Mixing
Adding changes or derivatives between two points
$begingroup$
I've playing with the limit definition of derivative and I've to somewhat confusing conclusions.
To clarify, I'm from an Engineering background so I don't think that an instantaneous rate of change makes any sense, for it to be a rate of change something has to change, right?!
So I think of derivatives as the ratio between a very small change in $y$ to a very small change in $x$. So It must be different at every point for many curves.
So I plugged in the formula to calculate the "ratio of change" between points $x+h$ and $x$
$lim _{hto :0}left(frac{left(x+hright)^2-x^2}{x}right) = 2x$
This is the expected answer
However If I try to plug in the points $x+6h$ and $x+5h$, both are distinct from $x+h$ and $x$, so at I expect to find a different answer but I found the same answer
$lim _{hto :0}left(frac{left(x+6hright)^2-left(x+5hright)^2}{x}right) = 2x$
It gets even more confusing If I try to compute the derivative between $x$ and a previous point $x-h$, it still gives me the same answer!
$lim _{hto :0}left(frac{left(xright)^2-left(x-hright)^2}{x}right) = 2x$
I can't make sense of it. It means If chose an arbitrary point $x = m$ then no matter how far I go far from it in any direction I get the same derivative which is quite patently wrong! If it is due to $h$ approaching zero, or it being an infinitesimal how then can we proceed from point to point if not by adding an infinitesimally lengthy line segment, or aren't all curves made up very small line segments?
I hope I made myself clear!
Thanks in advance!
real-analysis limits derivatives curves infinitesimals
edited Jan 18 at 22:30
idriskameni
745321
asked Jan 18 at 20:57
Raafat AbualazmRaafat Abualazm
12
$endgroup$
add a comment |
$begingroup$
I've playing with the limit definition of derivative and I've to somewhat confusing conclusions.
To clarify, I'm from an Engineering background so I don't think that an instantaneous rate of change makes any sense, for it to be a rate of change something has to change, right?!
So I think of derivatives as the ratio between a very small change in $y$ to a very small change in $x$. So It must be different at every point for many curves.
So I plugged in the formula to calculate the "ratio of change" between points $x+h$ and $x$
$lim _{hto :0}left(frac{left(x+hright)^2-x^2}{x}right) = 2x$
This is the expected answer
However If I try to plug in the points $x+6h$ and $x+5h$, both are distinct from $x+h$ and $x$, so at I expect to find a different answer but I found the same answer
$lim _{hto :0}left(frac{left(x+6hright)^2-left(x+5hright)^2}{x}right) = 2x$
It gets even more confusing If I try to compute the derivative between $x$ and a previous point $x-h$, it still gives me the same answer!
$lim _{hto :0}left(frac{left(xright)^2-left(x-hright)^2}{x}right) = 2x$
I can't make sense of it. It means If chose an arbitrary point $x = m$ then no matter how far I go far from it in any direction I get the same derivative which is quite patently wrong! If it is due to $h$ approaching zero, or it being an infinitesimal how then can we proceed from point to point if not by adding an infinitesimally lengthy line segment, or aren't all curves made up very small line segments?
I hope I made myself clear!
Thanks in advance!
real-analysis limits derivatives curves infinitesimals
edited Jan 18 at 22:30
idriskameni
745321
asked Jan 18 at 20:57
Raafat AbualazmRaafat Abualazm
12
$endgroup$
$begingroup$
You’ve got a consistent error in all of your difference quotients: the denominator should be $h$.
$endgroup$
– amd
Jan 19 at 1:01
$begingroup$
I wanted to write h. My bad, my bad!
$endgroup$
– Raafat Abualazm
Jan 19 at 9:36
add a comment |
$begingroup$
I've playing with the limit definition of derivative and I've to somewhat confusing conclusions.
To clarify, I'm from an Engineering background so I don't think that an instantaneous rate of change makes any sense, for it to be a rate of change something has to change, right?!
So I think of derivatives as the ratio between a very small change in $y$ to a very small change in $x$. So It must be different at every point for many curves.
So I plugged in the formula to calculate the "ratio of change" between points $x+h$ and $x$
$lim _{hto :0}left(frac{left(x+hright)^2-x^2}{x}right) = 2x$
This is the expected answer
However If I try to plug in the points $x+6h$ and $x+5h$, both are distinct from $x+h$ and $x$, so at I expect to find a different answer but I found the same answer
$lim _{hto :0}left(frac{left(x+6hright)^2-left(x+5hright)^2}{x}right) = 2x$
It gets even more confusing If I try to compute the derivative between $x$ and a previous point $x-h$, it still gives me the same answer!
