What is the tangent at a sharp point on a curve?
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How to know which line represents tangent to a curve $y=f(x)$ (in RED) ?From the diagram , I cannot decide which line to take as tangent , all seem to touch at a single point.
tangent-line
asked Mar 18 at 16:04
ADITYA PRAKASHADITYA PRAKASH
33818
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|
show 3 more comments
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How to know which line represents tangent to a curve $y=f(x)$ (in RED) ?From the diagram , I cannot decide which line to take as tangent , all seem to touch at a single point.
tangent-line
asked Mar 18 at 16:04
ADITYA PRAKASHADITYA PRAKASH
33818
$endgroup$
24
$begingroup$
It does not exist for the reason you give. The tangent, if it exists, is unique.
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– John Douma
Mar 18 at 16:06
1
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how about the bisector of the "angle" made of the right and left part of your nice sharp curve?
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– dmtri
Mar 18 at 16:14
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If the curve consists of 2 arcs, let's say parts of the circles with radius 1 and centers -1 and 1 then the axis $x=0$ might be the tangent .
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– dmtri
Mar 18 at 17:22
4
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A tangent is a line that approximates the curve near the point - such that if you zoom in close enough on the point, the tangent and curve become pratically indistinguishable. That is the property of tangents that makes them useful. Left and Right tangents, which approximate the curve on only one side of the point, are also useful concepts. A line that does not approximate the curve on either side is useful only in very rare and limited circumstances, unworthy of a special name.
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– Paul Sinclair
Mar 19 at 3:11
1
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@alephzero, I am not saying that such an angle should exist.The sketch of the OP is just a sketch and not mathematicaly described, so no strict results can be deduced. In my first comment I just wonder what the tangent would be if $f$ were something like $y=x^{2/3}+c$
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– dmtri
Mar 19 at 8:03
|
show 3 more comments
$begingroup$
How to know which line represents tangent to a curve $y=f(x)$ (in RED) ?From the diagram , I cannot decide which line to take as tangent , all seem to touch at a single point.
tangent-line
asked Mar 18 at 16:04
ADITYA PRAKASHADITYA PRAKASH
33818
$endgroup$
How to know which line represents tangent to a curve $y=f(x)$ (in RED) ?From the diagram , I cannot decide which line to take as tangent , all seem to touch at a single point.
tangent-line
tangent-line
asked Mar 18 at 16:04
ADITYA PRAKASHADITYA PRAKASH
33818
asked Mar 18 at 16:04
ADITYA PRAKASHADITYA PRAKASH
33818
asked Mar 18 at 16:04
ADITYA PRAKASHADITYA PRAKASH
33818
asked Mar 18 at 16:04
ADITYA PRAKASHADITYA PRAKASH
33818
asked Mar 18 at 16:04
ADITYA PRAKASHADITYA PRAKASH
33818
33818
24
$begingroup$
It does not exist for the reason you give. The tangent, if it exists, is unique.
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– John Douma
Mar 18 at 16:06
1
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how about the bisector of the "angle" made of the right and left part of your nice sharp curve?
$endgroup$
– dmtri
Mar 18 at 16:14
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If the curve consists of 2 arcs, let's say parts of the circles with radius 1 and centers -1 and 1 then the axis $x=0$ might be the tangent .
$endgroup$
– dmtri
Mar 18 at 17:22
4
$begingroup$
A tangent is a line that approximates the curve near the point - such that if you zoom in close enough on the point, the tangent and curve become pratically indistinguishable. That is the property of tangents that makes them useful. Left and Right tangents, which approximate the curve on only one side of the point, are also useful concepts. A line that does not approximate the curve on either side is useful only in very rare and limited circumstances, unworthy of a special name.
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– Paul Sinclair
Mar 19 at 3:11
1
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@alephzero, I am not saying that such an angle should exist.The sketch of the OP is just a sketch and not mathematicaly described, so no strict results can be deduced. In my first comment I just wonder what the tangent would be if $f$ were something like $y=x^{2/3}+c$
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– dmtri
Mar 19 at 8:03
|
show 3 more comments
24
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It does not exist for the reason you give. The tangent, if it exists, is unique.
$endgroup$
– John Douma
Mar 18 at 16:06
1
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how about the bisector of the "angle" made of the right and left part of your nice sharp curve?
$endgroup$
– dmtri
Mar 18 at 16:14
$begingroup$
If the curve consists of 2 arcs, let's say parts of the circles with radius 1 and centers -1 and 1 then the axis $x=0$ might be the tangent .
$endgroup$
– dmtri
Mar 18 at 17:22
4
$begingroup$
A tangent is a line that approximates the curve near the point - such that if you zoom in close enough on the point, the tangent and curve become pratically indistinguishable. That is the property of tangents that makes them useful. Left and Right tangents, which approximate the curve on only one side of the point, are also useful concepts. A line that does not approximate the curve on either side is useful only in very rare and limited circumstances, unworthy of a special name.
$endgroup$
– Paul Sinclair
Mar 19 at 3:11
1
$begingroup$
@alephzero, I am not saying that such an angle should exist.The sketch of the OP is just a sketch and not mathematicaly described, so no strict results can be deduced. In my first comment I just wonder what the tangent would be if $f$ were something like $y=x^{2/3}+c$
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– dmtri
Mar 19 at 8:03
24
24
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It does not exist for the reason you give. The tangent, if it exists, is unique.
$endgroup$
– John Douma
Mar 18 at 16:06
$begingroup$
It does not exist for the reason you give. The tangent, if it exists, is unique.
$endgroup$
– John Douma
Mar 18 at 16:06
1
1
$begingroup$
how about the bisector of the "angle" made of the right and left part of your nice sharp curve?
$endgroup$
– dmtri
Mar 18 at 16:14
$begingroup$
how about the bisector of the "angle" made of the right and left part of your nice sharp curve?
$endgroup$
– dmtri
Mar 18 at 16:14
$begingroup$
If the curve consists of 2 arcs, let's say parts of the circles with radius 1 and centers -1 and 1 then the axis $x=0$ might be the tangent .
$endgroup$
– dmtri
Mar 18 at 17:22
$begingroup$
If the curve consists of 2 arcs, let's say parts of the circles with radius 1 and centers -1 and 1 then the axis $x=0$ might be the tangent .
$endgroup$
– dmtri
Mar 18 at 17:22
4
4
$begingroup$
A tangent is a line that approximates the curve near the point - such that if you zoom in close enough on the point, the tangent and curve become pratically indistinguishable. That is the property of tangents that makes them useful. Left and Right tangents, which approximate the curve on only one side of the point, are also useful concepts. A line that does not approximate the curve on either side is useful only in very rare and limited circumstances, unworthy of a special name.
$endgroup$
– Paul Sinclair
Mar 19 at 3:11
$begingroup$
A tangent is a line that approximates the curve near the point - such that if you zoom in close enough on the point, the tangent and curve become pratically indistinguishable. That is the property of tangents that makes them useful. Left and Right tangents, which approximate the curve on only one side of the point, are also useful concepts. A line that does not approximate the curve on either side is useful only in very rare and limited circumstances, unworthy of a special name.
$endgroup$
– Paul Sinclair
Mar 19 at 3:11
1
1
$begingroup$
@alephzero, I am not saying that such an angle should exist.The sketch of the OP is just a sketch and not mathematicaly described, so no strict results can be deduced. In my first comment I just wonder what the tangent would be if $f$ were something like $y=x^{2/3}+c$
$endgroup$
– dmtri
Mar 19 at 8:03
$begingroup$
@alephzero, I am not saying that such an angle should exist.The sketch of the OP is just a sketch and not mathematicaly described, so no strict results can be deduced. In my first comment I just wonder what the tangent would be if $f$ were something like $y=x^{2/3}+c$
$endgroup$
– dmtri
Mar 19 at 8:03
|
show 3 more comments
7 Answers
7
active
oldest
votes
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One of the defining points (no pun!) of a tangent is the idea that as you look at a smaller and smaller section of the curve, the curve starts to more and more looks like a straight line.
