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Typical Calculus BC Separation of Variables Question



Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)A related rate questionFilling a conical tankPuzzled in unit stuff .. please helpTwo full 48-gallon tanks begin draining at t = 0.Related Rates Galore!Related Rate Question: Water is leaking out of an inverted conical tank at a rate of 9,500 cm3/minRelated Rates Conical Water Tank find rate of change of the water depthRates of change question involving water leaking out of a hemispherical tankGetting the rate of drain from a tankWater tank question










6












$begingroup$


I was told that I have a cylindrical water tank $10$ ft tall that can store $5000 $ ft$^3$ of water, and that the water drains from the bottom of the tank at a rate proportional to the instantaneous water level. After $6$ hours, half of the tank has drained out from the bottom. If the drain is opened and water is added to the tank at a constant rate of $100$ $fracft^3hr$, at what height above the drain will the elevation remain constant?



My attempt:



We are given that $$fracdVdt=kh$$ where $fracdVdt$ is the rate at which water leaves the tank, $V$ is the total change in water that has left the tank, and $h$ is the water's height above the drain.



If the elevation must remain constant in the tank, then I am merely finding the height in which the water that enters the tank equals to the water that leaves the tank, and therefore $fracdVdt=100fracft^3hr$.



To find $k$, I must do separation of variables and integrate both sides, so: $$intdV=intkhdt$$ $$V=kht+C$$
I know that $C=0$ because at $t=0$ hr, no water has left the tank ($V=0$), and at $t=6$ hr, $V=2500$ ft$^3$ and $h=5$ ft, since half the tank has drained out.



Solving for $k$: $$2500=k(5)(6)+0$$ $$k=frac2503$$
And solving for $h$ when $fracdVdt=100fracft^3hr$:$$100=frac2503h$$ $$h=1.2ft$$
However, the answer is actually about $1.73ft$. Where is my mistake?










share|cite|improve this question









$endgroup$
















    6












    $begingroup$


    I was told that I have a cylindrical water tank $10$ ft tall that can store $5000 $ ft$^3$ of water, and that the water drains from the bottom of the tank at a rate proportional to the instantaneous water level. After $6$ hours, half of the tank has drained out from the bottom. If the drain is opened and water is added to the tank at a constant rate of $100$ $fracft^3hr$, at what height above the drain will the elevation remain constant?



    My attempt:



    We are given that $$fracdVdt=kh$$ where $fracdVdt$ is the rate at which water leaves the tank, $V$ is the total change in water that has left the tank, and $h$ is the water's height above the drain.



    If the elevation must remain constant in the tank, then I am merely finding the height in which the water that enters the tank equals to the water that leaves the tank, and therefore $fracdVdt=100fracft^3hr$.



    To find $k$, I must do separation of variables and integrate both sides, so: $$intdV=intkhdt$$ $$V=kht+C$$
    I know that $C=0$ because at $t=0$ hr, no water has left the tank ($V=0$), and at $t=6$ hr, $V=2500$ ft$^3$ and $h=5$ ft, since half the tank has drained out.



    Solving for $k$: $$2500=k(5)(6)+0$$ $$k=frac2503$$
    And solving for $h$ when $fracdVdt=100fracft^3hr$:$$100=frac2503h$$ $$h=1.2ft$$
    However, the answer is actually about $1.73ft$. Where is my mistake?










    share|cite|improve this question









    $endgroup$














      6












      6








      6





      $begingroup$


      I was told that I have a cylindrical water tank $10$ ft tall that can store $5000 $ ft$^3$ of water, and that the water drains from the bottom of the tank at a rate proportional to the instantaneous water level. After $6$ hours, half of the tank has drained out from the bottom. If the drain is opened and water is added to the tank at a constant rate of $100$ $fracft^3hr$, at what height above the drain will the elevation remain constant?



      My attempt:



      We are given that $$fracdVdt=kh$$ where $fracdVdt$ is the rate at which water leaves the tank, $V$ is the total change in water that has left the tank, and $h$ is the water's height above the drain.



      If the elevation must remain constant in the tank, then I am merely finding the height in which the water that enters the tank equals to the water that leaves the tank, and therefore $fracdVdt=100fracft^3hr$.



      To find $k$, I must do separation of variables and integrate both sides, so: $$intdV=intkhdt$$ $$V=kht+C$$
      I know that $C=0$ because at $t=0$ hr, no water has left the tank ($V=0$), and at $t=6$ hr, $V=2500$ ft$^3$ and $h=5$ ft, since half the tank has drained out.



