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Why is $Delta (R_i^2)$ equal to $2R_i Delta (R_i)$
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So when deriving the equation for charge, I reached a point where
$fraczeta4epsilon sum_i=1^N fracDelta (R_i^2)(z^2+R_i^2)^frac32$
But my teacher explains that
$Delta (R_i^2)$
can be rearranged to
$2R_i Delta (R_i)$
but he cannot explain why.
He talks about taking a differential but then it becomes $dr$ and the equation turns into an integral in terms of R, so doesn't that just counter his claim?
differential-equations derivatives physics
asked 6 hours ago
A man with a hat
132
add a comment |Â
up vote
1
down vote
favorite
So when deriving the equation for charge, I reached a point where
$fraczeta4epsilon sum_i=1^N fracDelta (R_i^2)(z^2+R_i^2)^frac32$
But my teacher explains that
$Delta (R_i^2)$
can be rearranged to
$2R_i Delta (R_i)$
but he cannot explain why.
He talks about taking a differential but then it becomes $dr$ and the equation turns into an integral in terms of R, so doesn't that just counter his claim?
differential-equations derivatives physics
asked 6 hours ago
A man with a hat
132
I think it comes from the fact that $(y^2)'=2yy'$.
– Botond
5 hours ago
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
So when deriving the equation for charge, I reached a point where
$fraczeta4epsilon sum_i=1^N fracDelta (R_i^2)(z^2+R_i^2)^frac32$
But my teacher explains that
$Delta (R_i^2)$
can be rearranged to
$2R_i Delta (R_i)$
but he cannot explain why.
He talks about taking a differential but then it becomes $dr$ and the equation turns into an integral in terms of R, so doesn't that just counter his claim?
differential-equations derivatives physics
asked 6 hours ago
A man with a hat
132
So when deriving the equation for charge, I reached a point where
$fraczeta4epsilon sum_i=1^N fracDelta (R_i^2)(z^2+R_i^2)^frac32$
But my teacher explains that
$Delta (R_i^2)$
can be rearranged to
$2R_i Delta (R_i)$
but he cannot explain why.
He talks about taking a differential but then it becomes $dr$ and the equation turns into an integral in terms of R, so doesn't that just counter his claim?
differential-equations derivatives physics
asked 6 hours ago
A man with a hat
132
asked 6 hours ago
A man with a hat
132
asked 6 hours ago
A man with a hat
132
asked 6 hours ago
A man with a hat
132
132
I think it comes from the fact that $(y^2)'=2yy'$.
– Botond
5 hours ago
add a comment |Â
I think it comes from the fact that $(y^2)'=2yy'$.
– Botond
5 hours ago
I think it comes from the fact that $(y^2)'=2yy'$.
– Botond
5 hours ago
I think it comes from the fact that $(y^2)'=2yy'$.
– Botond
5 hours ago
add a comment |Â
2 Answers
2
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oldest
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up vote
1
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Unpacking the meaning of $Delta$ as ...
$$ Delta f(t)equiv f(t+Delta t)-f(t) $$
We can write ...
$$ beginarray \ Delta x^2(t)
\=x^2(t+Delta t) - x^2(t) \=(x(t+Delta t)+x(t))(x(t+Delta t)-x(t))
\=left [Delta x(t) +2x(t) right ] Delta x(t) endarray $$
So provided that $Delta x(t) << 2x(t)$ we have ...
$$Delta x^2(t) approx 2x(t) Delta x(t)
$$
answered 5 hours ago
WW1
6,2021712
I think it's a bit better to say something like "If we neglect the higher order $Delta$ terms, i.e. $(Delta x)^2$, because $(Delta x)^2 << Delta x$."
– Botond
5 hours ago
1
I guess the truth is that the statement is not so much a rearrangement as it is an approximation
– WW1
5 hours ago
What do you mean?
– Botond
5 hours ago
1
The OP used the phrase "can be rearranged to" , but we realize that there is an approximation being made for finite $Delta t$
– WW1
5 hours ago
Yes, it's an approximation, but it will be true when $N to infty$.