$lim _{hto :0}left(frac{left(xright)^2-left(x-hright)^2}{x}right) = 2x$
I can't make sense of it. It means If chose an arbitrary point $x = m$ then no matter how far I go far from it in any direction I get the same derivative which is quite patently wrong! If it is due to $h$ approaching zero, or it being an infinitesimal how then can we proceed from point to point if not by adding an infinitesimally lengthy line segment, or aren't all curves made up very small line segments?
I hope I made myself clear!
Thanks in advance!
real-analysis limits derivatives curves infinitesimals
edited Jan 18 at 22:30
idriskameni
745321
asked Jan 18 at 20:57
Raafat AbualazmRaafat Abualazm
12
$endgroup$
I've playing with the limit definition of derivative and I've to somewhat confusing conclusions.
To clarify, I'm from an Engineering background so I don't think that an instantaneous rate of change makes any sense, for it to be a rate of change something has to change, right?!
So I think of derivatives as the ratio between a very small change in $y$ to a very small change in $x$. So It must be different at every point for many curves.
So I plugged in the formula to calculate the "ratio of change" between points $x+h$ and $x$
$lim _{hto :0}left(frac{left(x+hright)^2-x^2}{x}right) = 2x$
This is the expected answer
However If I try to plug in the points $x+6h$ and $x+5h$, both are distinct from $x+h$ and $x$, so at I expect to find a different answer but I found the same answer
$lim _{hto :0}left(frac{left(x+6hright)^2-left(x+5hright)^2}{x}right) = 2x$
It gets even more confusing If I try to compute the derivative between $x$ and a previous point $x-h$, it still gives me the same answer!
$lim _{hto :0}left(frac{left(xright)^2-left(x-hright)^2}{x}right) = 2x$
I can't make sense of it. It means If chose an arbitrary point $x = m$ then no matter how far I go far from it in any direction I get the same derivative which is quite patently wrong! If it is due to $h$ approaching zero, or it being an infinitesimal how then can we proceed from point to point if not by adding an infinitesimally lengthy line segment, or aren't all curves made up very small line segments?
I hope I made myself clear!
Thanks in advance!
real-analysis limits derivatives curves infinitesimals
real-analysis limits derivatives curves infinitesimals
edited Jan 18 at 22:30
idriskameni
745321
asked Jan 18 at 20:57
Raafat AbualazmRaafat Abualazm
12
edited Jan 18 at 22:30
idriskameni
745321
asked Jan 18 at 20:57
Raafat AbualazmRaafat Abualazm
12
edited Jan 18 at 22:30
idriskameni
745321
edited Jan 18 at 22:30
idriskameni
745321
edited Jan 18 at 22:30
idriskameni
745321
745321
asked Jan 18 at 20:57
Raafat AbualazmRaafat Abualazm
12
asked Jan 18 at 20:57
Raafat AbualazmRaafat Abualazm
12
asked Jan 18 at 20:57
Raafat AbualazmRaafat Abualazm
12
12
$begingroup$
You’ve got a consistent error in all of your difference quotients: the denominator should be $h$.
$endgroup$
– amd
Jan 19 at 1:01
$begingroup$
I wanted to write h. My bad, my bad!
$endgroup$
– Raafat Abualazm
Jan 19 at 9:36
add a comment |
$begingroup$
You’ve got a consistent error in all of your difference quotients: the denominator should be $h$.
$endgroup$
– amd
Jan 19 at 1:01
$begingroup$
I wanted to write h. My bad, my bad!
$endgroup$
– Raafat Abualazm
Jan 19 at 9:36
$begingroup$
You’ve got a consistent error in all of your difference quotients: the denominator should be $h$.
$endgroup$
– amd
Jan 19 at 1:01
$begingroup$
You’ve got a consistent error in all of your difference quotients: the denominator should be $h$.
$endgroup$
– amd
Jan 19 at 1:01
$begingroup$
I wanted to write h. My bad, my bad!
$endgroup$
– Raafat Abualazm
Jan 19 at 9:36
$begingroup$
I wanted to write h. My bad, my bad!
$endgroup$
– Raafat Abualazm
Jan 19 at 9:36
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Let's imagine the curve of an arbitrary function $f$. We can say that two points on the curve are $(x_1, f(x_1))$ and $(x_2, f(x_2))$
We can draw a straight line between these two points, and it's gradient is given by:
$$m=frac{f(x_2)-f(x_1)}{x_2-x_1}$$
for any $x_1, x_2 in mathcal D(f)$ (the domain, essentially, wherever $f$ is defined)
We can assume that $x_2 > x_1$, or rather that $x_2 =x_1 +h$ for some arbitrary $h$ (In the case where it isn't, we can just switch $x_1$ and $x_2$ everywhere I've written them)
Then we have:
$$m=frac{f(x_1+h)-f(x_1)}{x_1+h-x_1}=frac{f(x_1+h)-f(x_1)}{h}$$
If $f(x)=x^2$, we have $m=frac{(x_1+h)^2-x_1^2}{h}=frac{h^2+2hx_1}{h}=2x_1+h$
What the derivative calculates is $m$ when $h$ is infinitesimal, that is, it measures the gradient between two points that are so close to each other they are basically the same point, and that forms the gradient of the tangent to $f$ at that point.
