What is the tangent at a sharp point on a curve?Find the number of tangent lines to a curveTangent line at point PProof of tangent lines to a curveFor a curve, find the unit tangent vector and parametric equation of the line tangent to the curve at the given pointI have the slope of a tangent line and the curve but I have to find the point(s)Finding the tangent to a point on the curveT(t)≠0 for all values of t and the tangent line at any given point of the curve always passes through point D. Show that r represents a straight lineIn Calculus, why must the gradient of a specific point of a curve be given by a line that locally touches the curve at one and only one point?Can there be more than two tangents to a curve from an external point? And can a tangent at some point cut the curve elsewhere?Find the tangent line to a curve
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What is the tangent at a sharp point on a curve?
Find the number of tangent lines to a curveTangent line at point PProof of tangent lines to a curveFor a curve, find the unit tangent vector and parametric equation of the line tangent to the curve at the given pointI have the slope of a tangent line and the curve but I have to find the point(s)Finding the tangent to a point on the curveT(t)≠0 for all values of t and the tangent line at any given point of the curve always passes through point D. Show that r represents a straight lineIn Calculus, why must the gradient of a specific point of a curve be given by a line that locally touches the curve at one and only one point?Can there be more than two tangents to a curve from an external point? And can a tangent at some point cut the curve elsewhere?Find the tangent line to 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
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add a comment |
$begingroup$
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
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7
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It does not exist for the reason you give. The tangent, if it exists, is unique.
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– John Douma
9 hours ago
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
9 hours ago
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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
8 hours ago
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@dmtri because the "angle" might not exist. Consider a function like $y = 2|x| + x sin(1/x)$, at $x = 0$. If you draw a graph, it "sort of" looks like it has a cusp at $x = 0$, but the tangent direction is "infinitely wiggly" on either side of the cusp. Trying to give a useful mathematical definition of what you mean by "a nice sharp curve" is hard!
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– alephzero
6 hours ago
add a comment |
$begingroup$
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
$endgroup$
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 9 hours ago
ADITYA PRAKASHADITYA PRAKASH
1127
1127
7
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It does not exist for the reason you give. The tangent, if it exists, is unique.
$endgroup$
– John Douma
9 hours ago
1
$begingroup$
how about the bisector of the "angle" made of the right and left part of your nice sharp curve?
$endgroup$
– dmtri
9 hours ago
$begingroup$
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
8 hours ago
$begingroup$
@dmtri because the "angle" might not exist. Consider a function like $y = 2|x| + x sin(1/x)$, at $x = 0$. If you draw a graph, it "sort of" looks like it has a cusp at $x = 0$, but the tangent direction is "infinitely wiggly" on either side of the cusp. Trying to give a useful mathematical definition of what you mean by "a nice sharp curve" is hard!
$endgroup$
– alephzero
6 hours ago
add a comment |
7
$begingroup$
It does not exist for the reason you give. The tangent, if it exists, is unique.
$endgroup$
– John Douma
9 hours ago
1
$begingroup$
how about the bisector of the "angle" made of the right and left part of your nice sharp curve?
$endgroup$
– dmtri
9 hours ago
$begingroup$
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
8 hours ago
$begingroup$
@dmtri because the "angle" might not exist. Consider a function like $y = 2|x| + x sin(1/x)$, at $x = 0$. If you draw a graph, it "sort of" looks like it has a cusp at $x = 0$, but the tangent direction is "infinitely wiggly" on either side of the cusp. Trying to give a useful mathematical definition of what you mean by "a nice sharp curve" is hard!
$endgroup$
– alephzero
6 hours ago
7
7
$begingroup$
It does not exist for the reason you give. The tangent, if it exists, is unique.
$endgroup$
– John Douma
9 hours ago
$begingroup$
It does not exist for the reason you give. The tangent, if it exists, is unique.
$endgroup$
– John Douma
9 hours ago
1
1
$begingroup$
how about the bisector of the "angle" made of the right and left part of your nice sharp curve?
