I am a physics student and my teacher told me, to find the instantaneous velocity of an object, reduce the time interval to a very small extent. May the time interval be very very very close to 0, then, the velocity you find is termed as the instantaneous velocity.
Ok. But no matter how many times you say ‘ very very very...’ , there are still going to be infinite points between the ‘ very very very’ close points!
So, how can you apply the approximation? Thanks!
On
As the time interval gets smaller and smaller, the average velocity you compute gets closer and closer to a particular number, and that number -- the "limit" of the average velocity as the time interval approaches $0$, is the instantaneous velocity. In other words, if $f(t)$ is the position of a car along a straight road at time $t$, then the car's instantaneous velocity at time $t$ is \begin{equation} v(t) = f'(t) = \lim_{\Delta t \to 0} \frac{f(t + \Delta t) - f(t)}{\Delta t}. \end{equation}
On
1) Choose some difference "h"
2) Find a general formula for the slope of a function between (x, f(x)) and (x+h, f(x+h))
3) Set "h" equal to $0$. Not just a number kind of close to $0$, exactly $0$
Let's start with a simple example $f(x) = x^2$
Slope = $\frac{rise}{run}=\frac{f(x+h)-f(x)}{(x+h)-x}=\frac{(x+h)^2-x^2}{h}=\frac{(x^2+2xh+h^2)-x^2}{h}=\frac{2xh+h^2}{h}=2x+h$
Now you can set "h" equal precisely to $0$ instead of just some number really close, giving you an answer of $2x$
The wonderful result of mathematics is that we can substitute the ill-defined concept of ''approximation with very, very very close points'' with the concept of limit, that can be defined in a computable way thanks to the $\epsilon-\delta$ definition.