We can say that $$lim_{hto 0}{(2x_1+h)}=2x_1$$
In fact:
$$lim_{hto0} (kh)=0$$
is true for any constant $k$
and that is why the derivative always comes out as $2x$, because the $h$, and its multiples thereof, all disappear as the limit is taken to $0$.
answered Jan 18 at 21:43
Rhys HughesRhys Hughes
6,9441530
$endgroup$
$begingroup$
But isn't it quite nonsensical to talk about a gradient at a point? What does it mean?
$endgroup$
– Raafat Abualazm
Jan 18 at 21:58
$begingroup$
The gradient of the tangent to the curve at that point.
$endgroup$
– Rhys Hughes
Jan 18 at 22:13
$begingroup$
But how then the Gradient of the tangent to the curve tells us about change ALONG the curve? Don't we add these small changes to get from one point to another? Such as we do with integration?
$endgroup$
– Raafat Abualazm
Jan 18 at 22:28
$begingroup$
For the first question, it tells us where the curve is about to go. For example, a negative derivative tells us the curve is about to decrease, and the quantity of that negative describes the scale (e.g. a derivative of $-5$ implies a much steeper descent than one of, say, $-2$).
$endgroup$
– Rhys Hughes
Jan 19 at 3:38
$begingroup$
For the second, no, you just calculate it from a different point. The reason why $2x$ was the derivative in all three cases was because $h to 0 implies kh to 0 forall kin Bbb R$. You were just adding multiples of $h$ in your other examples, hence they disappeared. With regards to integration, that finds the area under the curve between two points, and our approximation method for this, like the gradient method I described in my answer is an approximation for differentiation, is called the Trapezoidal Rule.
$endgroup$
– Rhys Hughes
Jan 19 at 3:44
|
show 4 more comments
$begingroup$
This is correct—you're just approaching $x$ in different ways as you take the limit. In the example with $f(x+6h)-f(x+5h)$ you're approximating the slope a little to the right of $x$, but as $hrightarrow 0$ you still end up at $x$ and get the slope there—which turns out to be $2x$ as it should. You're deriving the same correct answer in different ways.
answered Jan 18 at 21:48
timtfjtimtfj
2,458420
$endgroup$
$begingroup$
How can they be correct? How am I approaching the same point x? Then how can I define another point? How can I go from X0 to X1?
$endgroup$
– Raafat Abualazm
Jan 18 at 22:00
$begingroup$
Isn't the interpretation of derivative is the rate of change between x and infinitesimally close point? Shouldn't they be different for each set if two points? What is the slope at a point? What is changing at a point which is a static entity?
$endgroup$
– Raafat Abualazm
Jan 18 at 22:04
$begingroup$
@RaafatAbualazm It's nowadays very carefully not defined in terms of infinitesimal distances but as a limit—since a point can't have a slope. Geometrically, it's the slope of a tangent to the curve at the point. Another way to imagine it is that it's the direction you'd be facing if you were travelling along the curve—which you can define at each instant.
$endgroup$
– timtfj
Jan 18 at 22:18
$begingroup$
@RaafatAbualazm You're approaching $x$ each time because $lim_{hrightarrow 0} (x+5h) =x$ just like $lim_{hrightarrow 0} (x+h)=x$.
$endgroup$
– timtfj
Jan 18 at 22:21
$begingroup$
Then how can we reach point b from point a? Don't we add small changes to go from a to b? If that change is not along the curve of the function then how can the above reasoning and Integration hold?
$endgroup$
– Raafat Abualazm
Jan 18 at 22:26
|
show 7 more comments
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let's imagine the curve of an arbitrary function $f$. We can say that two points on the curve are $(x_1, f(x_1))$ and $(x_2, f(x_2))$
We can draw a straight line between these two points, and it's gradient is given by:
$$m=frac{f(x_2)-f(x_1)}{x_2-x_1}$$
for any $x_1, x_2 in mathcal D(f)$ (the domain, essentially, wherever $f$ is defined)
We can assume that $x_2 > x_1$, or rather that $x_2 =x_1 +h$ for some arbitrary $h$ (In the case where it isn't, we can just switch $x_1$ and $x_2$ everywhere I've written them)
Then we have:
$$m=frac{f(x_1+h)-f(x_1)}{x_1+h-x_1}=frac{f(x_1+h)-f(x_1)}{h}$$
If $f(x)=x^2$, we have $m=frac{(x_1+h)^2-x_1^2}{h}=frac{h^2+2hx_1}{h}=2x_1+h$
What the derivative calculates is $m$ when $h$ is infinitesimal, that is, it measures the gradient between two points that are so close to each other they are basically the same point, and that forms the gradient of the tangent to $f$ at that point.
