Rotational volume and differential equation












0












$begingroup$


A container with the shape of some function y=f(x) is rotated around the y-axis. It's filled with a fluid and has a hole in the bottom where the fluid leaks out. The rate of liquid flowing out should be proportional to the square root of the height of liquid in the container.
Determine the function f(x) such that the height of liquid in the container has a constant negative slope.



$v(t)$ is the volume of water in the container at time t
$v'(t)$ is the rate of change in volume at time t
$h(t)$ is the height of the water level in the container.
$h'(t)$ is the rate of change of the water level and its constant.



$v'(t)=-k*sqrt{h(t)}$
$v(t)=piint^{h(t)}_{0}f^2(x) dx$

I feel as if maybe the fundamental theorem of calculus would get involved here.
$v'(t)=pi f^2(h(t))$



But I don't really get anywhere from here. Am I going at this from the wrong direction?



The answer is supposed to be $f(x)=Kx^4$, where K is a constant






calculus integration volume differential







share|cite|improve this question









$endgroup$












  • $begingroup$
    You've set this up as if the rotation is about the $x$-axis. I think your second equation should be integral of $2pi f.$
    $endgroup$
    – B. Goddard
    Jan 3 at 20:37










  • $begingroup$
    That's right, that's something that needs to change. But if I don't have an equation, how can I express x in forms of y? All I know is that there is a function y=f(x).
    $endgroup$
    – Micaele
    Jan 3 at 20:44
















0












$begingroup$


A container with the shape of some function y=f(x) is rotated around the y-axis. It's filled with a fluid and has a hole in the bottom where the fluid leaks out. The rate of liquid flowing out should be proportional to the square root of the height of liquid in the container.
Determine the function f(x) such that the height of liquid in the container has a constant negative slope.



$v(t)$ is the volume of water in the container at time t
$v'(t)$ is the rate of change in volume at time t
$h(t)$ is the height of the water level in the container.
$h'(t)$ is the rate of change of the water level and its constant.



$v'(t)=-k*sqrt{h(t)}$
$v(t)=piint^{h(t)}_{0}f^2(x) dx$

I feel as if maybe the fundamental theorem of calculus would get involved here.
$v'(t)=pi f^2(h(t))$



But I don't really get anywhere from here. Am I going at this from the wrong direction?



The answer is supposed to be $f(x)=Kx^4$, where K is a constant






calculus integration volume differential







share|cite|improve this question









$endgroup$












  • $begingroup$
    You've set this up as if the rotation is about the $x$-axis. I think your second equation should be integral of $2pi f.$
    $endgroup$
    – B. Goddard
    Jan 3 at 20:37










  • $begingroup$
    That's right, that's something that needs to change. But if I don't have an equation, how can I express x in forms of y? All I know is that there is a function y=f(x).
    $endgroup$
    – Micaele
    Jan 3 at 20:44














0












0








0





$begingroup$


A container with the shape of some function y=f(x) is rotated around the y-axis. It's filled with a fluid and has a hole in the bottom where the fluid leaks out. The rate of liquid flowing out should be proportional to the square root of the height of liquid in the container.
Determine the function f(x) such that the height of liquid in the container has a constant negative slope.



$v(t)$ is the volume of water in the container at time t
$v'(t)$ is the rate of change in volume at time t
$h(t)$ is the height of the water level in the container.
$h'(t)$ is the rate of change of the water level and its constant.



$v'(t)=-k*sqrt{h(t)}$
$v(t)=piint^{h(t)}_{0}f^2(x) dx$

I feel as if maybe the fundamental theorem of calculus would get involved here.
$v'(t)=pi f^2(h(t))$



But I don't really get anywhere from here. Am I going at this from the wrong direction?



The answer is supposed to be $f(x)=Kx^4$, where K is a constant






calculus integration volume differential







share|cite|improve this question









$endgroup$




A container with the shape of some function y=f(x) is rotated around the y-axis. It's filled with a fluid and has a hole in the bottom where the fluid leaks out. The rate of liquid flowing out should be proportional to the square root of the height of liquid in the container.
Determine the function f(x) such that the height of liquid in the container has a constant negative slope.



$v(t)$ is the volume of water in the container at time t
$v'(t)$ is the rate of change in volume at time t
$h(t)$ is the height of the water level in the container.
$h'(t)$ is the rate of change of the water level and its constant.



$v'(t)=-k*sqrt{h(t)}$
$v(t)=piint^{h(t)}_{0}f^2(x) dx$

I feel as if maybe the fundamental theorem of calculus would get involved here.
$v'(t)=pi f^2(h(t))$



But I don't really get anywhere from here. Am I going at this from the wrong direction?



The answer is supposed to be $f(x)=Kx^4$, where K is a constant






calculus integration volume differential




calculus integration volume differential






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 3 at 20:26









MicaeleMicaele

32




32












  • $begingroup$
    You've set this up as if the rotation is about the $x$-axis. I think your second equation should be integral of $2pi f.$
    $endgroup$
    – B. Goddard
    Jan 3 at 20:37










  • $begingroup$
    That's right, that's something that needs to change. But if I don't have an equation, how can I express x in forms of y? All I know is that there is a function y=f(x).
    $endgroup$
    – Micaele
    Jan 3 at 20:44


















  • $begingroup$
    You've set this up as if the rotation is about the $x$-axis. I think your second equation should be integral of $2pi f.$
    $endgroup$
    – B. Goddard
    Jan 3 at 20:37










  • $begingroup$
    That's right, that's something that needs to change. But if I don't have an equation, how can I express x in forms of y? All I know is that there is a function y=f(x).
    $endgroup$
    – Micaele
    Jan 3 at 20:44
















