Science > Physics > please help a girl again: An integral is an antidervative, so can'twe integrate the derivative to get back to the original equation?
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Science > Physics |
| User: |
"seven7sisters" |
| Date: |
01 Dec 2007 05:10:28 PM |
| Object: |
please help a girl again: An integral is an antidervative, so can'twe integrate the derivative to get back to the original equation? |
Suppose we have a function,
y=Cx
where C is a constant.
we take the derivative of both sides:
dy/dx = d/dx (Cx) = C
so dy/dx = C
Then supposed we want to get back to the original equation y=Cx via
integration.
How do we write this out?
An integral is an antidervative, so can't we integrate the derivative
to get back to the original equation?
.
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| User: "Androcles" |
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| Title: Re: please help a girl again: An integral is an antidervative, so can't we integrate the derivative to get back to the original equation? |
01 Dec 2007 07:28:12 PM |
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"seven7sisters" <seven7sisters@gmail.com> wrote in message
news:3dea29ef-25c5-4572-a935-ed40f8b8fb2f@a35g2000prf.googlegroups.com...
: Suppose we have a function,
:
: y=Cx
:
: where C is a constant.
:
: we take the derivative of both sides:
:
: dy/dx = d/dx (Cx) = C
:
: so dy/dx = C
:
: Then supposed we want to get back to the original equation y=Cx via
: integration.
:
: How do we write this out?
:
: An integral is an antidervative, so can't we integrate the derivative
: to get back to the original equation?
The original may have an offset, that is lost when you find the slope.
let y = mx + 2
dy/dx = m + 0
integral(dy/dx) = y = mx + (unknown constant)
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| User: "N:dlzc D:aol T:com \dlzc" |
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| Title: Re: please help a girl again: An integral is an antidervative, so can't we integrate the derivative to get back to the original equation? |
01 Dec 2007 05:22:30 PM |
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Dear seven7sisters:
"seven7sisters" <seven7sisters@gmail.com> wrote in message
news:3dea29ef-25c5-4572-a935-ed40f8b8fb2f@a35g2000prf.googlegroups.com...
Suppose we have a function,
y=Cx
where C is a constant.
we take the derivative of both sides:
dy/dx = d/dx (Cx) = C
so dy/dx = C
Then supposed we want to get back to the original
equation y=Cx via integration.
How do we write this out?
int( dy/dx * dx ) = int( C * dx | 0 <= x <= x )
An integral is an antidervative, so can't we integrate
the derivative to get back to the original equation?
Only if you place constraints on the "constant of integration" so
that it is equal to what you started with.
Note that the derivative-with-respect-to-x of
y = Cx
y = Cx + 1,000,000
y = Cx - 1,000,000
.... are all the same.
David A. Smith
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| User: "boho" |
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| Title: Re: please help a girl again: An integral is an antidervative, socan't we integrate the derivative to get back to the original equation? |
02 Dec 2007 04:14:17 AM |
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On Dec 2, 12:22 am, "N:dlzc D:aol T:com \(dlzc\)" <dl...@cox.net>
wrote:
Dear seven7sisters:
"seven7sisters" <seven7sist...@gmail.com> wrote in message
news:3dea29ef-25c5-4572-a935-ed40f8b8fb2f@a35g2000prf.googlegroups.com...
Suppose we have a function,
y=Cx
where C is a constant.
we take the derivative of both sides:
dy/dx = d/dx (Cx) = C
so dy/dx = C
Then supposed we want to get back to the original
equation y=Cx via integration.
How do we write this out?
int( dy/dx * dx ) = int( C * dx | 0 <= x <= x )
what is this crap
what is int
put it in code
An integral is an antidervative, so can't we integrate
the derivative to get back to the original equation?
Only if you place constraints on the "constant of integration" so
that it is equal to what you started with.
Note that the derivative-with-respect-to-x of
y = Cx
y = Cx + 1,000,000
y = Cx - 1,000,000
... are all the same.
David A. Smith
.
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| User: "" |
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| Title: Re: please help a girl again: An integral is an antidervative, socan't we integrate the derivative to get back to the original equation? |
01 Dec 2007 05:37:30 PM |
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On Dec 1, 11:10 pm, seven7sisters <seven7sist...@gmail.com> wrote:
Suppose we have a function,
y=Cx
where C is a constant.
we take the derivative of both sides:
dy/dx = d/dx (Cx) = C
so dy/dx = C
Then supposed we want to get back to the original equation y=Cx via
integration.
How do we write this out?
The usual way would be
dy/dx = C
dy = C*dx
Int dy = Int C*dx
y = C*x + D
where D is an arbitrary constant (there should theoretically be one on
the left hand side too, but they just combine into one). So, you don't
get back exactly what you started with; when you differentiate you
lose any information about additive constants (imagine shifting the
graph of y versus x by any amount in the y direction... the slope, dy/
dx, remains the same everywhere).
