How to Solve It: A Rate Problem
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I am currently reading How to Solve It by G. Polya. I am confused by the conclusion in one of his examples. You can find it here in this pdf on page 29.
The problems states: "Water is flowing into a conical vessel at the rate r. The vessel has the shape of a right circular cone, with horizontal base, the vertex pointing downwards; the radius of the base is a, the altitude of the cone b. Find the rate at which the surface is rising when the depth of the water is y. Finally obtain the numerical value of the unknown supposing that a = 4 ft., b = 3 ft., r = 2 cu ft. per minute, and y = 1 ft."
picture from the book
The ultimate solution comes out to be: $$frac{dV}{dt}=frac{pi a^2y^2}{b^2}frac{dy}{dt}$$
I am confused why the denominator is b^2? I get that this is mostly about find the volume as a function of y. I am just glitching on how this equation fulfills the solution.
geometry volume
asked Dec 18 '18 at 2:57
AaronAaron
12
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add a comment |
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I am currently reading How to Solve It by G. Polya. I am confused by the conclusion in one of his examples. You can find it here in this pdf on page 29.
The problems states: "Water is flowing into a conical vessel at the rate r. The vessel has the shape of a right circular cone, with horizontal base, the vertex pointing downwards; the radius of the base is a, the altitude of the cone b. Find the rate at which the surface is rising when the depth of the water is y. Finally obtain the numerical value of the unknown supposing that a = 4 ft., b = 3 ft., r = 2 cu ft. per minute, and y = 1 ft."
picture from the book
The ultimate solution comes out to be: $$frac{dV}{dt}=frac{pi a^2y^2}{b^2}frac{dy}{dt}$$
I am confused why the denominator is b^2? I get that this is mostly about find the volume as a function of y. I am just glitching on how this equation fulfills the solution.
geometry volume
asked Dec 18 '18 at 2:57
AaronAaron
12
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Or... if you have a physicist's bent: Notice that the units in the numerator are $length^5$ so to have a volume, you must divide that by a $length^2$, i.e., $b^2$.
$endgroup$
– David G. Stork
Dec 18 '18 at 3:15
add a comment |
$begingroup$
I am currently reading How to Solve It by G. Polya. I am confused by the conclusion in one of his examples. You can find it here in this pdf on page 29.
The problems states: "Water is flowing into a conical vessel at the rate r. The vessel has the shape of a right circular cone, with horizontal base, the vertex pointing downwards; the radius of the base is a, the altitude of the cone b. Find the rate at which the surface is rising when the depth of the water is y. Finally obtain the numerical value of the unknown supposing that a = 4 ft., b = 3 ft., r = 2 cu ft. per minute, and y = 1 ft."
picture from the book
The ultimate solution comes out to be: $$frac{dV}{dt}=frac{pi a^2y^2}{b^2}frac{dy}{dt}$$
I am confused why the denominator is b^2? I get that this is mostly about find the volume as a function of y. I am just glitching on how this equation fulfills the solution.
geometry volume
asked Dec 18 '18 at 2:57
AaronAaron
12
$endgroup$
I am currently reading How to Solve It by G. Polya. I am confused by the conclusion in one of his examples. You can find it here in this pdf on page 29.
The problems states: "Water is flowing into a conical vessel at the rate r. The vessel has the shape of a right circular cone, with horizontal base, the vertex pointing downwards; the radius of the base is a, the altitude of the cone b. Find the rate at which the surface is rising when the depth of the water is y. Finally obtain the numerical value of the unknown supposing that a = 4 ft., b = 3 ft., r = 2 cu ft. per minute, and y = 1 ft."
picture from the book
The ultimate solution comes out to be: $$frac{dV}{dt}=frac{pi a^2y^2}{b^2}frac{dy}{dt}$$
I am confused why the denominator is b^2? I get that this is mostly about find the volume as a function of y. I am just glitching on how this equation fulfills the solution.
geometry volume
geometry volume
asked Dec 18 '18 at 2:57
AaronAaron
12
asked Dec 18 '18 at 2:57
AaronAaron
12
edited Dec 18 '18 at 17:41
Aaron
asked Dec 18 '18 at 2:57
AaronAaron
12
asked Dec 18 '18 at 2:57
AaronAaron
12
asked Dec 18 '18 at 2:57
AaronAaron
12
12
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Or... if you have a physicist's bent: Notice that the units in the numerator are $length^5$ so to have a volume, you must divide that by a $length^2$, i.e., $b^2$.
