Gradient of modulus the modulus function amongst other things:

Look here for a plot of the graph defined by your parametric equations (with $a=1$). You'll notice it has a sharp point at $(0,1)$ where $t=0$ -- this is similar to what happens to the graph $y=|x|$ at $x=0$. It's no coincidence that $\dfrac{dy}{dx}$ is undefined at sharp points, because the notion of a "gradient" doesn't make sense there.

Going back to first principles, the gradient of a function $f$ at a point $x$ is given by the limit of $\dfrac{f(x+h)-f(x)}{h}$ as $h \to 0$ (if this limit exists). In the example $y=|x|$ we have:

$\dfrac{dy}{dx} = \displaystyle \lim_{h \to 0} \dfrac{|x+h|-|x|}{h} = \lim_{h \to 0} \dfrac{(|x+h|-|x|)(|x+h|+|x|)}{h(|x+h|+|x|)}$

$= \displaystyle \lim_{h \to 0} \dfrac{(x+h)^2-x^2}{h(|x+h|-|x|)} = \lim_{h \to 0} \dfrac{2x-h}{|x+h|+|x|}$

Look here for a plot of the graph defined by your parametric equations (with $a=1$). You'll notice it has a sharp point at $(0,1)$ where $t=0$ -- this is similar to what happens to the graph $y=|x|$ at $x=0$. It's no coincidence that $\dfrac{dy}{dx}$ is undefined at sharp points, because the notion of a "gradient" doesn't make sense there.

Going back to first principles, the gradient of a function $f$ at a point $x$ is given by the limit of $\dfrac{f(x+h)-f(x)}{h}$ as $h \to 0$ (if this limit exists). In the example $y=|x|$ we have:

$\dfrac{dy}{dx} = \displaystyle \lim_{h \to 0} \dfrac{|x+h|-|x|}{h} = \lim_{h \to 0} \dfrac{(|x+h|-|x|)(|x+h|+|x|)}{h(|x+h|+|x|)}$

$= \displaystyle \lim_{h \to 0} \dfrac{(x+h)^2-x^2}{h(|x+h|+|x|)} = \lim_{h \to 0} \dfrac{2x-h}{|x+h|+|x|}$

This is all well and good when $x \ne 0$; taking the limit simply gives $\dfrac{x}{|x|}$ which is as we expect: 1 when x is positive and -1 when it's negative. But when x=0, we have to take the limit of $\dfrac{-h}{|h|}$ as $h \to 0$, which changes value depending on whether h is coming from above or below, meaning the gradient is undefined.

Another way of thinking about it is deciding where the tangent would go. If you think of the tangent to a curve at a point as being the limit of a cord on the curve as the two points of intersection of the cord and the curve tend to the point in question, you could essentially generate a "tangent" of any gradient between -1 and 1, simply by choosing cords which intersect the curve at different points and slide towards the origin at different speeds.

If any of the above makes no sense to you, don't worry too much about it

So your point is that, for the gradient of a curve to be defined at that point, the forward difference approximation limit for positive h and negative h must converge to the same number?

How would I show it is also not the case in terms of my parametric equations? And do you think my reasoning is correct for saying a is unequal to 0?

Furthermore, is it right to conclude that, because dt/dx is undefined for t=0, dy/dx is undefined for t=0, using that reasoning?

Yes; the idea is that a limit only exists if you get the same result regardless of how you approach it. It's for this very reason that we can't define $\dfrac{0}{0}$. To illustrate, consider the fraction $\dfrac{ax}{x}$ as $x \to 0$. Both the numerator and denominator tend to zero, but the fraction itself is equal to $a$ when $x \ne 0$ (and so must tend to a as x tends to 0); but we can choose $a$ to be whatever we like, so we can't define $\dfrac{0}{0}$ to have any one value.

In order to find $\dfrac{dy}{dx}$ you have to divide by $\dfrac{dx}{dt}$. Since $\dfrac{dx}{dt} = 0$ when $t \to 0$, the graph of y against x must become vertical when x=0, and so its gradient isn't finite. The graph (see link in my previous post) shows that the gradient is very large and positive before x=0 and very large and negative after x=0.

As for saying that $a \ne 0$, well $a$ is just a fixed constant. I think given the context you can assume that $a \ne 0$, since if it were equal to zero then the graph would, very boringly, just be a dot at the origin (and we obviously can't define the gradient of a point).


Sort of. I was maybe a bit misleading with the explanation in terms of the graph having a point, I think in this case it's more to do with the fact that the graph becomes vertical when x=0. It's similar to what happens with the graph of $y=\dfrac{1}{x}$ as $x \to 0$, with the main difference being that your function is defined at x=0 whereas 1/x is not.

Hey.

Today in a C4 paper I came across the parametric equations of a curve:
x=a(t^3), y=a/(1+(t^2)).

I found the gradient function dy/dx to be -2/(3t((1+(t^2))^2)).

However, finding this function forced me to assume that t was unequal to 0 although t=0 does create a point on the curve. (Because a=0 would only create a point and not a curve I was fairly confident a could not be 0 (please challenge this if you think it is wrong )).

Now, dy/dx=dy/dt x dt/dx. I can see that if dy/dx is undefined, then dy/dt or dt/dx must also be undefined. However, if dy/dt or dt/dx is undefined then how can I be sure that dy/dx is undefined? I cannot see clearly how the reverse logic works. If this second statement is indeed true then I can be confident in saying dy/dx is undefined when t=0 (which it appears to be when I consider the graph of the curve).

I think your logic so does not work.
Start with that if dy/dx is defined then dy/dt and dx/dt must be defined
because $\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$,
so this fraction is defined.
If You state the negation of above that is if dy/dx is undefined then dy/dt or dx/dt
must also be undefined is not works, because both maybe defined but when dx/dt is zero then dy/dx is undefined.
So you can state that
dy/dx exists imply that (dy/dt)/(dx/dt) exists , and if (dx/dt)/(dx/dt) does not exists
imply that dy/dx does not exists
(modus tollens in the logic)
Similarly if dx/dt or dy/dt does not exist then (dx/dt)/(dx/dt) does not exists
but in your example they exist for all t.
But there is another case here: when dx/dt=0. This means that the fraction of
(dx/dt)/(dx/dt) does not exists and this imply that dy/dx does not exist.
dx/dt =0 when t=0

Okay. I get some of that having studied a bit of logic.

If ((dy/dt)/(dx/dt)) is undefined, does that necessarily mean dy/dx is undefined? (I am sorry if this is a stupid/easy question but I truly do not know the answer).

Knowing, dy/dt and dx/dt can I be sure that the gradient is undefined when t=0?

(which I hate we're supposed to ignore)

i.e. dy/dx = dy/dt x dt/dx

If the right-hand side is undefined does the left-hand side necessarily have to be undefined?

Also, can a point be described as a curve (because that would also categorically tell me a is unequal to 0 due to the use of the word "curve" in the question)?

Yes, when true that if dy/dx defined means that (dy/dt)/(dx/dt) is defined (but this is true by the equation)
When $A \rightarrow B$ then $notB \rightarrow notA$
A: dy/dx defined
Bdy/dt)/(dx/dt) defined

I accept this is true if A implies B. Why does dy/dx= dy/dt x dt/dx always? I can see why it is true for a point where the gradient is defined, but why must it hold true if the gradient is not defined?

It neither does nor doesn't hold true when the gradient is not defined, because when the gradient isn't defined, neither side of the equation makes sense.