That behaviour lets you do a huge number of things:
- You can estimate where the curve goes in places near that point (if tangent has a gradient of 2, then the curve at X + 0.000001 will probably be close to Y+0.000002, or whatever).
- You can approximate the curve by tiny sections of straight lines - this is the basis of simple calculus.
- You can take limits in several ways, and expect them to converge as you use narrower sections of the curve
- ? Many other things, all stemming from these.
Not all curves behave like that, and yours doesn't. No matter how close you inspect the cusp of your curve (the pointy bit), it's never going to resemble anything like a straight line, on any scale. It'll always look like a cusp.
That's the fundamental reason there isn't a tangent at that point. Because the curve just doesn't resemble a straight line, even in a microscopic close-up view, it doesn't have a gradient or tangent at that point, and it isn't differentiable at that point (much the same thing at a very simple level), and so on.
That isn't in fact unusual. In fact more curves don't have a gradient than do - its just that we don't study curves at random, so you mainly look at curves that do - at least, at this level of maths!
Other examples of curves that don't have a gradient at some or all points -
- the "step" curve (defined as y=0 if x<0, y=1 if x >= 0)
- a "curve" defined as y=1 if x contains a "1" when written in decimal, and y=0 otherwise. Because there are infinitely many numbers that simply don't have a 1 in their decimal expansion, slotted in between those that do!
- the curve y = 1/x at x=0
- the "blancmange curve", which doesn't have any breaks in it, and looks like it should be a nice straightforward curve, but actually "wobbles" so much at any microscopic level (however closely you look at it,and wherever you look!) that in the end, it doesn't have a gradient anywhere.
answered Mar 18 at 23:49
StilezStilez
776411
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add a comment |
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None of them, because it's not differentiable there, so there is no tangent line at that point.
answered Mar 18 at 16:06
BotondBotond
6,48331034
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If I have suppose a plot $(y^2-4x)(y^2-|-4x|)=0$ , I cannot differentiate this for all values of x?
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– ADITYA PRAKASH
Mar 18 at 16:16
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@ADITYAPRAKASH $(y^2-4x)(y^2-4|x|)=0$ is not a function, but a relation. How/where would you like to differentiate it?
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– Botond
Mar 18 at 16:23
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I put the wrong one , sorry . I wanted the graph of $y^2=4x and y^2=-4x$ together as a joint equation , positive parts only and to differentiate at x=0
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– ADITYA PRAKASH
Mar 18 at 16:27
6
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@ADITYAPRAKASH That's a relation too. But, for example, if you define $f$ as $f(x)=2sqrt{|x|}$, then $f$ is differentiable for all $x neq 0$. At $x=0$, the limit $lim_{x to 0} frac{sqrt{|x|}}{x}$ does not exists because when $x>0$, $frac{sqrt{|x|}}{x}=frac{sqrt{x}}{x}=frac{1}{sqrt{x}} to +infty$ and when $x < 0$, $frac{sqrt{|x|}}{x}=frac{sqrt{-x}}{x}=-frac{1}{sqrt{-x}} to -infty$, so the limit cannot exist.
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– Botond
Mar 18 at 16:33
3
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@dmtri How do you define $x^{2/3}$? If the domain is $mathbb{R}_{+} cup {0}$, then $0$ is not in the interior of the domain, so you can't differentiate it there. If you define it on $mathbb{R}$, then it won't be differentiable at $0$, so you won't have a tangent line there.
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– Botond
Mar 19 at 8:31
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show 6 more comments
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If the cuspid you draw is actually a cuspid (roughly speaking: if microscopically it is not "rounded") you can speak of a right and left tangent, same as you do with limits.
Refer to this wiki article.
Also note that, if the semi-derivative on one side has the value of $m$ and the other $-,m$, then the two lines geometrically coincide into a single line (or, maybe better, half line or ray).
answered Mar 18 at 16:13
G CabG Cab
20.4k31341
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To this I might add that for one point, left derivative = right derivative (and they both exist) $iff$ differentiable $iff$ has a (unique) tangent. And also cite wikipedia
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– asky
Mar 18 at 23:17
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@asky: good suggestion, thanks, added the ref.
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– G Cab
Mar 18 at 23:45
1
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@asky as the answer points out, unique tangent is equivalent to left derivative $=pm$ right derivative rather than differentiable.
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– jgon
Mar 19 at 0:22
2
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@jgon that depends how you define tangents being equal. From an analytic geometer's perspective, they have opposite slopes, so are not equal, but from a classical geometer's perspective, they look the same, so are equal. I think there is no canonical equivalence relation on lines, and both views accept an equivalence relation so are equally valid.
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– asky
Mar 19 at 1:43
2
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@asky I suppose that's a reasonable viewpoint. I'm suppose I'm just much more used to thinking of lines as unoriented.
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– jgon
Mar 19 at 1:48
add a comment |
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The "classical" tangent does not exist indeed.
However for a convex function with a sharp point, you can use a generalization of the derivative called the subderivative. It is used frequently in optimization.
answered Mar 18 at 16:39
poloCpoloC
912
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While I thikn mentioning the subderivative is definitely relevant here, the function of OP seems to be not convex at all. However, in this case you can take the generalized Clarke gradient (which reduces to subderivative when the function is convex).
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– Surb
Mar 18 at 20:20
add a comment |
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If $y=F(x)$ is not differentiable then the differential coefficient /derivative does not exist and there will be no unique tangent defined.
Ignore the following if it makes no sense now. There are situations when if a third space variable $z$ like in $(x,y,z) $ is defined involving a fourth common parameter $t$ like it is for this four branched curve between four sharp cusps:
$$ x= cos^3t, y= sin^3t, z= frac12 cos^2t $$
At $ x-z$ or $ y-z$ projections ( a shadow, if you will) exist tangents that suddenly rotate by $90^{circ}$ at cusp points and tangent slope includes all slopes in between.. in a situation where there is differentiability at these critical points.
answered Mar 18 at 16:40
NarasimhamNarasimham
21.1k62158
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Since we're doing geometry, let's think kinematically, too. Imagine a point moving along your curve, in the direction of increasing $x.$ Furthermore, let a ray project from the point, tangent to the curve there (you might think of your curve as a lane, and the point a car in the night; imagine the headlights beaming straight ahead as the tangent ray at any location on the curve).
Then as you approach the cuspoid from the left, the ray points down; at that point therefore, we can say it points down from the left. Now, let the point turn without leaving the point; then the tangent ray points up to the right -- at the same point! That happens nowhere else on the curve. Thus, we have two possible candidates for the tangent using this definition of tangent. If we require the tangent at a point to be unique, then there is no way to pick one of the two at the cuspoid without making further assumptions (which will be more or less arbitrary). The same thing happens for a point approaching the offending point from the right along the curve.
Thus, the curve has no unique tangent at that point. The reason is that you can think of the two branches of the curve as separate curves intersecting at a nonzero angle at that point. Suppose, for example, that we have two semicircular arcs instead, of the same radius and touching externally. For more clarity, take the positive parts of $(x+1)^2+y^2=1$ and $(x-1)^2+y^2=1.$ Then if we do not consider out tangents as directed, we can have a unique tangent at the origin in this case, since the two arcs are parallel at that place. However, if we think in terms of functions, there is still no derivative at that point, since the derivative measures the slope of the tangent line wrt increasing $x.$ However, if you imagine our point moving along this curve from left to right and crossing the origin, then the tangent ray points in opposite directions immediately before and after the point. Thus, there is no derivative there. However, purely geometrically, it is possible to speak of a tangent in this case.
I hope I have contributed to your understanding.
answered Mar 19 at 18:27
AllawonderAllawonder
1
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There are two possible definitions of "tangent" at a point $vec{p}$ in terms of the equation $f(vec{v})=0$ of the curve as follows.