We can say that $$lim_{hto 0}{(2x_1+h)}=2x_1$$
In fact:
$$lim_{hto0} (kh)=0$$
is true for any constant $k$
and that is why the derivative always comes out as $2x$, because the $h$, and its multiples thereof, all disappear as the limit is taken to $0$.
answered Jan 18 at 21:43
Rhys HughesRhys Hughes
6,9441530
$endgroup$
$begingroup$
But isn't it quite nonsensical to talk about a gradient at a point? What does it mean?
$endgroup$
– Raafat Abualazm
Jan 18 at 21:58
$begingroup$
The gradient of the tangent to the curve at that point.
$endgroup$
– Rhys Hughes
Jan 18 at 22:13
$begingroup$
But how then the Gradient of the tangent to the curve tells us about change ALONG the curve? Don't we add these small changes to get from one point to another? Such as we do with integration?
$endgroup$
– Raafat Abualazm
Jan 18 at 22:28
$begingroup$
For the first question, it tells us where the curve is about to go. For example, a negative derivative tells us the curve is about to decrease, and the quantity of that negative describes the scale (e.g. a derivative of $-5$ implies a much steeper descent than one of, say, $-2$).
$endgroup$
– Rhys Hughes
Jan 19 at 3:38
$begingroup$
For the second, no, you just calculate it from a different point. The reason why $2x$ was the derivative in all three cases was because $h to 0 implies kh to 0 forall kin Bbb R$. You were just adding multiples of $h$ in your other examples, hence they disappeared. With regards to integration, that finds the area under the curve between two points, and our approximation method for this, like the gradient method I described in my answer is an approximation for differentiation, is called the Trapezoidal Rule.
$endgroup$
– Rhys Hughes
Jan 19 at 3:44
|
show 4 more comments
$begingroup$
Let's imagine the curve of an arbitrary function $f$. We can say that two points on the curve are $(x_1, f(x_1))$ and $(x_2, f(x_2))$
We can draw a straight line between these two points, and it's gradient is given by:
$$m=frac{f(x_2)-f(x_1)}{x_2-x_1}$$
for any $x_1, x_2 in mathcal D(f)$ (the domain, essentially, wherever $f$ is defined)
We can assume that $x_2 > x_1$, or rather that $x_2 =x_1 +h$ for some arbitrary $h$ (In the case where it isn't, we can just switch $x_1$ and $x_2$ everywhere I've written them)
Then we have:
$$m=frac{f(x_1+h)-f(x_1)}{x_1+h-x_1}=frac{f(x_1+h)-f(x_1)}{h}$$
If $f(x)=x^2$, we have $m=frac{(x_1+h)^2-x_1^2}{h}=frac{h^2+2hx_1}{h}=2x_1+h$
What the derivative calculates is $m$ when $h$ is infinitesimal, that is, it measures the gradient between two points that are so close to each other they are basically the same point, and that forms the gradient of the tangent to $f$ at that point.
We can say that $$lim_{hto 0}{(2x_1+h)}=2x_1$$
In fact:
$$lim_{hto0} (kh)=0$$
is true for any constant $k$
and that is why the derivative always comes out as $2x$, because the $h$, and its multiples thereof, all disappear as the limit is taken to $0$.
answered Jan 18 at 21:43
Rhys HughesRhys Hughes
6,9441530
$endgroup$
$begingroup$
But isn't it quite nonsensical to talk about a gradient at a point? What does it mean?
$endgroup$
– Raafat Abualazm
Jan 18 at 21:58
$begingroup$
The gradient of the tangent to the curve at that point.
$endgroup$
– Rhys Hughes
Jan 18 at 22:13
$begingroup$
But how then the Gradient of the tangent to the curve tells us about change ALONG the curve? Don't we add these small changes to get from one point to another? Such as we do with integration?
$endgroup$
– Raafat Abualazm
Jan 18 at 22:28
$begingroup$
For the first question, it tells us where the curve is about to go. For example, a negative derivative tells us the curve is about to decrease, and the quantity of that negative describes the scale (e.g. a derivative of $-5$ implies a much steeper descent than one of, say, $-2$).
$endgroup$
– Rhys Hughes
Jan 19 at 3:38
$begingroup$
For the second, no, you just calculate it from a different point. The reason why $2x$ was the derivative in all three cases was because $h to 0 implies kh to 0 forall kin Bbb R$. You were just adding multiples of $h$ in your other examples, hence they disappeared. With regards to integration, that finds the area under the curve between two points, and our approximation method for this, like the gradient method I described in my answer is an approximation for differentiation, is called the Trapezoidal Rule.
