# How do you define a tangent to the curve y = f(x)?

Is x-axis tangent to y = x^3?

siamese_scythe

You provided good insight, but I still have a querry.

As you stated that the tangent line is a limiting postion of the secant which means that when the secant is moved parallel to itself, the point where its length becomes zero or in other words the points of intersection of the secant (two or more) merge into one it becomes a tangent line. In that case would not any line x = 0 to the curve y = lxl other than those having slopes in the interval [-1, 1] be a tangent line? This is because, a secant moved parallel to itself can become the tangent when tthe two points of intersection merge into one at x = 0.

Can we define tangent to the curve y = f(x) at the point (x0, f(x0)) as the line which intersects the curve at that point in two or more times? What I mean is that if solving the equation of the given line with that of the curve, if we get two or more identical solutions, then that line can be considered a tangent line.

bskelkar has properly clarified my doubt in the first para of additional details. I understood that a secant through the point P at which we are considering a tangent and which passes through Q cutting a chord PQ is to be rotated such that Q approaches P and not that a secant is moved parallel to itself.

About another clarification of the definition of tangent that I am considering "tangent is a line which intersects the curve at the point of tangency two or more times." For example, in y = x^3, solving with the equation of x-axis, y = 0 => x^3 = 0 which has three identical solutions all x = 0. The line y = x+1 is a tangent to the parabola y^2 = 4x because their soltion gives (x-1)^2 = 0, i.e., two roots x=1 and x = 1at the point of tangency. But y = 2 when solved with y^2 = 4x gices x = 1 which is only one root and hence y = 2 intersects and is not a tangent.

Another point is that a tangent to a curve which is a polynomial of 3 or higher degree may meet the curve again at some point, but is still a tangent, For example, x-axis is a tangent at x = 1 to the curve y = (x-1)^2 (x-2), as their solution is x = 1, x= 1 and x = 2 which means that x-axis is tangent at x = 1 and not at x = 2. However, x-axis will be tangent at both points x = 1 and x = 2 for y = (x-1)^2(x-2)^2.

husoski

Your concept of continuity in defining the tangent as a limiting position of secant is indeed thought provoking. Keeping this point in mind, can we now proceed to define the tangent as under.

For a tangent to a curve at a point to be defined, the curve must be continuous everywhere in some small neighbourhood of the point. Suppose, it is desired to define the tangent at x = xo in a small neighbourhood (a-xo, a+xo) such that the curve is continuous everywhere in that interval. When the secant passing through xo and a point in the interval when rotated about the point at x = xo reaches a limiting value, such that it does not intersect the curve at any point other than xo in that interval, it becomes the tangent to the curve at x = xo.

husoski

Your concept of continuity in defining the tangent as a limiting position of secant is indeed thought provoking. Keeping this point in mind, can we now proceed to define the tangent as under.

For a tangent to a curve at a point to be defined, the curve must be continuous everywhere in some small neighbourhood of the point. Suppose, it is desired to define the tangent at x = xo in a small neighbourhood (a-xo, a+xo) such that the curve is continuous everywhere in that interval. When the secant passing through xo and a point in the interval when rotated about the point at x = xo reaches a limiting value, such that it does not intersect the curve at any point other than xo in that interval, it becomes the tangent to the curve at x = xo.

### 16 Answers

- QuadrilleratorLv 59 years agoBest Answer
Yikes! Are we really consigning the ancient Greeks and everyone without calculus to not be able to understand the definition for a tangent?! Somehow, calculus oughtn't to make us forget geometry. Don't we have a notion that a line is tangent to a curve if it approaches the curve, intersects with it just barely, and then "glances" off? More formally:

Line L is tangent to curve C at point P if both C and L pass through P and there exists a circle B about P such that no pair of points of C within P lie on opposite sides of L (In other words, in a local neighborhood around P, all of C's points lie on the same side of L).

"Pass through" ensures that P is not isolated on C. Something like: C passes through P if for every circle B' around P contained in B, there is a line L' through P such that there are at least two points of C - P in B' on opposite sides of L'.

Calling y=0 tangent to y=x³ comes from taking parts of two curves (such as -|x³| and |x³|) whose tangents agree at a given point but where the concavity flips. I have always been unhappy with calling this line tangent when it clearly isn't.

Also, by my definition, lines are tangent to themselves, and

any line through (0,0) whose slope was <= 1 would be a tangent to y=|x|.

- bskelkarLv 79 years ago
The function is y = f(x) mean we are in 2D set up.

If P and Q are 2 points on the curve The tangent at P is the limiting posititon of secant PQ as Q tends to P ALONG THE CURVE. So first, the question that a secant shifted // to itself does not arise. Second, secant is a line so we can not talk about its "length"

or a tanget at P to exist, the function must have derivative defined at P. Geometrically the curve must be smooth. X-axis is in fact tangent to y = x^3 at the origin.

The curves like circle and other (smooth) conic sectios give an impression that the curve should lie entirely on the same side of the tangent. But it not so!