– Botond
5 hours ago
add a comment |Â
up vote
0
down vote
When taking differentials, $$x = f(t) implies mathrm dx =
fracmathrm dfmathrm dt , mathrm dt$$
Let's apply the above statement. Let $u_i = R_i^2$. As $Delta u_i$ approaches $0$, it becomes a differential $mathrm du_i$. So we have
$$u_i = f(R_i) = R_i^2 implies mathrm du_i =
fracmathrm dfmathrm dR_i , mathrm dR_i$$
Note that $fracmathrm dfmathrm dR_i = fracmathrm dmathrm dR_i[R_i^2] = 2R_i$. Hence
$$mathrm du_i =
(2R_i) , mathrm dR_i$$
And, remembering that $u_i = R_i^2$:
$$mathrm d(R_i^2) = 2 R_i , mathrm dR_i$$
Which means that for small changes in $R_i^2$ (i.e. as we take the limit approaching $0$), $$Delta(R_i^2) approx 2 R_i Delta R_i^2$$
answered 5 hours ago
Tiwa Aina
2,526319
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Unpacking the meaning of $Delta$ as ...
$$ Delta f(t)equiv f(t+Delta t)-f(t) $$
We can write ...
$$ beginarray \ Delta x^2(t)
\=x^2(t+Delta t) - x^2(t) \=(x(t+Delta t)+x(t))(x(t+Delta t)-x(t))
\=left [Delta x(t) +2x(t) right ] Delta x(t) endarray $$
So provided that $Delta x(t) << 2x(t)$ we have ...
$$Delta x^2(t) approx 2x(t) Delta x(t)
$$
answered 5 hours ago
WW1
6,2021712
I think it's a bit better to say something like "If we neglect the higher order $Delta$ terms, i.e. $(Delta x)^2$, because $(Delta x)^2 << Delta x$."
– Botond
5 hours ago
1
I guess the truth is that the statement is not so much a rearrangement as it is an approximation
– WW1
5 hours ago
What do you mean?
– Botond
5 hours ago
1
The OP used the phrase "can be rearranged to" , but we realize that there is an approximation being made for finite $Delta t$
– WW1
5 hours ago
Yes, it's an approximation, but it will be true when $N to infty$.
– Botond
5 hours ago
add a comment |Â
up vote
1
down vote
Unpacking the meaning of $Delta$ as ...
$$ Delta f(t)equiv f(t+Delta t)-f(t) $$
We can write ...
$$ beginarray \ Delta x^2(t)
\=x^2(t+Delta t) - x^2(t) \=(x(t+Delta t)+x(t))(x(t+Delta t)-x(t))
\=left [Delta x(t) +2x(t) right ] Delta x(t) endarray $$
So provided that $Delta x(t) << 2x(t)$ we have ...
$$Delta x^2(t) approx 2x(t) Delta x(t)
$$
answered 5 hours ago
WW1
6,2021712
I think it's a bit better to say something like "If we neglect the higher order $Delta$ terms, i.e. $(Delta x)^2$, because $(Delta x)^2 << Delta x$."
– Botond
5 hours ago
1
I guess the truth is that the statement is not so much a rearrangement as it is an approximation
– WW1
5 hours ago
What do you mean?
– Botond
5 hours ago
1
The OP used the phrase "can be rearranged to" , but we realize that there is an approximation being made for finite $Delta t$
– WW1
5 hours ago
Yes, it's an approximation, but it will be true when $N to infty$.