$endgroup$
– Rhys Hughes
Jan 19 at 3:44
|
show 4 more comments
$begingroup$
Let's imagine the curve of an arbitrary function $f$. We can say that two points on the curve are $(x_1, f(x_1))$ and $(x_2, f(x_2))$
We can draw a straight line between these two points, and it's gradient is given by:
$$m=frac{f(x_2)-f(x_1)}{x_2-x_1}$$
for any $x_1, x_2 in mathcal D(f)$ (the domain, essentially, wherever $f$ is defined)
We can assume that $x_2 > x_1$, or rather that $x_2 =x_1 +h$ for some arbitrary $h$ (In the case where it isn't, we can just switch $x_1$ and $x_2$ everywhere I've written them)
Then we have:
$$m=frac{f(x_1+h)-f(x_1)}{x_1+h-x_1}=frac{f(x_1+h)-f(x_1)}{h}$$
If $f(x)=x^2$, we have $m=frac{(x_1+h)^2-x_1^2}{h}=frac{h^2+2hx_1}{h}=2x_1+h$
What the derivative calculates is $m$ when $h$ is infinitesimal, that is, it measures the gradient between two points that are so close to each other they are basically the same point, and that forms the gradient of the tangent to $f$ at that point.
We can say that $$lim_{hto 0}{(2x_1+h)}=2x_1$$
In fact:
$$lim_{hto0} (kh)=0$$
is true for any constant $k$
and that is why the derivative always comes out as $2x$, because the $h$, and its multiples thereof, all disappear as the limit is taken to $0$.
answered Jan 18 at 21:43
Rhys HughesRhys Hughes
6,9441530
$endgroup$
Let's imagine the curve of an arbitrary function $f$. We can say that two points on the curve are $(x_1, f(x_1))$ and $(x_2, f(x_2))$
We can draw a straight line between these two points, and it's gradient is given by:
$$m=frac{f(x_2)-f(x_1)}{x_2-x_1}$$
for any $x_1, x_2 in mathcal D(f)$ (the domain, essentially, wherever $f$ is defined)
We can assume that $x_2 > x_1$, or rather that $x_2 =x_1 +h$ for some arbitrary $h$ (In the case where it isn't, we can just switch $x_1$ and $x_2$ everywhere I've written them)
Then we have:
$$m=frac{f(x_1+h)-f(x_1)}{x_1+h-x_1}=frac{f(x_1+h)-f(x_1)}{h}$$
If $f(x)=x^2$, we have $m=frac{(x_1+h)^2-x_1^2}{h}=frac{h^2+2hx_1}{h}=2x_1+h$
What the derivative calculates is $m$ when $h$ is infinitesimal, that is, it measures the gradient between two points that are so close to each other they are basically the same point, and that forms the gradient of the tangent to $f$ at that point.
We can say that $$lim_{hto 0}{(2x_1+h)}=2x_1$$
In fact:
$$lim_{hto0} (kh)=0$$
is true for any constant $k$
and that is why the derivative always comes out as $2x$, because the $h$, and its multiples thereof, all disappear as the limit is taken to $0$.
answered Jan 18 at 21:43
Rhys HughesRhys Hughes
6,9441530
answered Jan 18 at 21:43
Rhys HughesRhys Hughes
6,9441530
answered Jan 18 at 21:43
Rhys HughesRhys Hughes
6,9441530
answered Jan 18 at 21:43
Rhys HughesRhys Hughes
6,9441530
6,9441530
$begingroup$
But isn't it quite nonsensical to talk about a gradient at a point? What does it mean?
$endgroup$
– Raafat Abualazm
Jan 18 at 21:58
$begingroup$
The gradient of the tangent to the curve at that point.
$endgroup$
– Rhys Hughes
Jan 18 at 22:13
$begingroup$
But how then the Gradient of the tangent to the curve tells us about change ALONG the curve? Don't we add these small changes to get from one point to another? Such as we do with integration?
$endgroup$
– Raafat Abualazm
Jan 18 at 22:28
$begingroup$
For the first question, it tells us where the curve is about to go. For example, a negative derivative tells us the curve is about to decrease, and the quantity of that negative describes the scale (e.g. a derivative of $-5$ implies a much steeper descent than one of, say, $-2$).
$endgroup$
– Rhys Hughes
Jan 19 at 3:38
$begingroup$
For the second, no, you just calculate it from a different point. The reason why $2x$ was the derivative in all three cases was because $h to 0 implies kh to 0 forall kin Bbb R$. You were just adding multiples of $h$ in your other examples, hence they disappeared. With regards to integration, that finds the area under the curve between two points, and our approximation method for this, like the gradient method I described in my answer is an approximation for differentiation, is called the Trapezoidal Rule.