I Guess this suffices, Madhukar Sir.

- husoskiLv 79 years ago
So far, I'd vote for siamese_scythe's answer, with one objection: What is the definition of the limiting position of a sequence of lines? Actually, it's more like a continuum of lines, given that the "moving" point has a continuous set of positions leading up to the limit. Don't get me wrong...I know what's meant by the definition. I just can't think of a rigorous definition of such a limit, offhand.

I'd like to point out that there is more than one kind of tangent. The original definition only applied to circles: "A tangent line intersects the circle at just one point." That's easily extended to arbitrary convex closed curves by adding ", if and only if it is the only line to do so." to the end; but I'm not so sure the addition is all that desirable. I see no particular problem with a convex polygon having multiple "tangents" at the vertices.

The reason I bring this up: It seems to me that "tangent" is a geometric concept and ought to have a geometric definition. Analytic geometry is fine, but it's tied to a coordinate system. The tangent line that I imaging when somebody says "tangent line" is independent of any coordinate system and doesn't blow up just because dy/dx may not be defined in the some coordinate system that's in use.

Mind you, I'd *use* the calculus version in practice. I certainly wouldn't want to do an "existence and uniqueness" proof every time I solved for a tangent line--which is what the convex-curve tangent above requires. I just like definitions to be rooted somewhere, without wandering off into possible circularity.

"What's a tangent?"

"Thats the line with the same derivative."

"Okay, but what's a derivative?"

"Oh, that's the slope of the tangent line."

Edit:

siamese_scythe has an acceptable definition for the limit of a secant lines, which will work so long as the curve doesn't intersect itself. Or, at least not in some neighborhood of the point of tangency. I'm still voting for that answer.

Quadrillerator has an undeserved thumbs-down. There are problems, (like which side of the straight line are the points of the straight line on?) but ones that can be cleared up. Rakesh Dubey also offers a "pure tangent" idea that's interesting.

From Merriam-Webster's 11th Collegiate Dictionary, "tangent" derives from the Latin tangere: "to touch". That fits with the definition from Euclid, Book III, Defintion 2: A straight line is said to touch a circle, when meeting it and being produced does not cut it.

Google Books has a digitized copy of a 1765 edit of an early translation. Page 450-451 contain remarks about how this definition is deficient for the idea of a tangent for curves other than conic sections.

http://books.google.com/books?id=kT44AAAAMAAJ&pg=P...

One of the points mentioned there is the same as Rakesh Dubey's "local tangent that is still a secant" idea, that the straight line can, in fact, just touch a curve in one place and yet cut it in another.

The tangent to a circle has a few useful properies:

1. It intersects the circle at a single point.

2. The remaining points of the tangent line are all outside the circle. (i.e. the tangent line does not "cut" the circle.)

3. The direction of the line is the same as the "instantaneous" direction of the circle at the point of tangency.

Euclid specifies properties 1 and 2 in his definition for circles, while calculus specifies property 3. You can't have all three for a general smooth curve.

For a calculus definition, I like the vector calculus approach. A curve is defined by parmetrization using the vector R(t) = (x(t), y(t)), where x and y are continuous, piecewise differentiable functions of t, and |R'(t)| is nonzero wherever both x'(t) and y'(t) exist. Then the tangent at some point R_0 = R(t0) is the line through R_0 with the direction given by R'(t0). If R'(t) is undefined at t0, there is no tangent defined by R at t=t0.

That last remark is a nod to Rita the Dog, who pointed out the problem of self intersecting curves. A simple example is a figure-8 that crosses itself at right angles. One parametrization may go smoothly around the curve, with well-defined tangents everywhere, including two different ones at the intersection point. Another my take right-angled turns at the intersection, and produce no definition of either tangent at the intersection.

It's messy, but it does avoid the problem of vertical tangents.

- MathematishanLv 69 years ago
Not sure about this but I guess a better definition of the function is required

if I take a function like y = mod x (abs value of x) i will not have a tangent at (0,0) becuase the line will have slope -1 and +1 just to teh left and right of (0,0)

so we first need to ensure that the curve has a smooth transitition and continuity. Otherwise the defination of tangent has 2 parts like others have answered

1) first order derivative = slope and the line meets the curve at one point

the point where x axis y= 0 meets y = x^3 is an infection point but is a tangent (special case) where teh lines crosses the curve

- How do you think about the answers? You can sign in to vote the answer.
- 9 years ago
According to the power rule

df(x)/dx=3*(X^2)

A tangent to a curve will have a slope equal to the derivative of f(x) with respect to x, but it will be a line in point slope form having x and f(x) as coordinates, but it does not make sense to use x, because you have to plug in an input and there are an infinite number of tangent lines, each with different points and slopes so I will plug in a variable Q to explain see below

When X=Q for the function f(x) the tangent line would be defined as

Coordinates for tangent line are [Q, (X-Q)*f ' (Q) + f(Q)]

