Volume of a leaky tank












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$begingroup$



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







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$endgroup$

















    0












    $begingroup$



    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







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$



      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







      share|cite|improve this question











      $endgroup$





      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






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Mar 14 '18 at 14:59









      Rodrigo de Azevedo

      12.9k41856




      12.9k41856










      asked Feb 26 '16 at 11:57









      user11128user11128

      1335




      1335






















          2 Answers
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          $begingroup$

          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.






          share|cite|improve this answer









          $endgroup$





















            0












            $begingroup$

            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.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              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











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            2 Answers
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            active

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            2 Answers
            2






            active

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            active

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            0












            $begingroup$

            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.






            share|cite|improve this answer









            $endgroup$


















              0












              $begingroup$

              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.






              share|cite|improve this answer









              $endgroup$
















                0












                0








                0





                $begingroup$

                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.






                share|cite|improve this answer









                $endgroup$



                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.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Feb 26 '16 at 12:06









                Bobson DugnuttBobson Dugnutt

                8,53831939




                8,53831939























                    0












                    $begingroup$

                    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.






                    share|cite|improve this answer









                    $endgroup$













                    • $begingroup$
                      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
















                    0












                    $begingroup$

                    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.






                    share|cite|improve this answer









                    $endgroup$













                    • $begingroup$
                      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














                    0












                    0








                    0





                    $begingroup$

                    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.






                    share|cite|improve this answer









                    $endgroup$



                    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.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    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


















                    • $begingroup$
                      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


















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