      Solving for $k$: $$2500=k(5)(6)+0$$ $$k=frac2503$$
      And solving for $h$ when $fracdVdt=100fracft^3hr$:$$100=frac2503h$$ $$h=1.2ft$$
      However, the answer is actually about $1.73ft$. Where is my mistake?










      share|cite|improve this question









      $endgroup$




      I was told that I have a cylindrical water tank $10$ ft tall that can store $5000 $ ft$^3$ of water, and that the water drains from the bottom of the tank at a rate proportional to the instantaneous water level. After $6$ hours, half of the tank has drained out from the bottom. If the drain is opened and water is added to the tank at a constant rate of $100$ $fracft^3hr$, at what height above the drain will the elevation remain constant?



      My attempt:



      We are given that $$fracdVdt=kh$$ where $fracdVdt$ is the rate at which water leaves the tank, $V$ is the total change in water that has left the tank, and $h$ is the water's height above the drain.



      If the elevation must remain constant in the tank, then I am merely finding the height in which the water that enters the tank equals to the water that leaves the tank, and therefore $fracdVdt=100fracft^3hr$.



      To find $k$, I must do separation of variables and integrate both sides, so: $$intdV=intkhdt$$ $$V=kht+C$$
      I know that $C=0$ because at $t=0$ hr, no water has left the tank ($V=0$), and at $t=6$ hr, $V=2500$ ft$^3$ and $h=5$ ft, since half the tank has drained out.



      Solving for $k$: $$2500=k(5)(6)+0$$ $$k=frac2503$$
      And solving for $h$ when $fracdVdt=100fracft^3hr$:$$100=frac2503h$$ $$h=1.2ft$$
      However, the answer is actually about $1.73ft$. Where is my mistake?







      calculus






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




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      asked 3 hours ago









      Niwde AupNiwde Aup

      534




      534




















          2 Answers
          2






          active

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          1












          $begingroup$

          Your mistake is in the solution of differential equation and in the initial boundary condition.



          see the following solution:
          $$fracdVdt=kh$$
          $$Afracdhdt=kh$$
          $$500fracdhh=kdt$$
          integrate
          $$500log h=kt+C$$
          at $t=0, h=10ft$



          so $$C=1151.3$$
          at $t=6 hr, h=5 ft$
          $$k=-57.76$$
          now use that
          $$fracdVdt=-57.76h$$
          $$-100=-57.76h$$
          so
          $$h=1.73 ft$$






          share|cite|improve this answer











          $endgroup$




















            0












            $begingroup$

            We have a differential equation that defines the rate that water leaves the drain.



            $fracdxdt = -ax\
            x = C e^-at$



            We know the initial conditions and after 6 hours to find our constants.



            $x(0) = 10$



            This gives us $C$



            $x(6) = 5\
            5 = 10e^-6a\
            6a = ln 2\
            a = fracln26$



            What do we know about the dimensions of the tank?



            $v = pi r^2 x\
            pi r^2 = 500$



            The flow out the drain equals the flow into the tank.



            $fracdvdt = frac dvdxfracdxdt\
            500 frac dxdt = 100\
            500(fracln 26)x = 100\
            x = frac65ln 2$






            share|cite|improve this answer









            $endgroup$













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              2 Answers
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              2 Answers
              2






              active

              oldest

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              active

              oldest

              votes






              active

              oldest

              votes









              1












              $begingroup$

              Your mistake is in the solution of differential equation and in the initial boundary condition.



              see the following solution:
              $$fracdVdt=kh$$
              $$Afracdhdt=kh$$
              $$500fracdhh=kdt$$
              integrate
              $$500log h=kt+C$$
              at $t=0, h=10ft$



              so $$C=1151.3$$
              at $t=6 hr, h=5 ft$
              $$k=-57.76$$
              now use that
              $$fracdVdt=-57.76h$$
              $$-100=-57.76h$$
              so
              $$h=1.73 ft$$






              share|cite|improve this answer











              $endgroup$

















                1












                $begingroup$

                Your mistake is in the solution of differential equation and in the initial boundary condition.



                see the following solution:
                $$fracdVdt=kh$$
                $$Afracdhdt=kh$$
                $$500fracdhh=kdt$$
                integrate
                $$500log h=kt+C$$
                at $t=0, h=10ft$



                so $$C=1151.3$$
                at $t=6 hr, h=5 ft$
                $$k=-57.76$$
                now use that
                $$fracdVdt=-57.76h$$
                $$-100=-57.76h$$
                so
                $$h=1.73 ft$$






                share|cite|improve this answer











                $endgroup$















                  1












                  1








                  1





                  $begingroup$

                  Your mistake is in the solution of differential equation and in the initial boundary condition.