$begingroup$
You've set this up as if the rotation is about the $x$-axis. I think your second equation should be integral of $2pi f.$
$endgroup$
– B. Goddard
Jan 3 at 20:37




$begingroup$
You've set this up as if the rotation is about the $x$-axis. I think your second equation should be integral of $2pi f.$
$endgroup$
– B. Goddard
Jan 3 at 20:37












$begingroup$
That's right, that's something that needs to change. But if I don't have an equation, how can I express x in forms of y? All I know is that there is a function y=f(x).
$endgroup$
– Micaele
Jan 3 at 20:44




$begingroup$
That's right, that's something that needs to change. But if I don't have an equation, how can I express x in forms of y? All I know is that there is a function y=f(x).
$endgroup$
– Micaele
Jan 3 at 20:44










1 Answer
1






active

oldest

votes


















0












$begingroup$

Let $x = g(y)$ be the inverse function of $y=f(x)$. The volume of the liquid, w.r.t the height $y$ is given by



$$ V(y) = pi int_0^y g(s)^2 ds$$



By the fundamental theorem of calculus:



$$ frac{dV}{dy} = pi g(y)^2 $$



By the chain rule



$$ frac{dV}{dt} = frac{dV}{dy}frac{dy}{dt} = -cpi g(y)^2 $$



where $frac{dy}{dt}=-c$ is some negative constant



It is also given that



$$ frac{dV}{dt} = -ksqrt{y} $$



Therefore



$$ -cpi g(y)^2 = -ksqrt{y} implies x = g(y) = sqrt{frac{k}{cpi}}y^{1/4} $$



Inverting gives



$$ y = f(x) = left(frac{cpi}{k}right)^2x^4 = Kx^4 $$






share|cite|improve this answer









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    1 Answer
    1






    active

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    active

    oldest

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    active

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    0












    $begingroup$

    Let $x = g(y)$ be the inverse function of $y=f(x)$. The volume of the liquid, w.r.t the height $y$ is given by



    $$ V(y) = pi int_0^y g(s)^2 ds$$



    By the fundamental theorem of calculus:



    $$ frac{dV}{dy} = pi g(y)^2 $$



    By the chain rule



    $$ frac{dV}{dt} = frac{dV}{dy}frac{dy}{dt} = -cpi g(y)^2 $$



    where $frac{dy}{dt}=-c$ is some negative constant



    It is also given that



    $$ frac{dV}{dt} = -ksqrt{y} $$



    Therefore



    $$ -cpi g(y)^2 = -ksqrt{y} implies x = g(y) = sqrt{frac{k}{cpi}}y^{1/4} $$



    Inverting gives



    $$ y = f(x) = left(frac{cpi}{k}right)^2x^4 = Kx^4 $$






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      Let $x = g(y)$ be the inverse function of $y=f(x)$. The volume of the liquid, w.r.t the height $y$ is given by



      $$ V(y) = pi int_0^y g(s)^2 ds$$



      By the fundamental theorem of calculus:



      $$ frac{dV}{dy} = pi g(y)^2 $$



      By the chain rule



      $$ frac{dV}{dt} = frac{dV}{dy}frac{dy}{dt} = -cpi g(y)^2 $$



      where $frac{dy}{dt}=-c$ is some negative constant



      It is also given that



      $$ frac{dV}{dt} = -ksqrt{y} $$



      Therefore



      $$ -cpi g(y)^2 = -ksqrt{y} implies x = g(y) = sqrt{frac{k}{cpi}}y^{1/4} $$



      Inverting gives



      $$ y = f(x) = left(frac{cpi}{k}right)^2x^4 = Kx^4 $$






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        Let $x = g(y)$ be the inverse function of $y=f(x)$. The volume of the liquid, w.r.t the height $y$ is given by



        $$ V(y) = pi int_0^y g(s)^2 ds$$



        By the fundamental theorem of calculus:



        $$ frac{dV}{dy} = pi g(y)^2 $$



        By the chain rule



        $$ frac{dV}{dt} = frac{dV}{dy}frac{dy}{dt} = -cpi g(y)^2 $$



        where $frac{dy}{dt}=-c$ is some negative constant



        It is also given that



        $$ frac{dV}{dt} = -ksqrt{y} $$



        Therefore



        $$ -cpi g(y)^2 = -ksqrt{y} implies x = g(y) = sqrt{frac{k}{cpi}}y^{1/4} $$



        Inverting gives



        $$ y = f(x) = left(frac{cpi}{k}right)^2x^4 = Kx^4 $$






        share|cite|improve this answer









        $endgroup$



        Let $x = g(y)$ be the inverse function of $y=f(x)$. The volume of the liquid, w.r.t the height $y$ is given by



        $$ V(y) = pi int_0^y g(s)^2 ds$$



        By the fundamental theorem of calculus:



        $$ frac{dV}{dy} = pi g(y)^2 $$



        By the chain rule



        $$ frac{dV}{dt} = frac{dV}{dy}frac{dy}{dt} = -cpi g(y)^2 $$



        where $frac{dy}{dt}=-c$ is some negative constant



        It is also given that



        $$ frac{dV}{dt} = -ksqrt{y} $$



        Therefore



        $$ -cpi g(y)^2 = -ksqrt{y} implies x = g(y) = sqrt{frac{k}{cpi}}y^{1/4} $$



        Inverting gives



        $$ y = f(x) = left(frac{cpi}{k}right)^2x^4 = Kx^4 $$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 4 at 7:52









        DylanDylan

        14.2k31127




        14.2k31127






























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