Given a line $(1-t)vec{p}+tvec{q}$ (where $vec{p}$ and $vec{q}$ are distinct points on the line) we substitute this in the equation to obtain a function $g(t)$. The multiplicity of intersection of the curve and the line at $vec{p}$ is the order of vanishing of $g(t)$ at $t=0$ (which could be "infinite").
Definition 1
:
A tangent line is a line which meets with multiplicity at least 2.
This is the most common definition. With this definition, there are infinitely many tangents at the point in your figure.
Definition 2
:
A tangent line is a line which meets with the maximal multiplicity.
With this definition, there is a single such line in your example. However, we can easily make a figure with two or more cusps joined together and thus create an example with more than one tangent under this sense of tangent.
To avoid confusion, the second kind of tangent is usually not called a tangent but an "infinitely near point" in algebraic geometry. (However, this notion makes sense for more general situations than the case where $f$ is a algebraic.)
answered Mar 19 at 4:00
KapilKapil
39115
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7 Answers
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$begingroup$
One of the defining points (no pun!) of a tangent is the idea that as you look at a smaller and smaller section of the curve, the curve starts to more and more looks like a straight line.
That behaviour lets you do a huge number of things:
- You can estimate where the curve goes in places near that point (if tangent has a gradient of 2, then the curve at X + 0.000001 will probably be close to Y+0.000002, or whatever).
- You can approximate the curve by tiny sections of straight lines - this is the basis of simple calculus.
- You can take limits in several ways, and expect them to converge as you use narrower sections of the curve
- ? Many other things, all stemming from these.
Not all curves behave like that, and yours doesn't. No matter how close you inspect the cusp of your curve (the pointy bit), it's never going to resemble anything like a straight line, on any scale. It'll always look like a cusp.
That's the fundamental reason there isn't a tangent at that point. Because the curve just doesn't resemble a straight line, even in a microscopic close-up view, it doesn't have a gradient or tangent at that point, and it isn't differentiable at that point (much the same thing at a very simple level), and so on.
That isn't in fact unusual. In fact more curves don't have a gradient than do - its just that we don't study curves at random, so you mainly look at curves that do - at least, at this level of maths!
Other examples of curves that don't have a gradient at some or all points -
- the "step" curve (defined as y=0 if x<0, y=1 if x >= 0)
- a "curve" defined as y=1 if x contains a "1" when written in decimal, and y=0 otherwise. Because there are infinitely many numbers that simply don't have a 1 in their decimal expansion, slotted in between those that do!
- the curve y = 1/x at x=0
- the "blancmange curve", which doesn't have any breaks in it, and looks like it should be a nice straightforward curve, but actually "wobbles" so much at any microscopic level (however closely you look at it,and wherever you look!) that in the end, it doesn't have a gradient anywhere.
answered Mar 18 at 23:49
StilezStilez
776411
$endgroup$
add a comment |
$begingroup$
One of the defining points (no pun!) of a tangent is the idea that as you look at a smaller and smaller section of the curve, the curve starts to more and more looks like a straight line.
That behaviour lets you do a huge number of things:
- You can estimate where the curve goes in places near that point (if tangent has a gradient of 2, then the curve at X + 0.000001 will probably be close to Y+0.000002, or whatever).
- You can approximate the curve by tiny sections of straight lines - this is the basis of simple calculus.
- You can take limits in several ways, and expect them to converge as you use narrower sections of the curve
- ? Many other things, all stemming from these.
Not all curves behave like that, and yours doesn't. No matter how close you inspect the cusp of your curve (the pointy bit), it's never going to resemble anything like a straight line, on any scale. It'll always look like a cusp.
That's the fundamental reason there isn't a tangent at that point. Because the curve just doesn't resemble a straight line, even in a microscopic close-up view, it doesn't have a gradient or tangent at that point, and it isn't differentiable at that point (much the same thing at a very simple level), and so on.
That isn't in fact unusual. In fact more curves don't have a gradient than do - its just that we don't study curves at random, so you mainly look at curves that do - at least, at this level of maths!
Other examples of curves that don't have a gradient at some or all points -
- the "step" curve (defined as y=0 if x<0, y=1 if x >= 0)
- a "curve" defined as y=1 if x contains a "1" when written in decimal, and y=0 otherwise. Because there are infinitely many numbers that simply don't have a 1 in their decimal expansion, slotted in between those that do!
- the curve y = 1/x at x=0
- the "blancmange curve", which doesn't have any breaks in it, and looks like it should be a nice straightforward curve, but actually "wobbles" so much at any microscopic level (however closely you look at it,and wherever you look!) that in the end, it doesn't have a gradient anywhere.
answered Mar 18 at 23:49
StilezStilez
776411
$endgroup$
add a comment |
$begingroup$
One of the defining points (no pun!) of a tangent is the idea that as you look at a smaller and smaller section of the curve, the curve starts to more and more looks like a straight line.
That behaviour lets you do a huge number of things:
- You can estimate where the curve goes in places near that point (if tangent has a gradient of 2, then the curve at X + 0.000001 will probably be close to Y+0.000002, or whatever).
- You can approximate the curve by tiny sections of straight lines - this is the basis of simple calculus.
- You can take limits in several ways, and expect them to converge as you use narrower sections of the curve
- ? Many other things, all stemming from these.
Not all curves behave like that, and yours doesn't. No matter how close you inspect the cusp of your curve (the pointy bit), it's never going to resemble anything like a straight line, on any scale. It'll always look like a cusp.
That's the fundamental reason there isn't a tangent at that point. Because the curve just doesn't resemble a straight line, even in a microscopic close-up view, it doesn't have a gradient or tangent at that point, and it isn't differentiable at that point (much the same thing at a very simple level), and so on.
That isn't in fact unusual. In fact more curves don't have a gradient than do - its just that we don't study curves at random, so you mainly look at curves that do - at least, at this level of maths!
Other examples of curves that don't have a gradient at some or all points -
- the "step" curve (defined as y=0 if x<0, y=1 if x >= 0)
- a "curve" defined as y=1 if x contains a "1" when written in decimal, and y=0 otherwise. Because there are infinitely many numbers that simply don't have a 1 in their decimal expansion, slotted in between those that do!
- the curve y = 1/x at x=0
- the "blancmange curve", which doesn't have any breaks in it, and looks like it should be a nice straightforward curve, but actually "wobbles" so much at any microscopic level (however closely you look at it,and wherever you look!) that in the end, it doesn't have a gradient anywhere.
answered Mar 18 at 23:49
StilezStilez
776411
$endgroup$
One of the defining points (no pun!) of a tangent is the idea that as you look at a smaller and smaller section of the curve, the curve starts to more and more looks like a straight line.
That behaviour lets you do a huge number of things:
- You can estimate where the curve goes in places near that point (if tangent has a gradient of 2, then the curve at X + 0.000001 will probably be close to Y+0.000002, or whatever).
- You can approximate the curve by tiny sections of straight lines - this is the basis of simple calculus.
- You can take limits in several ways, and expect them to converge as you use narrower sections of the curve
- ? Many other things, all stemming from these.
Not all curves behave like that, and yours doesn't. No matter how close you inspect the cusp of your curve (the pointy bit), it's never going to resemble anything like a straight line, on any scale. It'll always look like a cusp.
That's the fundamental reason there isn't a tangent at that point. Because the curve just doesn't resemble a straight line, even in a microscopic close-up view, it doesn't have a gradient or tangent at that point, and it isn't differentiable at that point (much the same thing at a very simple level), and so on.
That isn't in fact unusual. In fact more curves don't have a gradient than do - its just that we don't study curves at random, so you mainly look at curves that do - at least, at this level of maths!
Other examples of curves that don't have a gradient at some or all points -
- the "step" curve (defined as y=0 if x<0, y=1 if x >= 0)
- a "curve" defined as y=1 if x contains a "1" when written in decimal, and y=0 otherwise. Because there are infinitely many numbers that simply don't have a 1 in their decimal expansion, slotted in between those that do!