Notice that I wrote F prime of Q when I wrote f ' Q

So in this specific case

Coordinates for tangent line are [Q, (X-Q)*3*(Q^2)+Q^3 ]

that is when x=Q it would have an x coordinate of Q and a Y coordinate of f(Q)

but it's slope would be 3*(Q^2) because the derivative of f(x) with respect to X is 3*(X^2)

x-axis is only tangent to that function when X=0 because that is the only time that 3*(X^2)=0

for all other X values the x-axis is NOT tangent to that f(x)

By the way I am assuming only 2 dimensions

Derivative is used because it makes the instantaneous slope the same as the instantaneous slope of the function at that point

Source(s): Took at least 5 math classes using Calculus - Rakesh DubeyLv 69 years ago
Nice answers by experts.TUs. Here is some points by me.

For a tangent to a curve y = ƒ(x), where ƒ(x) is continuous in [a, b] and is differentiable at c ∈ (a, b) { (c, ƒ(c) ) is the point at which the tangent is desired } the derivative ƒ'(c) represents the slope of the tangent to the graph of the function y = ƒ(x) at (c, ƒ(c) ).

A tangent at a particular point on the graph of the curve may be secant at some other point on the graph also.

For a circle (which is not a function ) any tangent will be a pure tangent (means never can be a secant).

For the cubic parabola y = x³, X-axis is the only a pure tangent at the origin and Y-axis is the normal. All the other tangents at other points on the curve will also be secants (it will meet again to the curve).

- Rita the dogLv 79 years ago
Just some food for thought: it seems to me various definitions are possible, which would agree in most situations but probably not in all. For example, the unit circle should have x=1 has a tangent line at (1,0), but it is not a function near that point. Most functions do not have graphs that are curves, e.g. f(x) = 0 if x is irrational and 1 if x is rational, so "curve" needs some definition. The "double point" or higher notion, leads to (or is related to) intersection theory (see http://en.wikipedia.org/wiki/Intersection_theory_%... and Bezout's theorem (see http://en.wikipedia.org/wiki/B%C3%A9zout%27s_theor...

Many people are defining tangent lines in terms of derivatives. But there are some situations where the derivative does not exist and the tangent line does (or ought to, depending on definition). For example, consider f(x) =√(-x(x+2)) for -2≤x≤0 and f(x) = -√(-x(x-2)) for 0≤x≤2. This function is not differentiable at x=0 but surely the line x=0 should be a tangent line there. What about the function whose graph is two semicircles of radius 1, centers on the x axis, lying above the x-axis, and tangent at (0,0)? I would say x=0 is a tangent line at (0,0) but chords not passing through (0,0) do not have a limiting position as their endpoints approach (0,0).

What about curves that lie between y=x^2 and the x-axis near (0,0)? Should they have a horizontal tangent line at (0,0)? It might seem so, but consider y=x^2 sin(1/x). It fills those requirements and is continuous, and dy/dx exists everywhere even at the origin (where its value is 0) [thanks gianlino for pointing out my error here], but the x-axis crosses the curve y=x^2 sin(1/x) infinitely many times in every interval containing 0, so should it be considered a tangent line to the curve at (0,0), or not?

[Edit: gianlino kindly points out that y=x^2 sin(1/x) is the standard example of a differentiable function whose derivative is not continuous at a point, a fact which takes me back more years than I care to admit. I corrected my last example above, accordingly.]

- ShyLv 69 years ago
Hi Madhukar

I would imagine that y' is the slope of the tangent at the point on the curve y=x^3

y' = 3x^2

so at x = 0, y' = 0 and y = 0

So y = 0 is the tangent to the curve at (0,0)

YES x axis is the tangent to the curve at(0,0)

At x = 1 say

y' = 3 and y =1 and

c in y = mx+c (the equation of the tangent)

1 =3*1 + c

c = -2

So y = 3x -2 is the eqn of tangent at (1,1) on the curve y =x^3

Shy

- supastremphLv 69 years ago
I would have to suggest that not only the curve be continuous, but also be differentiable, or have a continuous derivative.

Using the definition of the limiting value of secant, we have a tangent line at x0 iff the limit of the secant between ( x0,f(x0)) and (x0-epsilon, f(x0-epsilon)) equals the same line as the secant (x0,f(x0)) and (x0+epsilon, f(x0 + epsilon)) as epsilon approaches 0. That is a secant to the curve left of the point of interest will equal the secant to the curve right of the point of interest as their lengths go to zero.

This condition is not satisfied by y = |x| for x0=0, meaning that there exist no tangent at that point.

- intc_escapeeLv 79 years ago
A straight line is tangent to a given curve f(x) at a point x₀ on the curve if the line passes through the point (x₀,f(x₀)) on the curve and has slope f'(x₀), where f'(x) is the derivative of f(x).

Yes, y=0 is tangent to f(x) = x^3 at x = 0

Edit: your follow-up is an interesting idea. please elaborate. I'm not sure I understand it completely.

Answer: see above