$endgroup$
– dmtri
9 hours ago
$begingroup$
how about the bisector of the "angle" made of the right and left part of your nice sharp curve?
$endgroup$
– dmtri
9 hours ago
$begingroup$
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
8 hours ago
$begingroup$
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
8 hours ago
$begingroup$
@dmtri because the "angle" might not exist. Consider a function like $y = 2|x| + x sin(1/x)$, at $x = 0$. If you draw a graph, it "sort of" looks like it has a cusp at $x = 0$, but the tangent direction is "infinitely wiggly" on either side of the cusp. Trying to give a useful mathematical definition of what you mean by "a nice sharp curve" is hard!
$endgroup$
– alephzero
6 hours ago
$begingroup$
@dmtri because the "angle" might not exist. Consider a function like $y = 2|x| + x sin(1/x)$, at $x = 0$. If you draw a graph, it "sort of" looks like it has a cusp at $x = 0$, but the tangent direction is "infinitely wiggly" on either side of the cusp. Trying to give a useful mathematical definition of what you mean by "a nice sharp curve" is hard!
$endgroup$
– alephzero
6 hours ago
add a comment |
5 Answers
5
active
oldest
votes
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None of them, because it's not differentiable there, so there is no tangent line at that point.
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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
9 hours ago
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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
9 hours ago
$begingroup$
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
9 hours ago
4
$begingroup$
@ADITYAPRAKASH That's a relation too. But, for example, if you define $f$ as $f(x)=2sqrt$, then $f$ is differentiable for all $x neq 0$. At $x=0$, the limit $lim_x to 0 fracsqrtx$ does not exists because when $x>0$, $fracsqrtx=fracsqrtxx=frac1sqrtx to +infty$ and when $x < 0$, $fracsqrtx=fracsqrt-xx=-frac1sqrt-x to -infty$, so the limit cannot exist.
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– Botond
9 hours ago
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).
$endgroup$
2
$begingroup$
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
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– asky
2 hours ago
$begingroup$
@asky: good suggestion, thanks, added the ref.
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– G Cab
2 hours ago
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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
1 hour ago
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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
3 mins ago
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.
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1
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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
5 hours ago
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:
$$ x= cos^3t, y= sin^3t, z= frac12 cos^2t $$
then at $ x-z$ or $ y-z$ projections ( a shadow if you will) shows a tangent that suddenly rotates 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.
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add a comment |
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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.
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add a comment |
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5 Answers
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active
oldest
votes
5 Answers
5
active
oldest
votes
active
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active
oldest
votes
$begingroup$
None of them, because it's not differentiable there, so there is no tangent line at that point.
$endgroup$
$begingroup$
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
9 hours ago
$begingroup$
@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
9 hours ago
$begingroup$
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
9 hours ago
4
$begingroup$
@ADITYAPRAKASH That's a relation too. But, for example, if you define $f$ as $f(x)=2sqrt$, then $f$ is differentiable for all $x neq 0$. At $x=0$, the limit $lim_x to 0 fracsqrtx$ does not exists because when $x>0$, $fracsqrtx=fracsqrtxx=frac1sqrtx to +infty$ and when $x < 0$, $fracsqrtx=fracsqrt-xx=-frac1sqrt-x to -infty$, so the limit cannot exist.
$endgroup$
– Botond
9 hours ago
add a comment |
$begingroup$
None of them, because it's not differentiable there, so there is no tangent line at that point.
$endgroup$
$begingroup$
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
9 hours ago
$begingroup$
@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
9 hours ago
$begingroup$
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
9 hours ago
4
$begingroup$
@ADITYAPRAKASH That's a relation too. But, for example, if you define $f$ as $f(x)=2sqrt$, then $f$ is differentiable for all $x neq 0$. At $x=0$, the limit $lim_x to 0 fracsqrtx$ does not exists because when $x>0$, $fracsqrtx=fracsqrtxx=frac1sqrtx to +infty$ and when $x < 0$, $fracsqrtx=fracsqrt-xx=-frac1sqrt-x to -infty$, so the limit cannot exist.