$endgroup$
– Rhys Hughes
Jan 19 at 3:44
|
show 4 more comments
$begingroup$
But isn't it quite nonsensical to talk about a gradient at a point? What does it mean?
$endgroup$
– Raafat Abualazm
Jan 18 at 21:58
$begingroup$
The gradient of the tangent to the curve at that point.
$endgroup$
– Rhys Hughes
Jan 18 at 22:13
$begingroup$
But how then the Gradient of the tangent to the curve tells us about change ALONG the curve? Don't we add these small changes to get from one point to another? Such as we do with integration?
$endgroup$
– Raafat Abualazm
Jan 18 at 22:28
$begingroup$
For the first question, it tells us where the curve is about to go. For example, a negative derivative tells us the curve is about to decrease, and the quantity of that negative describes the scale (e.g. a derivative of $-5$ implies a much steeper descent than one of, say, $-2$).
$endgroup$
– Rhys Hughes
Jan 19 at 3:38
$begingroup$
For the second, no, you just calculate it from a different point. The reason why $2x$ was the derivative in all three cases was because $h to 0 implies kh to 0 forall kin Bbb R$. You were just adding multiples of $h$ in your other examples, hence they disappeared. With regards to integration, that finds the area under the curve between two points, and our approximation method for this, like the gradient method I described in my answer is an approximation for differentiation, is called the Trapezoidal Rule.
$endgroup$
– Rhys Hughes
Jan 19 at 3:44
$begingroup$
But isn't it quite nonsensical to talk about a gradient at a point? What does it mean?
$endgroup$
– Raafat Abualazm
Jan 18 at 21:58
$begingroup$
But isn't it quite nonsensical to talk about a gradient at a point? What does it mean?
$endgroup$
– Raafat Abualazm
Jan 18 at 21:58
$begingroup$
The gradient of the tangent to the curve at that point.
$endgroup$
– Rhys Hughes
Jan 18 at 22:13
$begingroup$
The gradient of the tangent to the curve at that point.
$endgroup$
– Rhys Hughes
Jan 18 at 22:13
$begingroup$
But how then the Gradient of the tangent to the curve tells us about change ALONG the curve? Don't we add these small changes to get from one point to another? Such as we do with integration?
$endgroup$
– Raafat Abualazm
Jan 18 at 22:28
$begingroup$
But how then the Gradient of the tangent to the curve tells us about change ALONG the curve? Don't we add these small changes to get from one point to another? Such as we do with integration?
$endgroup$
– Raafat Abualazm
Jan 18 at 22:28
$begingroup$
For the first question, it tells us where the curve is about to go. For example, a negative derivative tells us the curve is about to decrease, and the quantity of that negative describes the scale (e.g. a derivative of $-5$ implies a much steeper descent than one of, say, $-2$).
$endgroup$
– Rhys Hughes
Jan 19 at 3:38
$begingroup$
For the first question, it tells us where the curve is about to go. For example, a negative derivative tells us the curve is about to decrease, and the quantity of that negative describes the scale (e.g. a derivative of $-5$ implies a much steeper descent than one of, say, $-2$).
$endgroup$
– Rhys Hughes
Jan 19 at 3:38
$begingroup$
For the second, no, you just calculate it from a different point. The reason why $2x$ was the derivative in all three cases was because $h to 0 implies kh to 0 forall kin Bbb R$. You were just adding multiples of $h$ in your other examples, hence they disappeared. With regards to integration, that finds the area under the curve between two points, and our approximation method for this, like the gradient method I described in my answer is an approximation for differentiation, is called the Trapezoidal Rule.
$endgroup$
– Rhys Hughes
Jan 19 at 3:44
$begingroup$
For the second, no, you just calculate it from a different point. The reason why $2x$ was the derivative in all three cases was because $h to 0 implies kh to 0 forall kin Bbb R$. You were just adding multiples of $h$ in your other examples, hence they disappeared. With regards to integration, that finds the area under the curve between two points, and our approximation method for this, like the gradient method I described in my answer is an approximation for differentiation, is called the Trapezoidal Rule.
$endgroup$
– Rhys Hughes
Jan 19 at 3:44
|
show 4 more comments
$begingroup$
This is correct—you're just approaching $x$ in different ways as you take the limit. In the example with $f(x+6h)-f(x+5h)$ you're approximating the slope a little to the right of $x$, but as $hrightarrow 0$ you still end up at $x$ and get the slope there—which turns out to be $2x$ as it should. You're deriving the same correct answer in different ways.
answered Jan 18 at 21:48
timtfjtimtfj
2,458420
$endgroup$
$begingroup$
How can they be correct? How am I approaching the same point x? Then how can I define another point? How can I go from X0 to X1?
$endgroup$
– Raafat Abualazm
Jan 18 at 22:00
$begingroup$
Isn't the interpretation of derivative is the rate of change between x and infinitesimally close point? Shouldn't they be different for each set if two points? What is the slope at a point? What is changing at a point which is a static entity?
