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Does differentiating volume give surface area?
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Hi there,
In this question, you have to find the rate at which the surface area of the water is increasing.
I understand up to the step where you find the volume of the truncated pyramid.
But how come the surface area formula is the same for the differentiated formula for the pyramid? Are they supposed to be the same?
Thanks
calculus differential-equations
edited Jul 28 at 0:33
user573497
2009
asked Jul 28 at 0:04
user21808
123
add a comment |Â
up vote
0
down vote
favorite
Hi there,
In this question, you have to find the rate at which the surface area of the water is increasing.
I understand up to the step where you find the volume of the truncated pyramid.
But how come the surface area formula is the same for the differentiated formula for the pyramid? Are they supposed to be the same?
Thanks
calculus differential-equations
edited Jul 28 at 0:33
user573497
2009
asked Jul 28 at 0:04
user21808
123
For volumes, the derivative of the volume of a sphere gives you exactly the surface of it. For areas, the derivative of the area of the circle is its circunference. However, It doesn't work for every linear measure you use as independent variable of the volume function. Take a cube for example, its volume is $s^3$ for its side $s$. The derivative of the volume(w.r.t side) is $3s^2$, which is only half the real surface area.
– Cristhian Grundmann
Jul 28 at 0:16
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Hi there,
In this question, you have to find the rate at which the surface area of the water is increasing.
I understand up to the step where you find the volume of the truncated pyramid.
But how come the surface area formula is the same for the differentiated formula for the pyramid? Are they supposed to be the same?
Thanks
calculus differential-equations
edited Jul 28 at 0:33
user573497
2009
asked Jul 28 at 0:04
user21808
123
Hi there,
In this question, you have to find the rate at which the surface area of the water is increasing.
I understand up to the step where you find the volume of the truncated pyramid.
But how come the surface area formula is the same for the differentiated formula for the pyramid? Are they supposed to be the same?
Thanks
calculus differential-equations
edited Jul 28 at 0:33
user573497
2009
asked Jul 28 at 0:04
user21808
123
edited Jul 28 at 0:33
user573497
2009
edited Jul 28 at 0:33
user573497
2009
edited Jul 28 at 0:33
user573497
2009
2009
asked Jul 28 at 0:04
user21808
123
asked Jul 28 at 0:04
user21808
123
asked Jul 28 at 0:04
user21808
123
123
For volumes, the derivative of the volume of a sphere gives you exactly the surface of it. For areas, the derivative of the area of the circle is its circunference. However, It doesn't work for every linear measure you use as independent variable of the volume function. Take a cube for example, its volume is $s^3$ for its side $s$. The derivative of the volume(w.r.t side) is $3s^2$, which is only half the real surface area.
– Cristhian Grundmann
Jul 28 at 0:16
add a comment |Â
For volumes, the derivative of the volume of a sphere gives you exactly the surface of it. For areas, the derivative of the area of the circle is its circunference. However, It doesn't work for every linear measure you use as independent variable of the volume function. Take a cube for example, its volume is $s^3$ for its side $s$. The derivative of the volume(w.r.t side) is $3s^2$, which is only half the real surface area.
– Cristhian Grundmann
Jul 28 at 0:16
For volumes, the derivative of the volume of a sphere gives you exactly the surface of it. For areas, the derivative of the area of the circle is its circunference. However, It doesn't work for every linear measure you use as independent variable of the volume function. Take a cube for example, its volume is $s^3$ for its side $s$. The derivative of the volume(w.r.t side) is $3s^2$, which is only half the real surface area.
– Cristhian Grundmann
Jul 28 at 0:16
For volumes, the derivative of the volume of a sphere gives you exactly the surface of it. For areas, the derivative of the area of the circle is its circunference. However, It doesn't work for every linear measure you use as independent variable of the volume function. Take a cube for example, its volume is $s^3$ for its side $s$. The derivative of the volume(w.r.t side) is $3s^2$, which is only half the real surface area.
– Cristhian Grundmann
Jul 28 at 0:16
add a comment |Â
1 Answer
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Your question relates to the fundamental theorem of calculus, which you probably haven't learned yet.
As you add water to the vessel you can think of it filling out that red layer on the surface. The thickness of the surface is small. So, the change in volume is proportional to the area of the surface.
I won't say that the change in volume is always the surface area, because it does depend a little bit on how you set it up and which surfaces you are comparing to. i.e. The lateral surfaces are not be considered here.
But frequently the change in volume is the surface area.
answered Jul 28 at 0:34
Doug M
39.1k31749
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Your question relates to the fundamental theorem of calculus, which you probably haven't learned yet.
As you add water to the vessel you can think of it filling out that red layer on the surface. The thickness of the surface is small. So, the change in volume is proportional to the area of the surface.
I won't say that the change in volume is always the surface area, because it does depend a little bit on how you set it up and which surfaces you are comparing to. i.e. The lateral surfaces are not be considered here.
But frequently the change in volume is the surface area.
answered Jul 28 at 0:34
Doug M
39.1k31749
add a comment |Â
up vote
1
down vote
accepted
Your question relates to the fundamental theorem of calculus, which you probably haven't learned yet.
As you add water to the vessel you can think of it filling out that red layer on the surface. The thickness of the surface is small. So, the change in volume is proportional to the area of the surface.
I won't say that the change in volume is always the surface area, because it does depend a little bit on how you set it up and which surfaces you are comparing to. i.e. The lateral surfaces are not be considered here.
But frequently the change in volume is the surface area.
answered Jul 28 at 0:34
Doug M
39.1k31749
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Your question relates to the fundamental theorem of calculus, which you probably haven't learned yet.
As you add water to the vessel you can think of it filling out that red layer on the surface. The thickness of the surface is small. So, the change in volume is proportional to the area of the surface.
I won't say that the change in volume is always the surface area, because it does depend a little bit on how you set it up and which surfaces you are comparing to. i.e. The lateral surfaces are not be considered here.
But frequently the change in volume is the surface area.
answered Jul 28 at 0:34
Doug M
39.1k31749
Your question relates to the fundamental theorem of calculus, which you probably haven't learned yet.
As you add water to the vessel you can think of it filling out that red layer on the surface. The thickness of the surface is small. So, the change in volume is proportional to the area of the surface.
I won't say that the change in volume is always the surface area, because it does depend a little bit on how you set it up and which surfaces you are comparing to. i.e. The lateral surfaces are not be considered here.
But frequently the change in volume is the surface area.
answered Jul 28 at 0:34
Doug M
39.1k31749
answered Jul 28 at 0:34
Doug M
39.1k31749
answered Jul 28 at 0:34
Doug M
39.1k31749
answered Jul 28 at 0:34
Doug M
39.1k31749
39.1k31749
add a comment |Â
add a comment |Â
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For volumes, the derivative of the volume of a sphere gives you exactly the surface of it. For areas, the derivative of the area of the circle is its circunference. However, It doesn't work for every linear measure you use as independent variable of the volume function. Take a cube for example, its volume is $s^3$ for its side $s$. The derivative of the volume(w.r.t side) is $3s^2$, which is only half the real surface area.
– Cristhian Grundmann
Jul 28 at 0:16