$endgroup$
– Botond
9 hours ago
add a comment |
$begingroup$
None of them, because it's not differentiable there, so there is no tangent line at that point.
$endgroup$
None of them, because it's not differentiable there, so there is no tangent line at that point.
answered 9 hours ago
BotondBotond
6,2832933
6,2832933
$begingroup$
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
9 hours ago
$begingroup$
@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
9 hours ago
$begingroup$
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
9 hours ago
4
$begingroup$
@ADITYAPRAKASH That's a relation too. But, for example, if you define $f$ as $f(x)=2sqrt$, then $f$ is differentiable for all $x neq 0$. At $x=0$, the limit $lim_x to 0 fracsqrtx$ does not exists because when $x>0$, $fracsqrtx=fracsqrtxx=frac1sqrtx to +infty$ and when $x < 0$, $fracsqrtx=fracsqrt-xx=-frac1sqrt-x to -infty$, so the limit cannot exist.
$endgroup$
– Botond
9 hours ago
add a comment |
$begingroup$
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
9 hours ago
$begingroup$
@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
9 hours ago
$begingroup$
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
9 hours ago
4
$begingroup$
@ADITYAPRAKASH That's a relation too. But, for example, if you define $f$ as $f(x)=2sqrt$, then $f$ is differentiable for all $x neq 0$. At $x=0$, the limit $lim_x to 0 fracsqrtx$ does not exists because when $x>0$, $fracsqrtx=fracsqrtxx=frac1sqrtx to +infty$ and when $x < 0$, $fracsqrtx=fracsqrt-xx=-frac1sqrt-x to -infty$, so the limit cannot exist.
$endgroup$
– Botond
9 hours ago
$begingroup$
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
9 hours ago
$begingroup$
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
9 hours ago
$begingroup$
@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
9 hours ago
$begingroup$
@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
9 hours ago
$begingroup$
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
9 hours ago
$begingroup$
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
9 hours ago
4
4
$begingroup$
@ADITYAPRAKASH That's a relation too. But, for example, if you define $f$ as $f(x)=2sqrt$, then $f$ is differentiable for all $x neq 0$. At $x=0$, the limit $lim_x to 0 fracsqrtx$ does not exists because when $x>0$, $fracsqrtx=fracsqrtxx=frac1sqrtx to +infty$ and when $x < 0$, $fracsqrtx=fracsqrt-xx=-frac1sqrt-x to -infty$, so the limit cannot exist.
$endgroup$
– Botond
9 hours ago
$begingroup$
@ADITYAPRAKASH That's a relation too. But, for example, if you define $f$ as $f(x)=2sqrt$, then $f$ is differentiable for all $x neq 0$. At $x=0$, the limit $lim_x to 0 fracsqrtx$ does not exists because when $x>0$, $fracsqrtx=fracsqrtxx=frac1sqrtx to +infty$ and when $x < 0$, $fracsqrtx=fracsqrt-xx=-frac1sqrt-x to -infty$, so the limit cannot exist.
$endgroup$
– Botond
9 hours ago
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).
$endgroup$
2
$begingroup$
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
2 hours ago
$begingroup$
@asky: good suggestion, thanks, added the ref.
$endgroup$
– G Cab
2 hours ago
$begingroup$
@asky as the answer points out, unique tangent is equivalent to left derivative $=pm$ right derivative rather than differentiable.
$endgroup$
– jgon
1 hour ago
$begingroup$
@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
3 mins ago
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).
$endgroup$
2
$begingroup$
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
2 hours ago
$begingroup$
@asky: good suggestion, thanks, added the ref.
$endgroup$
– G Cab
2 hours ago
$begingroup$
@asky as the answer points out, unique tangent is equivalent to left derivative $=pm$ right derivative rather than differentiable.
$endgroup$
– jgon
1 hour ago
$begingroup$
@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
3 mins ago
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).
