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

How do Java 8 default methods hеlp with lambdas?

One-one communication

How to make triangles with rounded sides and corners? (squircle with 3 sides)

How can I prevent/balance waiting and turtling as a response to cooldown mechanics

Why does BitLocker not use RSA?

malloc in main() or malloc in another function: allocating memory for a struct and its members

Does the main washing effect of soap come from foam?

Understanding piped commands in GNU/Linux

Should man-made satellites feature an intelligent inverted "cow catcher"?

Are there any irrational/transcendental numbers for which the distribution of decimal digits is not uniform?

Vertical ranges of Column Plots in 12

Is this Half-dragon Quaggoth boss monster balanced?

Noise in Eigenvalues plot

What is a more techy Technical Writer job title that isn't cutesy or confusing?

NIntegrate on a solution of a matrix ODE

Is it OK to use the testing sample to compare algorithms?

How does the body cool itself in a stillsuit?

Pointing to problems without suggesting solutions

Did pre-Columbian Americans know the spherical shape of the Earth?

Flight departed from the gate 5 min before scheduled departure time. Refund options

How could a hydrazine and N2O4 cloud (or it's reactants) show up in weather radar?

As a dual citizen, my US passport will expire one day after traveling to the US. Will this work?

Twin's vs. Twins'

The test team as an enemy of development? And how can this be avoided?



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




      share|cite|improve this question










      asked 3 hours ago









      Niwde AupNiwde Aup

      534




      534




















          2 Answers
          2






          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











          $endgroup$




















            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









            $endgroup$













              Your Answer








              StackExchange.ready(function()
              var channelOptions =
              tags: "".split(" "),
              id: "69"
              ;
              initTagRenderer("".split(" "), "".split(" "), channelOptions);

              StackExchange.using("externalEditor", function()
              // Have to fire editor after snippets, if snippets enabled
              if (StackExchange.settings.snippets.snippetsEnabled)
              StackExchange.using("snippets", function()
              createEditor();
              );

              else
              createEditor();

              );

              function createEditor()
              StackExchange.prepareEditor(
              heartbeatType: 'answer',
              autoActivateHeartbeat: false,
              convertImagesToLinks: true,
              noModals: true,
              showLowRepImageUploadWarning: true,
              reputationToPostImages: 10,
              bindNavPrevention: true,
              postfix: "",
              imageUploader:
              brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
              contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
              allowUrls: true
              ,
              noCode: true, onDemand: true,
              discardSelector: ".discard-answer"
              ,immediatelyShowMarkdownHelp:true
              );



              );













              draft saved

              draft discarded


















              StackExchange.ready(
              function ()
              StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3196337%2ftypical-calculus-bc-separation-of-variables-question%23new-answer', 'question_page');

              );

              Post as a guest















              Required, but never shown

























              2 Answers
              2






              active

              oldest

              votes








              2 Answers
              2






              active

              oldest

              votes









              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



























                              draft saved

                              draft discarded
















































                              Thanks for contributing an answer to Mathematics Stack Exchange!


                              • Please be sure to answer the question. Provide details and share your research!

                              But avoid


                              • Asking for help, clarification, or responding to other answers.

                              • Making statements based on opinion; back them up with references or personal experience.

                              Use MathJax to format equations. MathJax reference.


                              To learn more, see our tips on writing great answers.




                              draft saved


                              draft discarded














                              StackExchange.ready(
                              function ()
                              StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3196337%2ftypical-calculus-bc-separation-of-variables-question%23new-answer', 'question_page');

                              );

                              Post as a guest















                              Required, but never shown





















































                              Required, but never shown














                              Required, but never shown












                              Required, but never shown







                              Required, but never shown

































                              Required, but never shown














                              Required, but never shown












                              Required, but never shown







                              Required, but never shown







                              Popular posts from this blog

                              Oświęcim Innehåll Historia | Källor | Externa länkar | Navigeringsmeny50°2′18″N 19°13′17″Ö / 50.03833°N 19.22139°Ö / 50.03833; 19.2213950°2′18″N 19°13′17″Ö / 50.03833°N 19.22139°Ö / 50.03833; 19.221393089658Nordisk familjebok, AuschwitzInsidan tro och existensJewish Community i OświęcimAuschwitz Jewish Center: MuseumAuschwitz Jewish Center

                              Valle di Casies Indice Geografia fisica | Origini del nome | Storia | Società | Amministrazione | Sport | Note | Bibliografia | Voci correlate | Altri progetti | Collegamenti esterni | Menu di navigazione46°46′N 12°11′E / 46.766667°N 12.183333°E46.766667; 12.183333 (Valle di Casies)46°46′N 12°11′E / 46.766667°N 12.183333°E46.766667; 12.183333 (Valle di Casies)Sito istituzionaleAstat Censimento della popolazione 2011 - Determinazione della consistenza dei tre gruppi linguistici della Provincia Autonoma di Bolzano-Alto Adige - giugno 2012Numeri e fattiValle di CasiesDato IstatTabella dei gradi/giorno dei Comuni italiani raggruppati per Regione e Provincia26 agosto 1993, n. 412Heraldry of the World: GsiesStatistiche I.StatValCasies.comWikimedia CommonsWikimedia CommonsValle di CasiesSito ufficialeValle di CasiesMM14870458910042978-6

                              Typsetting diagram chases (with TikZ?) Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)How to define the default vertical distance between nodes?Draw edge on arcNumerical conditional within tikz keys?TikZ: Drawing an arc from an intersection to an intersectionDrawing rectilinear curves in Tikz, aka an Etch-a-Sketch drawingLine up nested tikz enviroments or how to get rid of themHow to place nodes in an absolute coordinate system in tikzCommutative diagram with curve connecting between nodesTikz with standalone: pinning tikz coordinates to page cmDrawing a Decision Diagram with Tikz and layout manager