- the curve y = 1/x at x=0
- the "blancmange curve", which doesn't have any breaks in it, and looks like it should be a nice straightforward curve, but actually "wobbles" so much at any microscopic level (however closely you look at it,and wherever you look!) that in the end, it doesn't have a gradient anywhere.
answered Mar 18 at 23:49
StilezStilez
776411
edited Mar 20 at 10:43
answered Mar 18 at 23:49
StilezStilez
776411
answered Mar 18 at 23:49
StilezStilez
776411
answered Mar 18 at 23:49
StilezStilez
776411
776411
add a comment |
add a comment |
$begingroup$
None of them, because it's not differentiable there, so there is no tangent line at that point.
answered Mar 18 at 16:06
BotondBotond
6,48331034
$endgroup$
$begingroup$
If I have suppose a plot $(y^2-4x)(y^2-|-4x|)=0$ , I cannot differentiate this for all values of x?
$endgroup$
– ADITYA PRAKASH
Mar 18 at 16:16
$begingroup$
@ADITYAPRAKASH $(y^2-4x)(y^2-4|x|)=0$ is not a function, but a relation. How/where would you like to differentiate it?
$endgroup$
– Botond
Mar 18 at 16:23
$begingroup$
I put the wrong one , sorry . I wanted the graph of $y^2=4x and y^2=-4x$ together as a joint equation , positive parts only and to differentiate at x=0
$endgroup$
– ADITYA PRAKASH
Mar 18 at 16:27
6
$begingroup$
@ADITYAPRAKASH That's a relation too. But, for example, if you define $f$ as $f(x)=2sqrt{|x|}$, then $f$ is differentiable for all $x neq 0$. At $x=0$, the limit $lim_{x to 0} frac{sqrt{|x|}}{x}$ does not exists because when $x>0$, $frac{sqrt{|x|}}{x}=frac{sqrt{x}}{x}=frac{1}{sqrt{x}} to +infty$ and when $x < 0$, $frac{sqrt{|x|}}{x}=frac{sqrt{-x}}{x}=-frac{1}{sqrt{-x}} to -infty$, so the limit cannot exist.
$endgroup$
– Botond
Mar 18 at 16:33
3
$begingroup$
@dmtri How do you define $x^{2/3}$? If the domain is $mathbb{R}_{+} cup {0}$, then $0$ is not in the interior of the domain, so you can't differentiate it there. If you define it on $mathbb{R}$, then it won't be differentiable at $0$, so you won't have a tangent line there.
$endgroup$
– Botond
Mar 19 at 8:31
|
show 6 more comments
$begingroup$
None of them, because it's not differentiable there, so there is no tangent line at that point.
answered Mar 18 at 16:06
BotondBotond
6,48331034
$endgroup$
$begingroup$
If I have suppose a plot $(y^2-4x)(y^2-|-4x|)=0$ , I cannot differentiate this for all values of x?
$endgroup$
– ADITYA PRAKASH
Mar 18 at 16:16
$begingroup$
@ADITYAPRAKASH $(y^2-4x)(y^2-4|x|)=0$ is not a function, but a relation. How/where would you like to differentiate it?
$endgroup$
– Botond
Mar 18 at 16:23
$begingroup$
I put the wrong one , sorry . I wanted the graph of $y^2=4x and y^2=-4x$ together as a joint equation , positive parts only and to differentiate at x=0
$endgroup$
– ADITYA PRAKASH
Mar 18 at 16:27
6
$begingroup$
@ADITYAPRAKASH That's a relation too. But, for example, if you define $f$ as $f(x)=2sqrt{|x|}$, then $f$ is differentiable for all $x neq 0$. At $x=0$, the limit $lim_{x to 0} frac{sqrt{|x|}}{x}$ does not exists because when $x>0$, $frac{sqrt{|x|}}{x}=frac{sqrt{x}}{x}=frac{1}{sqrt{x}} to +infty$ and when $x < 0$, $frac{sqrt{|x|}}{x}=frac{sqrt{-x}}{x}=-frac{1}{sqrt{-x}} to -infty$, so the limit cannot exist.
$endgroup$
– Botond
Mar 18 at 16:33
3
$begingroup$
@dmtri How do you define $x^{2/3}$? If the domain is $mathbb{R}_{+} cup {0}$, then $0$ is not in the interior of the domain, so you can't differentiate it there. If you define it on $mathbb{R}$, then it won't be differentiable at $0$, so you won't have a tangent line there.
$endgroup$
– Botond
Mar 19 at 8:31
|
show 6 more comments
$begingroup$
None of them, because it's not differentiable there, so there is no tangent line at that point.
answered Mar 18 at 16:06
BotondBotond
6,48331034
$endgroup$
None of them, because it's not differentiable there, so there is no tangent line at that point.
answered Mar 18 at 16:06
BotondBotond
6,48331034
answered Mar 18 at 16:06
BotondBotond
6,48331034
answered Mar 18 at 16:06
BotondBotond
6,48331034
answered Mar 18 at 16:06
BotondBotond
6,48331034
6,48331034
$begingroup$
If I have suppose a plot $(y^2-4x)(y^2-|-4x|)=0$ , I cannot differentiate this for all values of x?
$endgroup$
– ADITYA PRAKASH
Mar 18 at 16:16
$begingroup$
@ADITYAPRAKASH $(y^2-4x)(y^2-4|x|)=0$ is not a function, but a relation. How/where would you like to differentiate it?
$endgroup$
– Botond
Mar 18 at 16:23
$begingroup$
I put the wrong one , sorry . I wanted the graph of $y^2=4x and y^2=-4x$ together as a joint equation , positive parts only and to differentiate at x=0
$endgroup$
– ADITYA PRAKASH
Mar 18 at 16:27
6
$begingroup$
@ADITYAPRAKASH That's a relation too. But, for example, if you define $f$ as $f(x)=2sqrt{|x|}$, then $f$ is differentiable for all $x neq 0$. At $x=0$, the limit $lim_{x to 0} frac{sqrt{|x|}}{x}$ does not exists because when $x>0$, $frac{sqrt{|x|}}{x}=frac{sqrt{x}}{x}=frac{1}{sqrt{x}} to +infty$ and when $x < 0$, $frac{sqrt{|x|}}{x}=frac{sqrt{-x}}{x}=-frac{1}{sqrt{-x}} to -infty$, so the limit cannot exist.
$endgroup$
– Botond
Mar 18 at 16:33
3
$begingroup$
@dmtri How do you define $x^{2/3}$? If the domain is $mathbb{R}_{+} cup {0}$, then $0$ is not in the interior of the domain, so you can't differentiate it there. If you define it on $mathbb{R}$, then it won't be differentiable at $0$, so you won't have a tangent line there.
$endgroup$
– Botond
Mar 19 at 8:31
|
show 6 more comments
$begingroup$
If I have suppose a plot $(y^2-4x)(y^2-|-4x|)=0$ , I cannot differentiate this for all values of x?
$endgroup$
– ADITYA PRAKASH
Mar 18 at 16:16
$begingroup$
@ADITYAPRAKASH $(y^2-4x)(y^2-4|x|)=0$ is not a function, but a relation. How/where would you like to differentiate it?
$endgroup$
– Botond
Mar 18 at 16:23
$begingroup$
I put the wrong one , sorry . I wanted the graph of $y^2=4x and y^2=-4x$ together as a joint equation , positive parts only and to differentiate at x=0
$endgroup$
– ADITYA PRAKASH
Mar 18 at 16:27
6
$begingroup$
@ADITYAPRAKASH That's a relation too. But, for example, if you define $f$ as $f(x)=2sqrt{|x|}$, then $f$ is differentiable for all $x neq 0$. At $x=0$, the limit $lim_{x to 0} frac{sqrt{|x|}}{x}$ does not exists because when $x>0$, $frac{sqrt{|x|}}{x}=frac{sqrt{x}}{x}=frac{1}{sqrt{x}} to +infty$ and when $x < 0$, $frac{sqrt{|x|}}{x}=frac{sqrt{-x}}{x}=-frac{1}{sqrt{-x}} to -infty$, so the limit cannot exist.
