0
$\begingroup$

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!

$\endgroup$
  • 2
    $\begingroup$ Calculus exists exactly to address your concerns! $\endgroup$ – qbert Apr 23 '16 at 18:50
  • $\begingroup$ There are an infinite number of points between $.0000000000001$ and $.00000000000011$. Are they significant? $\endgroup$ – John Douma Apr 23 '16 at 20:18
2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ There are also extensions to the real number system by the inclusion of infinitesimals to form the hyper-real number system. This approach, which avoids $\varepsilon - \delta $ arguments is based on Leibniz's calculus, and was developed by Prof. Robinson - see his 'Non-Standard Analysis'. $\endgroup$ – user328032 Apr 23 '16 at 18:34
1
$\begingroup$

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 $2x$

$\endgroup$
  • $\begingroup$ Step 2) won't always work though, so we should have a definition of the derivative that makes sense even if step 2 fails. $\endgroup$ – littleO Apr 23 '16 at 18:48
  • $\begingroup$ ... Isn't it the same answer you gave? $\endgroup$ – Simpson17866 Apr 24 '16 at 4:37
  • $\begingroup$ Actually it's not the same. Taking the limit as $h$ approaches $0$ is not the same thing as plugging in $h = 0$. Here's an example to keep in mind: let $f(x) = \sin(x)/x$. Then $\lim_{x \to 0} f(x) = 1$, and yet $f$ is undefined at $0$, so plugging in $0$ is not an option. Another example: Define $f$ so that $f(x) = x$ when $x \neq 0$, and $f(0) = 1$. Then $\lim_{x \to 0} f(x) = 0$, but $f(0) = 1$, so we can't simply plug in $x = 0$ (in fact the value of $f$ when $x = 0$ is irrelevant when we're taking the limit as $x \to 0$.) Often, there is simply no nice formula for $f$ at all. $\endgroup$ – littleO Apr 24 '16 at 5:14
0
$\begingroup$

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}

$\endgroup$

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.