$endgroup$
– David G. Stork
Dec 18 '18 at 3:15
add a comment |
$begingroup$
Or... if you have a physicist's bent: Notice that the units in the numerator are $length^5$ so to have a volume, you must divide that by a $length^2$, i.e., $b^2$.
$endgroup$
– David G. Stork
Dec 18 '18 at 3:15
$begingroup$
Or... if you have a physicist's bent: Notice that the units in the numerator are $length^5$ so to have a volume, you must divide that by a $length^2$, i.e., $b^2$.
$endgroup$
– David G. Stork
Dec 18 '18 at 3:15
$begingroup$
Or... if you have a physicist's bent: Notice that the units in the numerator are $length^5$ so to have a volume, you must divide that by a $length^2$, i.e., $b^2$.
$endgroup$
– David G. Stork
Dec 18 '18 at 3:15
add a comment |
1 Answer
1
active
oldest
votes
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We have that $V=frac{1}{3}pi r^2h$. By similar triangles, $frac{h}{r}=frac{b}{a}$, so this is just $V=frac{1}{3}pifrac{a^2h^3}{b^2}$. Differentiating, $frac{dV}{dt}=frac{pi a^2h^2}{b^2}frac{dh}{dt}$, as desired.
answered Dec 18 '18 at 3:02
william122william122
54912
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So it is just a means of rewriting it without the 1/3 ?
$endgroup$
– Aaron
Dec 18 '18 at 17:37
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
We have that $V=frac{1}{3}pi r^2h$. By similar triangles, $frac{h}{r}=frac{b}{a}$, so this is just $V=frac{1}{3}pifrac{a^2h^3}{b^2}$. Differentiating, $frac{dV}{dt}=frac{pi a^2h^2}{b^2}frac{dh}{dt}$, as desired.
answered Dec 18 '18 at 3:02
william122william122
54912
$endgroup$
$begingroup$
So it is just a means of rewriting it without the 1/3 ?
$endgroup$
– Aaron
Dec 18 '18 at 17:37
add a comment |
$begingroup$
We have that $V=frac{1}{3}pi r^2h$. By similar triangles, $frac{h}{r}=frac{b}{a}$, so this is just $V=frac{1}{3}pifrac{a^2h^3}{b^2}$. Differentiating, $frac{dV}{dt}=frac{pi a^2h^2}{b^2}frac{dh}{dt}$, as desired.
answered Dec 18 '18 at 3:02
william122william122
54912
$endgroup$
$begingroup$
So it is just a means of rewriting it without the 1/3 ?
$endgroup$
– Aaron
Dec 18 '18 at 17:37
add a comment |
$begingroup$
We have that $V=frac{1}{3}pi r^2h$. By similar triangles, $frac{h}{r}=frac{b}{a}$, so this is just $V=frac{1}{3}pifrac{a^2h^3}{b^2}$. Differentiating, $frac{dV}{dt}=frac{pi a^2h^2}{b^2}frac{dh}{dt}$, as desired.
answered Dec 18 '18 at 3:02
william122william122
54912
$endgroup$
We have that $V=frac{1}{3}pi r^2h$. By similar triangles, $frac{h}{r}=frac{b}{a}$, so this is just $V=frac{1}{3}pifrac{a^2h^3}{b^2}$. Differentiating, $frac{dV}{dt}=frac{pi a^2h^2}{b^2}frac{dh}{dt}$, as desired.
answered Dec 18 '18 at 3:02
william122william122
54912
answered Dec 18 '18 at 3:02
william122william122
54912
answered Dec 18 '18 at 3:02
william122william122
54912
answered Dec 18 '18 at 3:02
william122william122
54912
54912
$begingroup$
So it is just a means of rewriting it without the 1/3 ?
$endgroup$
– Aaron
Dec 18 '18 at 17:37
add a comment |
$begingroup$
So it is just a means of rewriting it without the 1/3 ?
$endgroup$
– Aaron
Dec 18 '18 at 17:37
$begingroup$
So it is just a means of rewriting it without the 1/3 ?
$endgroup$
– Aaron
Dec 18 '18 at 17:37
$begingroup$
So it is just a means of rewriting it without the 1/3 ?
$endgroup$
– Aaron
Dec 18 '18 at 17:37
add a comment |
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$begingroup$
Or... if you have a physicist's bent: Notice that the units in the numerator are $length^5$ so to have a volume, you must divide that by a $length^2$, i.e., $b^2$.
$endgroup$
– David G. Stork
Dec 18 '18 at 3:15