$endgroup$
– Botond
Mar 18 at 16:33
3
$begingroup$
@dmtri How do you define $x^{2/3}$? If the domain is $mathbb{R}_{+} cup {0}$, then $0$ is not in the interior of the domain, so you can't differentiate it there. If you define it on $mathbb{R}$, then it won't be differentiable at $0$, so you won't have a tangent line there.
$endgroup$
– Botond
Mar 19 at 8:31
$begingroup$
If I have suppose a plot $(y^2-4x)(y^2-|-4x|)=0$ , I cannot differentiate this for all values of x?
$endgroup$
– ADITYA PRAKASH
Mar 18 at 16:16
$begingroup$
If I have suppose a plot $(y^2-4x)(y^2-|-4x|)=0$ , I cannot differentiate this for all values of x?
$endgroup$
– ADITYA PRAKASH
Mar 18 at 16:16
$begingroup$
@ADITYAPRAKASH $(y^2-4x)(y^2-4|x|)=0$ is not a function, but a relation. How/where would you like to differentiate it?
$endgroup$
– Botond
Mar 18 at 16:23
$begingroup$
@ADITYAPRAKASH $(y^2-4x)(y^2-4|x|)=0$ is not a function, but a relation. How/where would you like to differentiate it?
$endgroup$
– Botond
Mar 18 at 16:23
$begingroup$
I put the wrong one , sorry . I wanted the graph of $y^2=4x and y^2=-4x$ together as a joint equation , positive parts only and to differentiate at x=0
$endgroup$
– ADITYA PRAKASH
Mar 18 at 16:27
$begingroup$
I put the wrong one , sorry . I wanted the graph of $y^2=4x and y^2=-4x$ together as a joint equation , positive parts only and to differentiate at x=0
$endgroup$
– ADITYA PRAKASH
Mar 18 at 16:27
6
6
$begingroup$
@ADITYAPRAKASH That's a relation too. But, for example, if you define $f$ as $f(x)=2sqrt{|x|}$, then $f$ is differentiable for all $x neq 0$. At $x=0$, the limit $lim_{x to 0} frac{sqrt{|x|}}{x}$ does not exists because when $x>0$, $frac{sqrt{|x|}}{x}=frac{sqrt{x}}{x}=frac{1}{sqrt{x}} to +infty$ and when $x < 0$, $frac{sqrt{|x|}}{x}=frac{sqrt{-x}}{x}=-frac{1}{sqrt{-x}} to -infty$, so the limit cannot exist.
$endgroup$
– Botond
Mar 18 at 16:33
$begingroup$
@ADITYAPRAKASH That's a relation too. But, for example, if you define $f$ as $f(x)=2sqrt{|x|}$, then $f$ is differentiable for all $x neq 0$. At $x=0$, the limit $lim_{x to 0} frac{sqrt{|x|}}{x}$ does not exists because when $x>0$, $frac{sqrt{|x|}}{x}=frac{sqrt{x}}{x}=frac{1}{sqrt{x}} to +infty$ and when $x < 0$, $frac{sqrt{|x|}}{x}=frac{sqrt{-x}}{x}=-frac{1}{sqrt{-x}} to -infty$, so the limit cannot exist.
$endgroup$
– Botond
Mar 18 at 16:33
3
3
$begingroup$
@dmtri How do you define $x^{2/3}$? If the domain is $mathbb{R}_{+} cup {0}$, then $0$ is not in the interior of the domain, so you can't differentiate it there. If you define it on $mathbb{R}$, then it won't be differentiable at $0$, so you won't have a tangent line there.
$endgroup$
– Botond
Mar 19 at 8:31
$begingroup$
@dmtri How do you define $x^{2/3}$? If the domain is $mathbb{R}_{+} cup {0}$, then $0$ is not in the interior of the domain, so you can't differentiate it there. If you define it on $mathbb{R}$, then it won't be differentiable at $0$, so you won't have a tangent line there.
$endgroup$
– Botond
Mar 19 at 8:31
|
show 6 more comments
$begingroup$
If the cuspid you draw is actually a cuspid (roughly speaking: if microscopically it is not "rounded") you can speak of a right and left tangent, same as you do with limits.
Refer to this wiki article.
Also note that, if the semi-derivative on one side has the value of $m$ and the other $-,m$, then the two lines geometrically coincide into a single line (or, maybe better, half line or ray).
answered Mar 18 at 16:13
G CabG Cab
20.4k31341
$endgroup$
5
$begingroup$
To this I might add that for one point, left derivative = right derivative (and they both exist) $iff$ differentiable $iff$ has a (unique) tangent. And also cite wikipedia
$endgroup$
– asky
Mar 18 at 23:17
$begingroup$
@asky: good suggestion, thanks, added the ref.
$endgroup$
– G Cab
Mar 18 at 23:45
1
$begingroup$
@asky as the answer points out, unique tangent is equivalent to left derivative $=pm$ right derivative rather than differentiable.
$endgroup$
– jgon
Mar 19 at 0:22
2
$begingroup$
@jgon that depends how you define tangents being equal. From an analytic geometer's perspective, they have opposite slopes, so are not equal, but from a classical geometer's perspective, they look the same, so are equal. I think there is no canonical equivalence relation on lines, and both views accept an equivalence relation so are equally valid.
$endgroup$
– asky
Mar 19 at 1:43
2
$begingroup$
@asky I suppose that's a reasonable viewpoint. I'm suppose I'm just much more used to thinking of lines as unoriented.
$endgroup$
– jgon
Mar 19 at 1:48
add a comment |
$begingroup$
If the cuspid you draw is actually a cuspid (roughly speaking: if microscopically it is not "rounded") you can speak of a right and left tangent, same as you do with limits.
Refer to this wiki article.
Also note that, if the semi-derivative on one side has the value of $m$ and the other $-,m$, then the two lines geometrically coincide into a single line (or, maybe better, half line or ray).
answered Mar 18 at 16:13
G CabG Cab
20.4k31341
$endgroup$
5
$begingroup$
To this I might add that for one point, left derivative = right derivative (and they both exist) $iff$ differentiable $iff$ has a (unique) tangent. And also cite wikipedia
$endgroup$
– asky
Mar 18 at 23:17
$begingroup$
@asky: good suggestion, thanks, added the ref.
$endgroup$
– G Cab
Mar 18 at 23:45
1
$begingroup$
@asky as the answer points out, unique tangent is equivalent to left derivative $=pm$ right derivative rather than differentiable.
$endgroup$
– jgon
Mar 19 at 0:22
2
$begingroup$
@jgon that depends how you define tangents being equal. From an analytic geometer's perspective, they have opposite slopes, so are not equal, but from a classical geometer's perspective, they look the same, so are equal. I think there is no canonical equivalence relation on lines, and both views accept an equivalence relation so are equally valid.
$endgroup$
– asky
Mar 19 at 1:43
2
$begingroup$
@asky I suppose that's a reasonable viewpoint. I'm suppose I'm just much more used to thinking of lines as unoriented.
$endgroup$
– jgon
Mar 19 at 1:48
add a comment |
$begingroup$
If the cuspid you draw is actually a cuspid (roughly speaking: if microscopically it is not "rounded") you can speak of a right and left tangent, same as you do with limits.
Refer to this wiki article.
Also note that, if the semi-derivative on one side has the value of $m$ and the other $-,m$, then the two lines geometrically coincide into a single line (or, maybe better, half line or ray).
answered Mar 18 at 16:13
G CabG Cab
20.4k31341
$endgroup$
If the cuspid you draw is actually a cuspid (roughly speaking: if microscopically it is not "rounded") you can speak of a right and left tangent, same as you do with limits.
Refer to this wiki article.