$endgroup$
– Raafat Abualazm
Jan 18 at 22:04
$begingroup$
@RaafatAbualazm It's nowadays very carefully not defined in terms of infinitesimal distances but as a limit—since a point can't have a slope. Geometrically, it's the slope of a tangent to the curve at the point. Another way to imagine it is that it's the direction you'd be facing if you were travelling along the curve—which you can define at each instant.
$endgroup$
– timtfj
Jan 18 at 22:18
$begingroup$
@RaafatAbualazm You're approaching $x$ each time because $lim_{hrightarrow 0} (x+5h) =x$ just like $lim_{hrightarrow 0} (x+h)=x$.
$endgroup$
– timtfj
Jan 18 at 22:21
$begingroup$
Then how can we reach point b from point a? Don't we add small changes to go from a to b? If that change is not along the curve of the function then how can the above reasoning and Integration hold?
$endgroup$
– Raafat Abualazm
Jan 18 at 22:26
|
show 7 more comments
$begingroup$
This is correct—you're just approaching $x$ in different ways as you take the limit. In the example with $f(x+6h)-f(x+5h)$ you're approximating the slope a little to the right of $x$, but as $hrightarrow 0$ you still end up at $x$ and get the slope there—which turns out to be $2x$ as it should. You're deriving the same correct answer in different ways.
answered Jan 18 at 21:48
timtfjtimtfj
2,458420
$endgroup$
$begingroup$
How can they be correct? How am I approaching the same point x? Then how can I define another point? How can I go from X0 to X1?
$endgroup$
– Raafat Abualazm
Jan 18 at 22:00
$begingroup$
Isn't the interpretation of derivative is the rate of change between x and infinitesimally close point? Shouldn't they be different for each set if two points? What is the slope at a point? What is changing at a point which is a static entity?
$endgroup$
– Raafat Abualazm
Jan 18 at 22:04
$begingroup$
@RaafatAbualazm It's nowadays very carefully not defined in terms of infinitesimal distances but as a limit—since a point can't have a slope. Geometrically, it's the slope of a tangent to the curve at the point. Another way to imagine it is that it's the direction you'd be facing if you were travelling along the curve—which you can define at each instant.
$endgroup$
– timtfj
Jan 18 at 22:18
$begingroup$
@RaafatAbualazm You're approaching $x$ each time because $lim_{hrightarrow 0} (x+5h) =x$ just like $lim_{hrightarrow 0} (x+h)=x$.
$endgroup$
– timtfj
Jan 18 at 22:21
$begingroup$
Then how can we reach point b from point a? Don't we add small changes to go from a to b? If that change is not along the curve of the function then how can the above reasoning and Integration hold?
$endgroup$
– Raafat Abualazm
Jan 18 at 22:26
|
show 7 more comments
$begingroup$
This is correct—you're just approaching $x$ in different ways as you take the limit. In the example with $f(x+6h)-f(x+5h)$ you're approximating the slope a little to the right of $x$, but as $hrightarrow 0$ you still end up at $x$ and get the slope there—which turns out to be $2x$ as it should. You're deriving the same correct answer in different ways.
answered Jan 18 at 21:48
timtfjtimtfj
2,458420
$endgroup$
This is correct—you're just approaching $x$ in different ways as you take the limit. In the example with $f(x+6h)-f(x+5h)$ you're approximating the slope a little to the right of $x$, but as $hrightarrow 0$ you still end up at $x$ and get the slope there—which turns out to be $2x$ as it should. You're deriving the same correct answer in different ways.
answered Jan 18 at 21:48
timtfjtimtfj
2,458420
answered Jan 18 at 21:48
timtfjtimtfj
2,458420
answered Jan 18 at 21:48
timtfjtimtfj
2,458420
answered Jan 18 at 21:48
timtfjtimtfj
2,458420
2,458420
$begingroup$
How can they be correct? How am I approaching the same point x? Then how can I define another point? How can I go from X0 to X1?
$endgroup$
– Raafat Abualazm
Jan 18 at 22:00
$begingroup$
Isn't the interpretation of derivative is the rate of change between x and infinitesimally close point? Shouldn't they be different for each set if two points? What is the slope at a point? What is changing at a point which is a static entity?
$endgroup$
– Raafat Abualazm
Jan 18 at 22:04
$begingroup$
@RaafatAbualazm It's nowadays very carefully not defined in terms of infinitesimal distances but as a limit—since a point can't have a slope. Geometrically, it's the slope of a tangent to the curve at the point. Another way to imagine it is that it's the direction you'd be facing if you were travelling along the curve—which you can define at each instant.