– Botond
5 hours ago
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Unpacking the meaning of $Delta$ as ...
$$ Delta f(t)equiv f(t+Delta t)-f(t) $$
We can write ...
$$ beginarray \ Delta x^2(t)
\=x^2(t+Delta t) - x^2(t) \=(x(t+Delta t)+x(t))(x(t+Delta t)-x(t))
\=left [Delta x(t) +2x(t) right ] Delta x(t) endarray $$
So provided that $Delta x(t) << 2x(t)$ we have ...
$$Delta x^2(t) approx 2x(t) Delta x(t)
$$
answered 5 hours ago
WW1
6,2021712
Unpacking the meaning of $Delta$ as ...
$$ Delta f(t)equiv f(t+Delta t)-f(t) $$
We can write ...
$$ beginarray \ Delta x^2(t)
\=x^2(t+Delta t) - x^2(t) \=(x(t+Delta t)+x(t))(x(t+Delta t)-x(t))
\=left [Delta x(t) +2x(t) right ] Delta x(t) endarray $$
So provided that $Delta x(t) << 2x(t)$ we have ...
$$Delta x^2(t) approx 2x(t) Delta x(t)
$$
answered 5 hours ago
WW1
6,2021712
edited 5 hours ago
answered 5 hours ago
WW1
6,2021712
answered 5 hours ago
WW1
6,2021712
answered 5 hours ago
WW1
6,2021712
6,2021712
I think it's a bit better to say something like "If we neglect the higher order $Delta$ terms, i.e. $(Delta x)^2$, because $(Delta x)^2 << Delta x$."
– Botond
5 hours ago
1
I guess the truth is that the statement is not so much a rearrangement as it is an approximation
– WW1
5 hours ago
What do you mean?
– Botond
5 hours ago
1
The OP used the phrase "can be rearranged to" , but we realize that there is an approximation being made for finite $Delta t$
– WW1
5 hours ago
Yes, it's an approximation, but it will be true when $N to infty$.
– Botond
5 hours ago
add a comment |Â
I think it's a bit better to say something like "If we neglect the higher order $Delta$ terms, i.e. $(Delta x)^2$, because $(Delta x)^2 << Delta x$."
– Botond
5 hours ago
1
I guess the truth is that the statement is not so much a rearrangement as it is an approximation
– WW1
5 hours ago
What do you mean?
– Botond
5 hours ago
1
The OP used the phrase "can be rearranged to" , but we realize that there is an approximation being made for finite $Delta t$
– WW1
5 hours ago
Yes, it's an approximation, but it will be true when $N to infty$.
– Botond
5 hours ago
I think it's a bit better to say something like "If we neglect the higher order $Delta$ terms, i.e. $(Delta x)^2$, because $(Delta x)^2 << Delta x$."
– Botond
5 hours ago
I think it's a bit better to say something like "If we neglect the higher order $Delta$ terms, i.e. $(Delta x)^2$, because $(Delta x)^2 << Delta x$."
– Botond
5 hours ago
1
1
I guess the truth is that the statement is not so much a rearrangement as it is an approximation
– WW1
5 hours ago
I guess the truth is that the statement is not so much a rearrangement as it is an approximation
– WW1
5 hours ago
What do you mean?
– Botond
5 hours ago
What do you mean?
– Botond
5 hours ago
1
1
The OP used the phrase "can be rearranged to" , but we realize that there is an approximation being made for finite $Delta t$
– WW1
5 hours ago
The OP used the phrase "can be rearranged to" , but we realize that there is an approximation being made for finite $Delta t$
– WW1
5 hours ago
Yes, it's an approximation, but it will be true when $N to infty$.
– Botond
5 hours ago
Yes, it's an approximation, but it will be true when $N to infty$.