                  see the following solution:
                  $$fracdVdt=kh$$
                  $$Afracdhdt=kh$$
                  $$500fracdhh=kdt$$
                  integrate
                  $$500log h=kt+C$$
                  at $t=0, h=10ft$



                  so $$C=1151.3$$
                  at $t=6 hr, h=5 ft$
                  $$k=-57.76$$
                  now use that
                  $$fracdVdt=-57.76h$$
                  $$-100=-57.76h$$
                  so
                  $$h=1.73 ft$$






                  share|cite|improve this answer











                  $endgroup$



                  Your mistake is in the solution of differential equation and in the initial boundary condition.



                  see the following solution:
                  $$fracdVdt=kh$$
                  $$Afracdhdt=kh$$
                  $$500fracdhh=kdt$$
                  integrate
                  $$500log h=kt+C$$
                  at $t=0, h=10ft$



                  so $$C=1151.3$$
                  at $t=6 hr, h=5 ft$
                  $$k=-57.76$$
                  now use that
                  $$fracdVdt=-57.76h$$
                  $$-100=-57.76h$$
                  so
                  $$h=1.73 ft$$







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited 2 hours ago

























                  answered 2 hours ago









                  E.H.EE.H.E

                  17k11969




                  17k11969





















                      0












                      $begingroup$

                      We have a differential equation that defines the rate that water leaves the drain.



                      $fracdxdt = -ax\
                      x = C e^-at$



                      We know the initial conditions and after 6 hours to find our constants.



                      $x(0) = 10$



                      This gives us $C$



                      $x(6) = 5\
                      5 = 10e^-6a\
                      6a = ln 2\
                      a = fracln26$



                      What do we know about the dimensions of the tank?



                      $v = pi r^2 x\
                      pi r^2 = 500$



                      The flow out the drain equals the flow into the tank.



                      $fracdvdt = frac dvdxfracdxdt\
                      500 frac dxdt = 100\
                      500(fracln 26)x = 100\
                      x = frac65ln 2$






                      share|cite|improve this answer









                      $endgroup$

















                        0












                        $begingroup$

                        We have a differential equation that defines the rate that water leaves the drain.



                        $fracdxdt = -ax\
                        x = C e^-at$



                        We know the initial conditions and after 6 hours to find our constants.



                        $x(0) = 10$



                        This gives us $C$



                        $x(6) = 5\
                        5 = 10e^-6a\
                        6a = ln 2\
                        a = fracln26$



                        What do we know about the dimensions of the tank?



                        $v = pi r^2 x\
                        pi r^2 = 500$



                        The flow out the drain equals the flow into the tank.



                        $fracdvdt = frac dvdxfracdxdt\
                        500 frac dxdt = 100\
                        500(fracln 26)x = 100\
                        x = frac65ln 2$






                        share|cite|improve this answer









                        $endgroup$















                          0












                          0








                          0





                          $begingroup$

                          We have a differential equation that defines the rate that water leaves the drain.



                          $fracdxdt = -ax\
                          x = C e^-at$



                          We know the initial conditions and after 6 hours to find our constants.



                          $x(0) = 10$



                          This gives us $C$



                          $x(6) = 5\
                          5 = 10e^-6a\
                          6a = ln 2\
                          a = fracln26$



                          What do we know about the dimensions of the tank?



                          $v = pi r^2 x\
                          pi r^2 = 500$



                          The flow out the drain equals the flow into the tank.



                          $fracdvdt = frac dvdxfracdxdt\
                          500 frac dxdt = 100\
                          500(fracln 26)x = 100\
                          x = frac65ln 2$






                          share|cite|improve this answer









                          $endgroup$



                          We have a differential equation that defines the rate that water leaves the drain.



                          $fracdxdt = -ax\
                          x = C e^-at$



                          We know the initial conditions and after 6 hours to find our constants.



                          $x(0) = 10$



                          This gives us $C$



                          $x(6) = 5\
                          5 = 10e^-6a\
                          6a = ln 2\
                          a = fracln26$



                          What do we know about the dimensions of the tank?



                          $v = pi r^2 x\
                          pi r^2 = 500$



                          The flow out the drain equals the flow into the tank.



                          $fracdvdt = frac dvdxfracdxdt\
                          500 frac dxdt = 100\
                          500(fracln 26)x = 100\
                          x = frac65ln 2$







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered 2 hours ago









                          Doug MDoug M

                          2,016512




                          2,016512



























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