Rotational volume and differential equation
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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
asked Jan 3 at 20:26
MicaeleMicaele
32
$endgroup$
add a comment |
$begingroup$
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
asked Jan 3 at 20:26
MicaeleMicaele
32
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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
asked Jan 3 at 20:26
MicaeleMicaele
32
$endgroup$
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
asked Jan 3 at 20:26
MicaeleMicaele
32
asked Jan 3 at 20:26
MicaeleMicaele
32
asked Jan 3 at 20:26
MicaeleMicaele
32
asked Jan 3 at 20:26
MicaeleMicaele
32
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
add a comment |
$begingroup$
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
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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
$$ y = f(x) = left(frac{cpi}{k}right)^2x^4 = Kx^4 $$
answered Jan 4 at 7:52
DylanDylan
14.2k31127
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add a comment |
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1 Answer
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votes
1 Answer
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active
oldest
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votes
$begingroup$
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
$$ y = f(x) = left(frac{cpi}{k}right)^2x^4 = Kx^4 $$
answered Jan 4 at 7:52
DylanDylan
14.2k31127
$endgroup$
add a comment |
$begingroup$
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
$$ y = f(x) = left(frac{cpi}{k}right)^2x^4 = Kx^4 $$
answered Jan 4 at 7:52
DylanDylan
14.2k31127
$endgroup$
add a comment |
$begingroup$
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
$$ y = f(x) = left(frac{cpi}{k}right)^2x^4 = Kx^4 $$
answered Jan 4 at 7:52
DylanDylan
14.2k31127
$endgroup$
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
$$ y = f(x) = left(frac{cpi}{k}right)^2x^4 = Kx^4 $$
answered Jan 4 at 7:52
DylanDylan
14.2k31127
answered Jan 4 at 7:52
DylanDylan
14.2k31127
answered Jan 4 at 7:52
DylanDylan
14.2k31127
answered Jan 4 at 7:52
DylanDylan
14.2k31127
14.2k31127
add a comment |
add a comment |
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$begingroup$
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