8
$\begingroup$

Suppose $x=t^2,y=t^3$ is a parametric curve. Here's a quote from my textbook:

The origin, which corresponds to $t=0$, is a singular point of the parametric curve, because $dx/dt=2t,dy/dt=3t^2$ are both zero when $t=0$.

So far so good.

But then they write:

However, the curve has a horizontal tangent line at the origin, because for all $t\neq 0$: $$\frac{dy}{dx}=\frac{dy/dt}{dx/dt}=\frac{3}{2}t$$ And thus: $$\lim_{t\to 0^+} \frac{dy}{dx}=\lim_{t\to 0^-} \frac{dy}{dx}=0$$

It looked a little odd for me. Nevertheless, I decided to use the same argument to show that the parametric curve $x=2\cos t - \cos (2t), y=2\sin t - \sin(2t)$ has a horizontal tangent line at $t=0$, that is at $(1,0)$.

However my professor said that this is wrong ("because the derivative is not zero" - indeed, $\frac{dy}{dx}\Big|_{t=0}$ is undefined - "$0/0$").

So who is right? Is the existence of the limit a sufficient condition for the (horizontal) tangent line to exist, as my textbook says, or not? I'm confused.

Thanks.

Short version of the question: can a parametric curve have a horizontal tangent line at a singular point?

I.e. is $\lim_{t\to 0} \frac{dy}{dx}=0$ a sufficient condition for a horizontal line to exist (at $t=0$)? (even if the derivative $\frac{dy}{dx}\Big |_{t=0}$ itself doesn't exist).

$\endgroup$
  • $\begingroup$ Reading your comments to the answers below, I think you are unduly concerned about the distinction between $f(x_0)$ and $\lim_{x\to x_0}f(x)$. Where we have an isolated point $x_0$ for which $f(x_0)$ does not exist, but $\lim_{x\to x_0}f(x)$ does exist, it is common for people to assume without comment that we define $f(x_0)$ as the limit. $\endgroup$ – almagest May 3 '16 at 18:31
  • $\begingroup$ what is your textbook ? $\endgroup$ – KonKan May 3 '16 at 18:44
  • $\begingroup$ @KonKan - Calculus by Anton (5th ed.) $\endgroup$ – user239753 May 3 '16 at 18:53
  • 1
    $\begingroup$ @almagest - unduly? This might be an obvious assumption for some, but not for me. The very fact that my professor dismissed it says that this is not trivial, and at least deserves a mention. $\endgroup$ – user239753 May 3 '16 at 19:02
  • $\begingroup$ @user239753: i'm a bit confused: I'm searching in your book (it happens to have a copy of the 10th edition, available right in front of me) but i cannot find the quote you are mentioning. I am actually looking at Ch.10, par.10.1, p.692-705. The example you are mentioning is example 6, p.697. However, the authors simply mention that $t=0$ is a singular point. I cannot find the limit computation you are mentioning. Can you indicate the exact page ? $\endgroup$ – KonKan May 3 '16 at 19:21
3
+25
$\begingroup$

Notice that both your curves are algebraic curves, of equations $x^3 - y^2 = 0$ and $(x^2+y^2-1)^2-4((x-1)^2+y^2)=0$ respectively.

We say that the line $ax + by + c = 0$ is tangent at the point $(x_0, y_0)$ to the curve given by $F(x,y) = 0$ if and only if $(x_0, y_0)$ is a multiple root of the system $\begin{cases} F(x,y) = 0 \\ ax + by + c = 0 \end{cases}$ (this is the definition that was originally used by algebraic geometers).

In your case, the points are $(0,0)$ in the first case and $(1, 0)$ in the second, and the line to be checked is $y=0$.

Plugging $y = 0$ into $x^3 - y^2 = 0$ gives $x^3 = 0$, which indeed has $x=0$ as a triple root, therefore $y=0$ is tangent to $x^3 - y^2 = 0$ at $(0,0)$.

Plugging $y = 0$ into $(x^2+y^2-1)^2-4((x-1)^2+y^2)=0$ gives $(x^2-1)^2 - 4(x-1)^2 = 0$, or equivalently $(x-1)^2 ((x+1)^2 - 4) = 0$, or again $(x-1)^3 (x+3) = 0$, which indeed has $x=1$ as a triple root (the root $-3$ being simple, meaning that at the point $(-3, 0)$ the line $y=0$ intersects the curve as a secant), therefore $y=0$ is tangent to $(x^2+y^2-1)^2-4((x-1)^2+y^2)=0$ at $(1,0)$.


Since the bounty giver is still unsure, let us follow Wikipedia and compute the intersection number in the first case. Wikipedia gives several methods, one of them is to consider the ideal generated by $y^2 - x^3$ and by $y$ in $\Bbb R[[x,y]]$ (the ring of formal series in $x$ and $y$) and compute the dimension of the quotient vector space $\Bbb R [[x,y]] / (x^3-y^2, y)$.