– Botond
5 hours ago
add a comment |Â
up vote
0
down vote
When taking differentials, $$x = f(t) implies mathrm dx =
fracmathrm dfmathrm dt , mathrm dt$$
Let's apply the above statement. Let $u_i = R_i^2$. As $Delta u_i$ approaches $0$, it becomes a differential $mathrm du_i$. So we have
$$u_i = f(R_i) = R_i^2 implies mathrm du_i =
fracmathrm dfmathrm dR_i , mathrm dR_i$$
Note that $fracmathrm dfmathrm dR_i = fracmathrm dmathrm dR_i[R_i^2] = 2R_i$. Hence
$$mathrm du_i =
(2R_i) , mathrm dR_i$$
And, remembering that $u_i = R_i^2$:
$$mathrm d(R_i^2) = 2 R_i , mathrm dR_i$$
Which means that for small changes in $R_i^2$ (i.e. as we take the limit approaching $0$), $$Delta(R_i^2) approx 2 R_i Delta R_i^2$$
answered 5 hours ago
Tiwa Aina
2,526319
add a comment |Â
up vote
0
down vote
When taking differentials, $$x = f(t) implies mathrm dx =
fracmathrm dfmathrm dt , mathrm dt$$
Let's apply the above statement. Let $u_i = R_i^2$. As $Delta u_i$ approaches $0$, it becomes a differential $mathrm du_i$. So we have
$$u_i = f(R_i) = R_i^2 implies mathrm du_i =
fracmathrm dfmathrm dR_i , mathrm dR_i$$
Note that $fracmathrm dfmathrm dR_i = fracmathrm dmathrm dR_i[R_i^2] = 2R_i$. Hence
$$mathrm du_i =
(2R_i) , mathrm dR_i$$
And, remembering that $u_i = R_i^2$:
$$mathrm d(R_i^2) = 2 R_i , mathrm dR_i$$
Which means that for small changes in $R_i^2$ (i.e. as we take the limit approaching $0$), $$Delta(R_i^2) approx 2 R_i Delta R_i^2$$
answered 5 hours ago
Tiwa Aina
2,526319
add a comment |Â
up vote
0
down vote
up vote
0
down vote
When taking differentials, $$x = f(t) implies mathrm dx =
fracmathrm dfmathrm dt , mathrm dt$$
Let's apply the above statement. Let $u_i = R_i^2$. As $Delta u_i$ approaches $0$, it becomes a differential $mathrm du_i$. So we have
$$u_i = f(R_i) = R_i^2 implies mathrm du_i =
fracmathrm dfmathrm dR_i , mathrm dR_i$$
Note that $fracmathrm dfmathrm dR_i = fracmathrm dmathrm dR_i[R_i^2] = 2R_i$. Hence
$$mathrm du_i =
(2R_i) , mathrm dR_i$$
And, remembering that $u_i = R_i^2$:
$$mathrm d(R_i^2) = 2 R_i , mathrm dR_i$$
Which means that for small changes in $R_i^2$ (i.e. as we take the limit approaching $0$), $$Delta(R_i^2) approx 2 R_i Delta R_i^2$$
answered 5 hours ago
Tiwa Aina
2,526319
When taking differentials, $$x = f(t) implies mathrm dx =
fracmathrm dfmathrm dt , mathrm dt$$
Let's apply the above statement. Let $u_i = R_i^2$. As $Delta u_i$ approaches $0$, it becomes a differential $mathrm du_i$. So we have
$$u_i = f(R_i) = R_i^2 implies mathrm du_i =
fracmathrm dfmathrm dR_i , mathrm dR_i$$
Note that $fracmathrm dfmathrm dR_i = fracmathrm dmathrm dR_i[R_i^2] = 2R_i$. Hence
$$mathrm du_i =
(2R_i) , mathrm dR_i$$
And, remembering that $u_i = R_i^2$:
$$mathrm d(R_i^2) = 2 R_i , mathrm dR_i$$
Which means that for small changes in $R_i^2$ (i.e. as we take the limit approaching $0$), $$Delta(R_i^2) approx 2 R_i Delta R_i^2$$
answered 5 hours ago
Tiwa Aina
2,526319
edited 5 hours ago
answered 5 hours ago
Tiwa Aina
2,526319
answered 5 hours ago
Tiwa Aina
2,526319
answered 5 hours ago
Tiwa Aina
2,526319
2,526319
add a comment |Â
add a comment |Â
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I think it comes from the fact that $(y^2)'=2yy'$.
– Botond
5 hours ago