$endgroup$
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).
edited 1 hour ago
answered 9 hours ago
G CabG Cab
20.3k31341
20.3k31341
2
$begingroup$
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
2 hours ago
$begingroup$
@asky: good suggestion, thanks, added the ref.
$endgroup$
– G Cab
2 hours ago
$begingroup$
@asky as the answer points out, unique tangent is equivalent to left derivative $=pm$ right derivative rather than differentiable.
$endgroup$
– jgon
1 hour ago
$begingroup$
@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
3 mins ago
add a comment |
2
$begingroup$
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
2 hours ago
$begingroup$
@asky: good suggestion, thanks, added the ref.
$endgroup$
– G Cab
2 hours ago
$begingroup$
@asky as the answer points out, unique tangent is equivalent to left derivative $=pm$ right derivative rather than differentiable.
$endgroup$
– jgon
1 hour ago
$begingroup$
@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
3 mins ago
2
2
$begingroup$
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
2 hours ago
$begingroup$
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
2 hours ago
$begingroup$
@asky: good suggestion, thanks, added the ref.
$endgroup$
– G Cab
2 hours ago
$begingroup$
@asky: good suggestion, thanks, added the ref.
$endgroup$
– G Cab
2 hours ago
$begingroup$
@asky as the answer points out, unique tangent is equivalent to left derivative $=pm$ right derivative rather than differentiable.
$endgroup$
– jgon
1 hour ago
$begingroup$
@asky as the answer points out, unique tangent is equivalent to left derivative $=pm$ right derivative rather than differentiable.
$endgroup$
– jgon
1 hour ago
$begingroup$
@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
3 mins ago
$begingroup$
@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
3 mins ago
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.
$endgroup$
1
$begingroup$
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
5 hours ago
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.
$endgroup$
1
$begingroup$
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
5 hours ago
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.
$endgroup$
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 9 hours ago
poloCpoloC
411
411
1
$begingroup$
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
5 hours ago
add a comment |
1
$begingroup$
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
5 hours ago
1
1
$begingroup$
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
5 hours ago
$begingroup$
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
5 hours ago
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:
$$ x= cos^3t, y= sin^3t, z= frac12 cos^2t $$
then at $ x-z$ or $ y-z$ projections ( a shadow if you will) shows a tangent that suddenly rotates 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.
$endgroup$
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:
$$ x= cos^3t, y= sin^3t, z= frac12 cos^2t $$
then at $ x-z$ or $ y-z$ projections ( a shadow if you will) shows a tangent that suddenly rotates 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.
$endgroup$
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:
$$ x= cos^3t, y= sin^3t, z= frac12 cos^2t $$
then at $ x-z$ or $ y-z$ projections ( a shadow if you will) shows a tangent that suddenly rotates 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.
$endgroup$
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:
$$ x= cos^3t, y= sin^3t, z= frac12 cos^2t $$
then at $ x-z$ or $ y-z$ projections ( a shadow if you will) shows a tangent that suddenly rotates 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.
edited 7 hours ago
answered 9 hours ago
NarasimhamNarasimham
21k62158
21k62158
add a comment |
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.
$endgroup$
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.
$endgroup$
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.
$endgroup$
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 1 hour ago
StilezStilez
42129
42129
add a comment |
add a comment |
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7
$begingroup$
It does not exist for the reason you give. The tangent, if it exists, is unique.
$endgroup$
– John Douma
9 hours ago
1
$begingroup$
how about the bisector of the "angle" made of the right and left part of your nice sharp curve?
$endgroup$
– dmtri
9 hours ago
$begingroup$
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
8 hours ago
$begingroup$
@dmtri because the "angle" might not exist. Consider a function like $y = 2|x| + x sin(1/x)$, at $x = 0$. If you draw a graph, it "sort of" looks like it has a cusp at $x = 0$, but the tangent direction is "infinitely wiggly" on either side of the cusp. Trying to give a useful mathematical definition of what you mean by "a nice sharp curve" is hard!
$endgroup$
– alephzero
6 hours ago