Also note that, if the semi-derivative on one side has the value of $m$ and the other $-,m$, then the two lines geometrically coincide into a single line (or, maybe better, half line or ray).
answered Mar 18 at 16:13
G CabG Cab
20.4k31341
edited Mar 18 at 23:54
answered Mar 18 at 16:13
G CabG Cab
20.4k31341
answered Mar 18 at 16:13
G CabG Cab
20.4k31341
answered Mar 18 at 16:13
G CabG Cab
20.4k31341
20.4k31341
5
$begingroup$
To this I might add that for one point, left derivative = right derivative (and they both exist) $iff$ differentiable $iff$ has a (unique) tangent. And also cite wikipedia
$endgroup$
– asky
Mar 18 at 23:17
$begingroup$
@asky: good suggestion, thanks, added the ref.
$endgroup$
– G Cab
Mar 18 at 23:45
1
$begingroup$
@asky as the answer points out, unique tangent is equivalent to left derivative $=pm$ right derivative rather than differentiable.
$endgroup$
– jgon
Mar 19 at 0:22
2
$begingroup$
@jgon that depends how you define tangents being equal. From an analytic geometer's perspective, they have opposite slopes, so are not equal, but from a classical geometer's perspective, they look the same, so are equal. I think there is no canonical equivalence relation on lines, and both views accept an equivalence relation so are equally valid.
$endgroup$
– asky
Mar 19 at 1:43
2
$begingroup$
@asky I suppose that's a reasonable viewpoint. I'm suppose I'm just much more used to thinking of lines as unoriented.
$endgroup$
– jgon
Mar 19 at 1:48
add a comment |
5
$begingroup$
To this I might add that for one point, left derivative = right derivative (and they both exist) $iff$ differentiable $iff$ has a (unique) tangent. And also cite wikipedia
$endgroup$
– asky
Mar 18 at 23:17
$begingroup$
@asky: good suggestion, thanks, added the ref.
$endgroup$
– G Cab
Mar 18 at 23:45
1
$begingroup$
@asky as the answer points out, unique tangent is equivalent to left derivative $=pm$ right derivative rather than differentiable.
$endgroup$
– jgon
Mar 19 at 0:22
2
$begingroup$
@jgon that depends how you define tangents being equal. From an analytic geometer's perspective, they have opposite slopes, so are not equal, but from a classical geometer's perspective, they look the same, so are equal. I think there is no canonical equivalence relation on lines, and both views accept an equivalence relation so are equally valid.
$endgroup$
– asky
Mar 19 at 1:43
2
$begingroup$
@asky I suppose that's a reasonable viewpoint. I'm suppose I'm just much more used to thinking of lines as unoriented.
$endgroup$
– jgon
Mar 19 at 1:48
5
5
$begingroup$
To this I might add that for one point, left derivative = right derivative (and they both exist) $iff$ differentiable $iff$ has a (unique) tangent. And also cite wikipedia
$endgroup$
– asky
Mar 18 at 23:17
$begingroup$
To this I might add that for one point, left derivative = right derivative (and they both exist) $iff$ differentiable $iff$ has a (unique) tangent. And also cite wikipedia
$endgroup$
– asky
Mar 18 at 23:17
$begingroup$
@asky: good suggestion, thanks, added the ref.
$endgroup$
– G Cab
Mar 18 at 23:45
$begingroup$
@asky: good suggestion, thanks, added the ref.
$endgroup$
– G Cab
Mar 18 at 23:45
1
1
$begingroup$
@asky as the answer points out, unique tangent is equivalent to left derivative $=pm$ right derivative rather than differentiable.
$endgroup$
– jgon
Mar 19 at 0:22
$begingroup$
@asky as the answer points out, unique tangent is equivalent to left derivative $=pm$ right derivative rather than differentiable.
$endgroup$
– jgon
Mar 19 at 0:22
2
2
$begingroup$
@jgon that depends how you define tangents being equal. From an analytic geometer's perspective, they have opposite slopes, so are not equal, but from a classical geometer's perspective, they look the same, so are equal. I think there is no canonical equivalence relation on lines, and both views accept an equivalence relation so are equally valid.
$endgroup$
– asky
Mar 19 at 1:43
$begingroup$
@jgon that depends how you define tangents being equal. From an analytic geometer's perspective, they have opposite slopes, so are not equal, but from a classical geometer's perspective, they look the same, so are equal. I think there is no canonical equivalence relation on lines, and both views accept an equivalence relation so are equally valid.
$endgroup$
– asky
Mar 19 at 1:43
2
2
$begingroup$
@asky I suppose that's a reasonable viewpoint. I'm suppose I'm just much more used to thinking of lines as unoriented.
$endgroup$
– jgon
Mar 19 at 1:48
$begingroup$
@asky I suppose that's a reasonable viewpoint. I'm suppose I'm just much more used to thinking of lines as unoriented.
$endgroup$
– jgon
Mar 19 at 1:48
add a comment |
$begingroup$
The "classical" tangent does not exist indeed.
However for a convex function with a sharp point, you can use a generalization of the derivative called the subderivative. It is used frequently in optimization.
answered Mar 18 at 16:39
poloCpoloC
912
$endgroup$
3
$begingroup$
While I thikn mentioning the subderivative is definitely relevant here, the function of OP seems to be not convex at all. However, in this case you can take the generalized Clarke gradient (which reduces to subderivative when the function is convex).
$endgroup$
– Surb
Mar 18 at 20:20
add a comment |
$begingroup$
The "classical" tangent does not exist indeed.
However for a convex function with a sharp point, you can use a generalization of the derivative called the subderivative. It is used frequently in optimization.
answered Mar 18 at 16:39
poloCpoloC
912
$endgroup$
3
$begingroup$
While I thikn mentioning the subderivative is definitely relevant here, the function of OP seems to be not convex at all. However, in this case you can take the generalized Clarke gradient (which reduces to subderivative when the function is convex).
$endgroup$
– Surb
Mar 18 at 20:20
add a comment |
$begingroup$
The "classical" tangent does not exist indeed.
However for a convex function with a sharp point, you can use a generalization of the derivative called the subderivative. It is used frequently in optimization.
answered Mar 18 at 16:39
poloCpoloC
912
$endgroup$
The "classical" tangent does not exist indeed.
However for a convex function with a sharp point, you can use a generalization of the derivative called the subderivative. It is used frequently in optimization.
answered Mar 18 at 16:39
poloCpoloC
912
answered Mar 18 at 16:39
poloCpoloC
912
answered Mar 18 at 16:39
poloCpoloC
912
answered Mar 18 at 16:39
poloCpoloC
912
912
3
$begingroup$
While I thikn mentioning the subderivative is definitely relevant here, the function of OP seems to be not convex at all. However, in this case you can take the generalized Clarke gradient (which reduces to subderivative when the function is convex).
$endgroup$
– Surb
Mar 18 at 20:20
add a comment |
3
$begingroup$
While I thikn mentioning the subderivative is definitely relevant here, the function of OP seems to be not convex at all. However, in this case you can take the generalized Clarke gradient (which reduces to subderivative when the function is convex).
$endgroup$
– Surb
Mar 18 at 20:20
3
3
$begingroup$
While I thikn mentioning the subderivative is definitely relevant here, the function of OP seems to be not convex at all. However, in this case you can take the generalized Clarke gradient (which reduces to subderivative when the function is convex).
$endgroup$
– Surb
Mar 18 at 20:20
$begingroup$
While I thikn mentioning the subderivative is definitely relevant here, the function of OP seems to be not convex at all. However, in this case you can take the generalized Clarke gradient (which reduces to subderivative when the function is convex).
$endgroup$
– Surb
Mar 18 at 20:20
add a comment |
$begingroup$
If $y=F(x)$ is not differentiable then the differential coefficient /derivative does not exist and there will be no unique tangent defined.