Strictly speaking you can't actually multiply through by dx in the
normal sense to get from dy/dx = C to dy = C*dx, because dx is not a
normal number. However, it turns out that this manipulation is valid
for these purposes, so, depending on how deeply you want to get into
the technicalities, you may feel that you can just accept it.
An integral is an antidervative, so can't we integrate the derivative
to get back to the original equation?
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| User: "boho" |
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| Title: Re: please help a girl again: An integral is an antidervative, socan't we integrate the derivative to get back to the original equation? |
02 Dec 2007 04:11:31 AM |
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On Dec 2, 12:37 am, wrote:
On Dec 1, 11:10 pm, seven7sisters <seven7sist...@gmail.com> wrote:
Suppose we have a function,
y=Cx
where C is a constant.
we take the derivative of both sides:
dy/dx = d/dx (Cx) = C
so dy/dx = C
Then supposed we want to get back to the original equation y=Cx via
integration.
How do we write this out?
The usual way would be
dy/dx = C
dy = C*dx
Int dy = Int C*dx
you use symbols you dont define
what is int?
put it in code, any language you want
pseudo code is okay
according to your *****, your lhs gets a dx, making
it a derivation, while your rhs dont gets anything
but transforms into a derivation, than you a int
therefore do some code in order for me to understan you
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| User: "J. M. Viani" |
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| Title: Re: please help a girl again: An integral is an antidervative, so can't we integrate the derivative to get back to the original equation? |
02 Dec 2007 06:31:12 AM |
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That's funny!
I really enjoy your candor.
-James
2007 02:11:31 -0800 (PST), boho <2g7y4h@contractor.net> wrote:
On Dec 2, 12:37 am, wrote:
On Dec 1, 11:10 pm, seven7sisters <seven7sist...@gmail.com> wrote:
Suppose we have a function,
y=Cx
where C is a constant.
we take the derivative of both sides:
dy/dx = d/dx (Cx) = C
so dy/dx = C
Then supposed we want to get back to the original equation y=Cx via
integration.
How do we write this out?
The usual way would be
dy/dx = C
dy = C*dx
Int dy = Int C*dx
you use symbols you dont define
what is int?
put it in code, any language you want
pseudo code is okay
according to your *****, your lhs gets a dx, making
it a derivation, while your rhs dont gets anything
but transforms into a derivation, than you a int
therefore do some code in order for me to understan you
.
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| User: "" |
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| Title: Re: please help a girl again: An integral is an antidervative, socan't we integrate the derivative to get back to the original equation? |
02 Dec 2007 03:22:11 AM |
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On Dec 2, 9:10 am, seven7sisters <seven7sist...@gmail.com> wrote:
Suppose we have a function,
y=Cx
Here's the correct way to look at it.
y = F(x) { y is a DEPENDENT VARIABLE}
F(x) = C*x { F is a function of x }
x = independent variable
The derivative of F with respect to x is the Newton Quotient of F
which gives
F'(x) = C
The infinitesimals dy and dx are related by
dy = F'(x) dx
The statement
dy/dx = F'(x) says he ratio of those two infinitesimal variables
equals to derivative of the function of x.
The statement
INT dy means INT (dy/dx) . dx which means INT F'(x) . dx which means
F(x) + K
where C is a constant.
we take the derivative of both sides:
dy/dx = d/dx (Cx) = C
so dy/dx = C
Then supposed we want to get back to the original equation y=Cx via
integration.
How do we write this out?
An integral is an antidervative, so can't we integrate the derivative
to get back to the original equation?
.
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| User: "boho" |
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| Title: Re: please help a girl again: An integral is an antidervative, socan't we integrate the derivative to get back to the original equation? |
02 Dec 2007 03:47:43 AM |
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On Dec 2, 10:22 am, wrote:
On Dec 2, 9:10 am, seven7sisters <seven7sist...@gmail.com> wrote:
Suppose we have a function,
y=Cx
Here's the correct way to look at it.
y = F(x) { y is a DEPENDENT VARIABLE}
F(x) = C*x { F is a function of x }
x = independent variable
The derivative of F with respect to x is the Newton Quotient of F
which gives
F'(x) = C
The infinitesimals dy and dx are related by
dy = F'(x) dx
The statement
dy/dx = F'(x) says he ratio of those two infinitesimal variables
equals to derivative of the function of x.
The statement
INT dy means INT (dy/dx) . dx which means INT F'(x) . dx which means
F(x) + K
where C is a constant.
we take the derivative of both sides:
dy/dx = d/dx (Cx) = C
so dy/dx = C
Then supposed we want to get back to the original equation y=Cx via
integration.
How do we write this out?
An integral is an antidervative, so can't we integrate the derivative
to get back to the original equation?
you lost me
.
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