In the quotient space, both $y$ and $x^3 - y^2$ will become $0$, which implies that $x^3$ also becomes $0$. Therefore, the only powers of $x$ and $y$ that survive in the quotient set are $x^0 = y^0 = 1$, and $x$ and $x^2$ - a total of $3$ linearly independent powers, so $3$ will be the intersection number of $y=0$ and $x^3-y^2$ at $(0,0)$. Since everything $\ge 2$ means tangency, this shows that the line $y=0$ is tangent to $x^3-y^2$ at $(0,0)$.

The same thing could be done with what Wikipedia denotes by $I_{(0,0)}$, letting $P = y$ and $Q = x^3 - y^2$. Applying the properties that you see on that page (the numbers above equal signs are the Wikipedia properties that I apply),

$$I_{(0,0)} (y, x^3-y^2) \overset 6 = I_{(0,0)} (y, x^3) \overset 5 = I_{(0,0)}(x,y) + I_{(0,0)}(x,y) + I_{(0,0)}(x,y) \overset 4 = 1 + 1 + 1 = 3 .$$


The same computations could be done for the second example, it is just that they are more tedious. First, let us translate the curve such that the cusp moves from $(1,0)$ to $(0,0)$ (because, for simplicity, Wikipedia's formulae are given only for $(0,0)$). To do this we shall make the change of variable $x = u + 1$, which will lead to $(u^2 +2u +y^2)^2 -4(u^2+y^2)$. Next,

$$I_{(0,0)} ((u^2 +2u +y^2)^2 -4(u^2+y^2), y) \overset 6 = I_{(0,0)} (u^4 + 4u^3, y) = I_{(0,0)} (u^3 (u+4), y) \overset 5 = I_{(0,0)} (u^3, y) + \\ I_{(0,0)} (u+4, y) = I_{(0,0)} (u, y) + I_{(0,0)} (u, y) + I_{(0,0)} (u, y) + I_{(0,0)} (u+4, y) \overset {3, 4} = 1+1+1+0 = 3 .$$

Again, we obtain an intersection number $\ge 2$, which means tangency.


What others have correctly stated is that the two plane curves given in the problem are not differentiable submanifolds of $\Bbb R^2$ at the singular points under discussion, therefore they do not fit into the framework of diferential geometry, therefore we may not speak of their tangent spaces at the singular points as defined in differential geometry (a proof that an almost identical curve cannot be a smooth submanifold can be found here). Counterintuitively, though, they do have tangent lines! There is no contradiction, the thing is that we use two meanings for "being tangent" that are not synonymous: one comes from differential geometry ("tangent space"), and it doesn't apply here, the other from algebraic geometry ("intersection number"), and it does apply here. In the latter framework, we may speak about "a given line being tangent (or not) to a curve in a point" without speaking of "the tangent space at that curve in that point".

$y=0$ is tangent to those curves at the specified points. Those curves do not have nicely defined tangent spaces at those points.

$\endgroup$
  • $\begingroup$ @user239753: Try these two: "Elementary Differential Geometry" by A.N. Pressley, and "Differential Geometry of Curves and Surfaces" by M.P. do Carmo. One last word: differential geometry in general is interested only in those spaces that are endowed with regular parametrizations (because one wants them to be invertible, with their inverses called "local charts"). Most theorems about parametrized curves assume the parametrization to be regular. $\endgroup$ – Alex M. Apr 28 '16 at 20:24
  • 2
    $\begingroup$ @AlexM. Because this is all about subtile nuances: You wrote, the cardiod would be a manifold, because the defining function $F$ is smooth. Aren't you applying the regular value theorem (which is a close relative to the implicite function theorem) here, which means 0 needs to be a regular value for $F$? (Anyway: great answer!) $\endgroup$ – hase_olaf Apr 28 '16 at 21:31
  • $\begingroup$ @AlexM. - thank you for your refined answer. A bit stupid question - what does "multiple root" mean? It has something to do with algebraic multiplicity? $\endgroup$ – user239753 May 3 '16 at 19:27
  • $\begingroup$ @user239753: Yes, exactly. We say that $a$ is a root of order $k$ of $P$ (where $P$ is a polynomial) if and only if $(x-a)^k \mid P$ and $(x-a)^{k+1} \nmid P$. The analogue for algebraic curves is the concept of "intersection multiplicity". $\endgroup$ – Alex M. May 3 '16 at 19:41
  • 1
    $\begingroup$ @AlexM: I have to confess .. I am the ignorant who downvoted your hard work ;) but give me a moment to explain. your answer (especially after your edit) seems valid in my eyes and you've obviously put hard work in it ! I give you credit for that. However, I feel it falls outside the spirit and the OP's question: it is a delicate question on a subtle point of parametric differentiation (from the calculus viewpoint) while you certainly view the situation from the algebraic/differential geometric eye. This is also why i rolled back your retagging. $\endgroup$ – KonKan May 5 '16 at 23:29
0
$\begingroup$