Ignore the following if it makes no sense now. There are situations when if a third space variable $z$ like in $(x,y,z) $ is defined involving a fourth common parameter $t$ like it is for this four branched curve between four sharp cusps:
$$ x= cos^3t, y= sin^3t, z= frac12 cos^2t $$
At $ x-z$ or $ y-z$ projections ( a shadow, if you will) exist tangents that suddenly rotate by $90^{circ}$ at cusp points and tangent slope includes all slopes in between.. in a situation where there is differentiability at these critical points.
answered Mar 18 at 16:40
NarasimhamNarasimham
21.1k62158
$endgroup$
add a comment |
$begingroup$
If $y=F(x)$ is not differentiable then the differential coefficient /derivative does not exist and there will be no unique tangent defined.
Ignore the following if it makes no sense now. There are situations when if a third space variable $z$ like in $(x,y,z) $ is defined involving a fourth common parameter $t$ like it is for this four branched curve between four sharp cusps:
$$ x= cos^3t, y= sin^3t, z= frac12 cos^2t $$
At $ x-z$ or $ y-z$ projections ( a shadow, if you will) exist tangents that suddenly rotate by $90^{circ}$ at cusp points and tangent slope includes all slopes in between.. in a situation where there is differentiability at these critical points.
answered Mar 18 at 16:40
NarasimhamNarasimham
21.1k62158
$endgroup$
add a comment |
$begingroup$
If $y=F(x)$ is not differentiable then the differential coefficient /derivative does not exist and there will be no unique tangent defined.
Ignore the following if it makes no sense now. There are situations when if a third space variable $z$ like in $(x,y,z) $ is defined involving a fourth common parameter $t$ like it is for this four branched curve between four sharp cusps:
$$ x= cos^3t, y= sin^3t, z= frac12 cos^2t $$
At $ x-z$ or $ y-z$ projections ( a shadow, if you will) exist tangents that suddenly rotate by $90^{circ}$ at cusp points and tangent slope includes all slopes in between.. in a situation where there is differentiability at these critical points.
answered Mar 18 at 16:40
NarasimhamNarasimham
21.1k62158
$endgroup$
If $y=F(x)$ is not differentiable then the differential coefficient /derivative does not exist and there will be no unique tangent defined.
Ignore the following if it makes no sense now. There are situations when if a third space variable $z$ like in $(x,y,z) $ is defined involving a fourth common parameter $t$ like it is for this four branched curve between four sharp cusps:
$$ x= cos^3t, y= sin^3t, z= frac12 cos^2t $$
At $ x-z$ or $ y-z$ projections ( a shadow, if you will) exist tangents that suddenly rotate by $90^{circ}$ at cusp points and tangent slope includes all slopes in between.. in a situation where there is differentiability at these critical points.
answered Mar 18 at 16:40
NarasimhamNarasimham
21.1k62158
edited Mar 19 at 4:58
answered Mar 18 at 16:40
NarasimhamNarasimham
21.1k62158
answered Mar 18 at 16:40
NarasimhamNarasimham
21.1k62158
answered Mar 18 at 16:40
NarasimhamNarasimham
21.1k62158
21.1k62158
add a comment |
add a comment |
$begingroup$
Since we're doing geometry, let's think kinematically, too. Imagine a point moving along your curve, in the direction of increasing $x.$ Furthermore, let a ray project from the point, tangent to the curve there (you might think of your curve as a lane, and the point a car in the night; imagine the headlights beaming straight ahead as the tangent ray at any location on the curve).
Then as you approach the cuspoid from the left, the ray points down; at that point therefore, we can say it points down from the left. Now, let the point turn without leaving the point; then the tangent ray points up to the right -- at the same point! That happens nowhere else on the curve. Thus, we have two possible candidates for the tangent using this definition of tangent. If we require the tangent at a point to be unique, then there is no way to pick one of the two at the cuspoid without making further assumptions (which will be more or less arbitrary). The same thing happens for a point approaching the offending point from the right along the curve.
Thus, the curve has no unique tangent at that point. The reason is that you can think of the two branches of the curve as separate curves intersecting at a nonzero angle at that point. Suppose, for example, that we have two semicircular arcs instead, of the same radius and touching externally. For more clarity, take the positive parts of $(x+1)^2+y^2=1$ and $(x-1)^2+y^2=1.$ Then if we do not consider out tangents as directed, we can have a unique tangent at the origin in this case, since the two arcs are parallel at that place. However, if we think in terms of functions, there is still no derivative at that point, since the derivative measures the slope of the tangent line wrt increasing $x.$ However, if you imagine our point moving along this curve from left to right and crossing the origin, then the tangent ray points in opposite directions immediately before and after the point. Thus, there is no derivative there. However, purely geometrically, it is possible to speak of a tangent in this case.
I hope I have contributed to your understanding.
answered Mar 19 at 18:27
AllawonderAllawonder
1
$endgroup$
add a comment |
$begingroup$
Since we're doing geometry, let's think kinematically, too. Imagine a point moving along your curve, in the direction of increasing $x.$ Furthermore, let a ray project from the point, tangent to the curve there (you might think of your curve as a lane, and the point a car in the night; imagine the headlights beaming straight ahead as the tangent ray at any location on the curve).
Then as you approach the cuspoid from the left, the ray points down; at that point therefore, we can say it points down from the left. Now, let the point turn without leaving the point; then the tangent ray points up to the right -- at the same point! That happens nowhere else on the curve. Thus, we have two possible candidates for the tangent using this definition of tangent. If we require the tangent at a point to be unique, then there is no way to pick one of the two at the cuspoid without making further assumptions (which will be more or less arbitrary). The same thing happens for a point approaching the offending point from the right along the curve.
Thus, the curve has no unique tangent at that point. The reason is that you can think of the two branches of the curve as separate curves intersecting at a nonzero angle at that point. Suppose, for example, that we have two semicircular arcs instead, of the same radius and touching externally. For more clarity, take the positive parts of $(x+1)^2+y^2=1$ and $(x-1)^2+y^2=1.$ Then if we do not consider out tangents as directed, we can have a unique tangent at the origin in this case, since the two arcs are parallel at that place. However, if we think in terms of functions, there is still no derivative at that point, since the derivative measures the slope of the tangent line wrt increasing $x.$ However, if you imagine our point moving along this curve from left to right and crossing the origin, then the tangent ray points in opposite directions immediately before and after the point. Thus, there is no derivative there. However, purely geometrically, it is possible to speak of a tangent in this case.
I hope I have contributed to your understanding.
answered Mar 19 at 18:27
AllawonderAllawonder
1
$endgroup$
add a comment |
$begingroup$
Since we're doing geometry, let's think kinematically, too. Imagine a point moving along your curve, in the direction of increasing $x.$ Furthermore, let a ray project from the point, tangent to the curve there (you might think of your curve as a lane, and the point a car in the night; imagine the headlights beaming straight ahead as the tangent ray at any location on the curve).
Then as you approach the cuspoid from the left, the ray points down; at that point therefore, we can say it points down from the left. Now, let the point turn without leaving the point; then the tangent ray points up to the right -- at the same point! That happens nowhere else on the curve. Thus, we have two possible candidates for the tangent using this definition of tangent. If we require the tangent at a point to be unique, then there is no way to pick one of the two at the cuspoid without making further assumptions (which will be more or less arbitrary). The same thing happens for a point approaching the offending point from the right along the curve.
Thus, the curve has no unique tangent at that point. The reason is that you can think of the two branches of the curve as separate curves intersecting at a nonzero angle at that point. Suppose, for example, that we have two semicircular arcs instead, of the same radius and touching externally. For more clarity, take the positive parts of $(x+1)^2+y^2=1$ and $(x-1)^2+y^2=1.$ Then if we do not consider out tangents as directed, we can have a unique tangent at the origin in this case, since the two arcs are parallel at that place. However, if we think in terms of functions, there is still no derivative at that point, since the derivative measures the slope of the tangent line wrt increasing $x.$ However, if you imagine our point moving along this curve from left to right and crossing the origin, then the tangent ray points in opposite directions immediately before and after the point. Thus, there is no derivative there. However, purely geometrically, it is possible to speak of a tangent in this case.