$endgroup$
– timtfj
Jan 18 at 22:18
$begingroup$
@RaafatAbualazm You're approaching $x$ each time because $lim_{hrightarrow 0} (x+5h) =x$ just like $lim_{hrightarrow 0} (x+h)=x$.
$endgroup$
– timtfj
Jan 18 at 22:21
$begingroup$
Then how can we reach point b from point a? Don't we add small changes to go from a to b? If that change is not along the curve of the function then how can the above reasoning and Integration hold?
$endgroup$
– Raafat Abualazm
Jan 18 at 22:26
|
show 7 more comments
$begingroup$
How can they be correct? How am I approaching the same point x? Then how can I define another point? How can I go from X0 to X1?
$endgroup$
– Raafat Abualazm
Jan 18 at 22:00
$begingroup$
Isn't the interpretation of derivative is the rate of change between x and infinitesimally close point? Shouldn't they be different for each set if two points? What is the slope at a point? What is changing at a point which is a static entity?
$endgroup$
– Raafat Abualazm
Jan 18 at 22:04
$begingroup$
@RaafatAbualazm It's nowadays very carefully not defined in terms of infinitesimal distances but as a limit—since a point can't have a slope. Geometrically, it's the slope of a tangent to the curve at the point. Another way to imagine it is that it's the direction you'd be facing if you were travelling along the curve—which you can define at each instant.
$endgroup$
– timtfj
Jan 18 at 22:18
$begingroup$
@RaafatAbualazm You're approaching $x$ each time because $lim_{hrightarrow 0} (x+5h) =x$ just like $lim_{hrightarrow 0} (x+h)=x$.
$endgroup$
– timtfj
Jan 18 at 22:21
$begingroup$
Then how can we reach point b from point a? Don't we add small changes to go from a to b? If that change is not along the curve of the function then how can the above reasoning and Integration hold?
$endgroup$
– Raafat Abualazm
Jan 18 at 22:26
$begingroup$
How can they be correct? How am I approaching the same point x? Then how can I define another point? How can I go from X0 to X1?
$endgroup$
– Raafat Abualazm
Jan 18 at 22:00
$begingroup$
How can they be correct? How am I approaching the same point x? Then how can I define another point? How can I go from X0 to X1?
$endgroup$
– Raafat Abualazm
Jan 18 at 22:00
$begingroup$
Isn't the interpretation of derivative is the rate of change between x and infinitesimally close point? Shouldn't they be different for each set if two points? What is the slope at a point? What is changing at a point which is a static entity?
$endgroup$
– Raafat Abualazm
Jan 18 at 22:04
$begingroup$
Isn't the interpretation of derivative is the rate of change between x and infinitesimally close point? Shouldn't they be different for each set if two points? What is the slope at a point? What is changing at a point which is a static entity?
$endgroup$
– Raafat Abualazm
Jan 18 at 22:04
$begingroup$
@RaafatAbualazm It's nowadays very carefully not defined in terms of infinitesimal distances but as a limit—since a point can't have a slope. Geometrically, it's the slope of a tangent to the curve at the point. Another way to imagine it is that it's the direction you'd be facing if you were travelling along the curve—which you can define at each instant.
$endgroup$
– timtfj
Jan 18 at 22:18
$begingroup$
@RaafatAbualazm It's nowadays very carefully not defined in terms of infinitesimal distances but as a limit—since a point can't have a slope. Geometrically, it's the slope of a tangent to the curve at the point. Another way to imagine it is that it's the direction you'd be facing if you were travelling along the curve—which you can define at each instant.
$endgroup$
– timtfj
Jan 18 at 22:18
$begingroup$
@RaafatAbualazm You're approaching $x$ each time because $lim_{hrightarrow 0} (x+5h) =x$ just like $lim_{hrightarrow 0} (x+h)=x$.
$endgroup$
– timtfj
Jan 18 at 22:21
$begingroup$
@RaafatAbualazm You're approaching $x$ each time because $lim_{hrightarrow 0} (x+5h) =x$ just like $lim_{hrightarrow 0} (x+h)=x$.
$endgroup$
– timtfj
Jan 18 at 22:21
$begingroup$
Then how can we reach point b from point a? Don't we add small changes to go from a to b? If that change is not along the curve of the function then how can the above reasoning and Integration hold?
$endgroup$
– Raafat Abualazm
Jan 18 at 22:26
$begingroup$
Then how can we reach point b from point a? Don't we add small changes to go from a to b? If that change is not along the curve of the function then how can the above reasoning and Integration hold?
$endgroup$
– Raafat Abualazm
Jan 18 at 22:26
|
show 7 more comments
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$begingroup$
You’ve got a consistent error in all of your difference quotients: the denominator should be $h$.
$endgroup$
– amd
Jan 19 at 1:01
$begingroup$
I wanted to write h. My bad, my bad!
$endgroup$
– Raafat Abualazm
Jan 19 at 9:36