In order to find the tangent at some special point you should not try to compute the limit of $y'$ but the actual limit of secant directions at that point. Therefore in your first example you get $$m_+:=\lim_{t\to0+}{y(t)-y(0)\over x(t)-x(0)}=\lim_{t\to0+}{t^3\over t^2}=0\ ,$$ and similarly $m_-=0$. In your second example you have $${y(t)-y(0)\over x(t)-x(0)}={2\sin t-\sin(2t)-0\over 2\cos t-\cos(2t)-1}={2\sin t(1-\cos t)\over2\cos t(1-\cos t)}=\tan t\qquad(t\ne0)\ ,$$ and therefore $$m_+=m_-=\lim_{t\to0}{y(t)-y(0)\over x(t)-x(0)}=0\ .$$

$\endgroup$
-1
$\begingroup$

You are referring to two different things here..

For the semi-cubical parabola curve $y=x^{3/2}$ there is horizontal tangent at the cusp y=0 or t=0.

For cardoid, depending on the cusp contact point chosen, slope of cusp tangent varies, look at all cusps of epicycloids:

EDIT 1:

Since it is passing through infinite curvature which is rate of change of tangent inclination, the slope is changing very fast and undefined.

It can be also seen at the point $(1,0)$ Changing the multiple angle

$$ x=2\cos t - \cos (n t), y=2\sin t - \sin( n t) $$

has the effect of influencing curvature. It changes from (positive for circle) to negative for looped case, via the present central cardoid case of infinite curvature.

The direction of tangent is vertical, indeterminate and vertical respectively

Cardoids

$\endgroup$
  • 5
    $\begingroup$ In both cases $\lim_{t\to 0} \frac{dy}{dx}=0$ yet $\frac{dy}{dx}\Big |_{t=0}$ is undefined. The question is whether in such a situation a horizontal tangent line exists. The textbook says it does, yet my professor claims that for a horizontal tangent line to exist the derivative must EXIST and be equal to zero, that is we must have $\frac{dy}{dx}\Big |_{t=0}=0$. $\endgroup$ – user239753 Apr 25 '16 at 6:53
  • 2
    $\begingroup$ please read my comment again. I'm talking about the value of the derivative which does NOT exist. There is a difference between the limit (which indeed can be found with L'Hospital) and the actual value. $\endgroup$ – user239753 Apr 25 '16 at 7:28
  • $\begingroup$ Note that you should not stop hunting for the L'Hospital limit if the first order evaluation fails. Do not conclude that it is undefined. At the cusp we are dealing with second orders and so limit of $ f''(t)/g^{''}(t)$ should be next evaluated. $\endgroup$ – Narasimham Apr 25 '16 at 7:31
  • 3
    $\begingroup$ unfortunately, you do not understand me. Here $dy/dx \neq 0$ even though $\lim_{t\to 0} \frac{dy}{dx}=0$. Do you know there is a difference between limit and value of the function? Take $$f(x)=\left\{\begin{matrix} 0 \text{ if } x\neq0\\ 1 \text{ if } x=0 \end{matrix}\right.$$. Here $$\lim_{x\to0}f(x)=0$$ yet $f(0)\neq 0$. The same goes here. $\lim_{t\to 0} \frac{dy}{dx}=0$ yet $\frac{dy}{dx} \Big |_{t=0}\neq 0$ ! Do you understand? $\endgroup$ – user239753 Apr 25 '16 at 7:37
  • $\begingroup$ Ok, we resume after sometime,else moderator may indicate chat room. $\endgroup$ – Narasimham Apr 25 '16 at 8:34
-1
$\begingroup$

the formula $$\frac{dy}{dx}=\frac{dy/dt}{dx/dt}$$ is valid for computing the derivative at points where $\frac{dx}{dt}\neq 0$. At singular points, i.e. points at which $$\frac{dy}{dt}=\frac{dx}{dt}=0$$ the above formula is not valid. However, this does not imply that the derivative does not exist; only that it is not computable by the above formula.

In such cases, a straightforward application of the definition of derivative (as already done by user Christian Blatter in his answer) can provide the answer: $$ \frac{dy}{dx}\bigg|_{(0,0)}=\lim_{t\to 0^+}\frac{y(t)-y(0)}{x(t)-x(0)}=\lim_{t\to 0^-}\frac{y(t)-y(0)}{x(t)-x(0)}=\lim_{x\to0^\pm}\frac{t^3}{t^2}=0 $$ which shows that the derivative exists at $(0,0)$ and so does the tangent line at $(0,0)$, which is no other than the horizontal axis itself.

