This is my first post on the Maths section of StackExchange.
I am a high school student from Greece studying for IAL Math and got a bit confused on derivatives while watching this video; specifically on minute 14:29.
The author gives an example of a car traveling on the curve $x^3$ where $x \in [0, 3]$. He then asks if the car is moving or not at $x = 0$
Let's use $x^2$ instead of $x^3$ here for simplicity.
Through the use of the derivative of $x^2$, $\frac{dy}{dx} = 2x$ we see that at $x=0$ the velocity of the car is $\frac{dy}{dx}=2(0)=0$ which suggests that the car is not moving.
I can see, however, the point the author is making, arguing that the car is actually moving, since the derivative formula for $x^2$ is derived as such:
$
\lim\limits_{h \to 0} \frac{f(x+h)-f(x)}{h}=
\lim\limits_{h \to 0} \frac{(x+h)^2-x^2}{h}=
\lim\limits_{h \to 0} \frac{x^2+2hx+h^2-x^2}{h}=
\lim\limits_{h \to 0} \frac{2hx+h^2}{h}=
\lim\limits_{h \to 0} 2x + h =
2x
$
where in the penultimate step we can see that the derivative formula is actually $2x$ plus some very very small quantity because $h$ doesn't actually ever reach zero. So is the car actually moving with its velocity being that very very small quantity?
How can someone first argue that $h$ approaching zero can be neglected from the equation(last step) and subsequently contradict himself and argue that $h$ approaching zero should actually be included? Is it safe to assume that for $\frac{dy}{dx}=0$ an object is stationary? (In the context where $y$ is the distance travelled and $x$ is time)
Moreover, for $x=0$ the derivative formula's result is $0$ which should also be the gradient of the line tangent to the parabola at $x=0$
And indeed, the gradient of the tangent line is $y=0$ which crosses $y=x^2$ only at point $0$ and not at point $0$ plus some very very small quantity
(Desmos)
I am very confused by this and would really appreciate help. I have also tried to make my questions as clear as possible but since I do not seem to understand calculus and limits all that well I will be more than willing to provide any clarifications.
Best Answer
This is the central paradox of calculus:
You are right in saying that "instantaneous velocity" does not truly exist, by the definition of velocity, that a time interval must pass.
However, in calculus, we are interested in finding this "instantaneous velocity" whereby we find the derivative at this very instant (even though in real life we cannot actually "have" such a quantity).
Regarding whether the car is stationary at that point: if t=0 is the instant right before the car moves, then it is stationary - but in a short instant later (dx) we see it is moving. Furthermore, you could argue it is beginning to move if we see that its acceleration, $\frac{d^2y}{dx^2}=2$ is more than zero hence its velocity is increasing.
The 2nd context where h should be approaching zero is how we find derivatives from "first principles" where we let dx be a very small (almost infinitesimal) number (ie finding velocity in a small instant denoted by $\frac{\delta y}{\delta x}$. However, in the 1st context where we remove h approaching zero is when we find the "instantaneous velocity": using first principles, then letting it approach zero, denoted by $\lim\limits_{\delta x \to 0} \, \frac{\delta y}{\delta x} = \frac{dy}{dx}$. Normally in calculus we "skip" the first principles using prederived formula.
Thus, the contradiction is actually the fact that there are two contexts: a small delta x vs delta x approaching zero. These are similar but still different contexts.
A "cheat" way to understand this is: what quantity would we choose when representing the gradient at 0? Furthermore, the gradient at 0 is a straight line not a variable which changes based on a quantity we choose.