There is one missing point - I assumed that the tank is initially full, but the question does not mention it anywhere.
$endgroup$
– GoodDeeds
Feb 26 '16 at 15:20
$begingroup$
Yes I think we are to assume this. Don't know what the problem was then.....alas
$endgroup$
– user11128
Feb 26 '16 at 15:51
$begingroup$
Hi thanks for this. Actually this was a mistake on my part when I re-did it today. I did it originally in a test a few days ago and got the same answer you have above and it was marked incorrect which got me puzzled. Oh well...
$endgroup$
– user11128
Feb 26 '16 at 14:23
$begingroup$
Hi thanks for this. Actually this was a mistake on my part when I re-did it today. I did it originally in a test a few days ago and got the same answer you have above and it was marked incorrect which got me puzzled. Oh well...
$endgroup$
– user11128
Feb 26 '16 at 14:23
$begingroup$
@user11128 There is one missing point - I assumed that the tank is initially full, but the question does not mention it anywhere.
$endgroup$
– GoodDeeds
Feb 26 '16 at 15:20
$begingroup$
@user11128 There is one missing point - I assumed that the tank is initially full, but the question does not mention it anywhere.
$endgroup$
– GoodDeeds
Feb 26 '16 at 15:20
$begingroup$
Yes I think we are to assume this. Don't know what the problem was then.....alas
$endgroup$
– user11128
Feb 26 '16 at 15:51
$begingroup$
Yes I think we are to assume this. Don't know what the problem was then.....alas
$endgroup$
– user11128
Feb 26 '16 at 15:51
add a comment |
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