P.S.1: Note that, when confined to each branch seperately: $$ y(x_0)=y(x(0))=y(0)=0, \ \ x_0=x(0)=0 \\ y(x)=y(x(t))=y(t)=t^3, \ \ x=x(t)=t^2 $$ So, on each one of the functions $y(x)=x^{3/2}$, $y(x)=−x^{3/2}$ i.e. on each one of the branches of the graph below, the following rates of change are identified: $$\frac{y\big(x(t)\big)−y(x_0)}{x(t)−x_0}=\frac{y(t)−y(0)}{x(t)−x(0)}$$ On the other hand, since $x=t^2\neq 0$ when $t\neq 0$ it should be clear that $$t\to 0\Leftrightarrow x\to 0$$ At this point, it remains to invoke a classical proposition on the limit of a composite function, claiming that:

Let the real functions $f,g$, for which $lim_{x\rightarrow x_0}g(x)=g_0\in \mathbb{R}$ and $lim_{g\rightarrow g_0}f(g)=l\in \mathbb{R}$.
If, furthermore $g(x)\neq g_0$ "close to $x_0$" (i.e. in some interval $(a,x_0))$, then $$ \lim_{x\rightarrow x_0}f\big(g(x)\big)=\lim_{g\rightarrow g_0}f(g)=l $$

Seting $f(x):=\frac{y(x)−y(0)}{x−0}$ and $g(x):=x(t)$ and applying the proposition we get that: $$\lim_{t\to 0}\frac{y(t)−y(0)}{x(t)−x(0)}\equiv \lim_{t\to 0}\frac{y\big(x(t)\big)−y(0)}{x(t)−x(0)}=\lim_{x\to 0}\frac{y(x)−y(0)}{x−0}=\frac{dy}{dx}\bigg|_0$$ (where the far left term is computed using side limits: for the upper branch $t>0$ thus $t\to0\Leftrightarrow t\to0^+$ while for the lower branch $t<0$ thus $t\to0\Leftrightarrow t\to0^-$).

P.S.2: Regarding your final question: I do not believe that the existence of the limit is a sufficient condition, because generally, a differentiable function at $x_0$ need not have a continuous derivative at $x_0$: In other words, there may well be a situation at which: $\lim_{x\to x_0}f'(x)\neq f'(x_0)$.

However, this is not the case for the semicubical parabola, as we can see from its graph enter image description here

which is probably what the author of your textbook had in his mind. He computes the limit of the derivative just as if we can imagine the tangent "slipping" accross either of the branches to "continuously" become horizontal. In my opinion, he implicitly uses the assumption that the derivative of the semicubical parabola is a continuous function; and he probably does that, based on the shape of the graph.

$\endgroup$
  • $\begingroup$ Intuitively I understand. But in the definition of the derivative we use $\frac{y(x)-y(x_0)}{x-x_0}$ and not $\frac{y(t)-y(0)}{x(t)-x(0)}$. That is, we must find $y(x)$ first. Otherwise, can you prove rigorously that it works with $t$? $\endgroup$ – user239753 May 4 '16 at 7:11
  • $\begingroup$ Please read carefully: There is nothing intuitive here. (Only the idea described in your textbook seems intuitive). When confined to each branch seperately: $$y(x_0)=y(x(0))=y(0), \ \ y(x)=y(x(t))=y(t) \\ x_0=x(0), \ \ x=x(t)$$ So, on each one of the functions $y=x^{3/2}$, $y=-x^{3/2}$ i.e. on each of the branches of the above graph, the rates of change you are writing are identified: $$\frac{y(x)-y(x_0)}{x-x_0}=\frac{y(t)-y(0)}{x(t)-x(0)}$$ On the other hand, since $x=t^2$ it should be clear that $$t\to0\Leftrightarrow x\to0$$ $\endgroup$ – KonKan May 4 '16 at 12:20
  • $\begingroup$ i've added the above comment to the post $\endgroup$ – KonKan May 4 '16 at 12:51
  • 1
    $\begingroup$ It should be $y(x(t))=(t^2)^3=t^6$, no? (I know that in this case it doesn't change the limit, but still). Thank you. P.S. - Also, I don't understand why somebody downvotes the answers here. $\endgroup$ – user239753 May 4 '16 at 14:28
  • $\begingroup$ actually, for $t>0$: $$y(x(t))=\pm (x(t))^{3/2}=\pm (t^2)^{3/2}=\pm t^3$$ So, for $t\in (-\infty, +\infty)$: $$y(t)=t^3$$ $\endgroup$ – KonKan May 4 '16 at 19:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.