I hope I have contributed to your understanding.
answered Mar 19 at 18:27
AllawonderAllawonder
1
$endgroup$
Since we're doing geometry, let's think kinematically, too. Imagine a point moving along your curve, in the direction of increasing $x.$ Furthermore, let a ray project from the point, tangent to the curve there (you might think of your curve as a lane, and the point a car in the night; imagine the headlights beaming straight ahead as the tangent ray at any location on the curve).
Then as you approach the cuspoid from the left, the ray points down; at that point therefore, we can say it points down from the left. Now, let the point turn without leaving the point; then the tangent ray points up to the right -- at the same point! That happens nowhere else on the curve. Thus, we have two possible candidates for the tangent using this definition of tangent. If we require the tangent at a point to be unique, then there is no way to pick one of the two at the cuspoid without making further assumptions (which will be more or less arbitrary). The same thing happens for a point approaching the offending point from the right along the curve.
Thus, the curve has no unique tangent at that point. The reason is that you can think of the two branches of the curve as separate curves intersecting at a nonzero angle at that point. Suppose, for example, that we have two semicircular arcs instead, of the same radius and touching externally. For more clarity, take the positive parts of $(x+1)^2+y^2=1$ and $(x-1)^2+y^2=1.$ Then if we do not consider out tangents as directed, we can have a unique tangent at the origin in this case, since the two arcs are parallel at that place. However, if we think in terms of functions, there is still no derivative at that point, since the derivative measures the slope of the tangent line wrt increasing $x.$ However, if you imagine our point moving along this curve from left to right and crossing the origin, then the tangent ray points in opposite directions immediately before and after the point. Thus, there is no derivative there. However, purely geometrically, it is possible to speak of a tangent in this case.
I hope I have contributed to your understanding.
answered Mar 19 at 18:27
AllawonderAllawonder
1
answered Mar 19 at 18:27
AllawonderAllawonder
1
answered Mar 19 at 18:27
AllawonderAllawonder
1
answered Mar 19 at 18:27
AllawonderAllawonder
1
1
add a comment |
add a comment |
$begingroup$
There are two possible definitions of "tangent" at a point $vec{p}$ in terms of the equation $f(vec{v})=0$ of the curve as follows.
Given a line $(1-t)vec{p}+tvec{q}$ (where $vec{p}$ and $vec{q}$ are distinct points on the line) we substitute this in the equation to obtain a function $g(t)$. The multiplicity of intersection of the curve and the line at $vec{p}$ is the order of vanishing of $g(t)$ at $t=0$ (which could be "infinite").
Definition 1
:
A tangent line is a line which meets with multiplicity at least 2.
This is the most common definition. With this definition, there are infinitely many tangents at the point in your figure.
Definition 2
:
A tangent line is a line which meets with the maximal multiplicity.
With this definition, there is a single such line in your example. However, we can easily make a figure with two or more cusps joined together and thus create an example with more than one tangent under this sense of tangent.
To avoid confusion, the second kind of tangent is usually not called a tangent but an "infinitely near point" in algebraic geometry. (However, this notion makes sense for more general situations than the case where $f$ is a algebraic.)
answered Mar 19 at 4:00
KapilKapil
39115
$endgroup$
add a comment |
$begingroup$
There are two possible definitions of "tangent" at a point $vec{p}$ in terms of the equation $f(vec{v})=0$ of the curve as follows.
Given a line $(1-t)vec{p}+tvec{q}$ (where $vec{p}$ and $vec{q}$ are distinct points on the line) we substitute this in the equation to obtain a function $g(t)$. The multiplicity of intersection of the curve and the line at $vec{p}$ is the order of vanishing of $g(t)$ at $t=0$ (which could be "infinite").
Definition 1
:
A tangent line is a line which meets with multiplicity at least 2.
This is the most common definition. With this definition, there are infinitely many tangents at the point in your figure.
Definition 2
:
A tangent line is a line which meets with the maximal multiplicity.
With this definition, there is a single such line in your example. However, we can easily make a figure with two or more cusps joined together and thus create an example with more than one tangent under this sense of tangent.
To avoid confusion, the second kind of tangent is usually not called a tangent but an "infinitely near point" in algebraic geometry. (However, this notion makes sense for more general situations than the case where $f$ is a algebraic.)
answered Mar 19 at 4:00
KapilKapil
39115
$endgroup$
add a comment |
$begingroup$
There are two possible definitions of "tangent" at a point $vec{p}$ in terms of the equation $f(vec{v})=0$ of the curve as follows.
Given a line $(1-t)vec{p}+tvec{q}$ (where $vec{p}$ and $vec{q}$ are distinct points on the line) we substitute this in the equation to obtain a function $g(t)$. The multiplicity of intersection of the curve and the line at $vec{p}$ is the order of vanishing of $g(t)$ at $t=0$ (which could be "infinite").
Definition 1
:
A tangent line is a line which meets with multiplicity at least 2.
This is the most common definition. With this definition, there are infinitely many tangents at the point in your figure.
Definition 2
:
A tangent line is a line which meets with the maximal multiplicity.
With this definition, there is a single such line in your example. However, we can easily make a figure with two or more cusps joined together and thus create an example with more than one tangent under this sense of tangent.
To avoid confusion, the second kind of tangent is usually not called a tangent but an "infinitely near point" in algebraic geometry. (However, this notion makes sense for more general situations than the case where $f$ is a algebraic.)
answered Mar 19 at 4:00
KapilKapil
39115
$endgroup$
There are two possible definitions of "tangent" at a point $vec{p}$ in terms of the equation $f(vec{v})=0$ of the curve as follows.
Given a line $(1-t)vec{p}+tvec{q}$ (where $vec{p}$ and $vec{q}$ are distinct points on the line) we substitute this in the equation to obtain a function $g(t)$. The multiplicity of intersection of the curve and the line at $vec{p}$ is the order of vanishing of $g(t)$ at $t=0$ (which could be "infinite").
Definition 1
:
A tangent line is a line which meets with multiplicity at least 2.
This is the most common definition. With this definition, there are infinitely many tangents at the point in your figure.
Definition 2
:
A tangent line is a line which meets with the maximal multiplicity.
With this definition, there is a single such line in your example. However, we can easily make a figure with two or more cusps joined together and thus create an example with more than one tangent under this sense of tangent.
To avoid confusion, the second kind of tangent is usually not called a tangent but an "infinitely near point" in algebraic geometry. (However, this notion makes sense for more general situations than the case where $f$ is a algebraic.)
answered Mar 19 at 4:00
KapilKapil
39115
edited Mar 20 at 1:46
answered Mar 19 at 4:00
KapilKapil
39115
answered Mar 19 at 4:00
KapilKapil
39115
answered Mar 19 at 4:00
KapilKapil
39115
39115
add a comment |
add a comment |
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$begingroup$
It does not exist for the reason you give. The tangent, if it exists, is unique.
$endgroup$
– John Douma
Mar 18 at 16:06
1
$begingroup$
how about the bisector of the "angle" made of the right and left part of your nice sharp curve?
$endgroup$
– dmtri
Mar 18 at 16:14
$begingroup$
If the curve consists of 2 arcs, let's say parts of the circles with radius 1 and centers -1 and 1 then the axis $x=0$ might be the tangent .
$endgroup$
– dmtri
Mar 18 at 17:22
4
$begingroup$
A tangent is a line that approximates the curve near the point - such that if you zoom in close enough on the point, the tangent and curve become pratically indistinguishable. That is the property of tangents that makes them useful. Left and Right tangents, which approximate the curve on only one side of the point, are also useful concepts. A line that does not approximate the curve on either side is useful only in very rare and limited circumstances, unworthy of a special name.
$endgroup$
– Paul Sinclair
Mar 19 at 3:11
1
$begingroup$
@alephzero, I am not saying that such an angle should exist.The sketch of the OP is just a sketch and not mathematicaly described, so no strict results can be deduced. In my first comment I just wonder what the tangent would be if $f$ were something like $y=x^{2/3}+c$
$endgroup$
– dmtri
Mar 19 at 8:03