Suppose we have a function $s=16t^2$ now if we find $\frac{ds}{dt}$ we get $v=28t$ but when we simply divide we know $s/t=v$ so $s/t=16t$. Why classical division fails?
$\endgroup$ 33 Answers
$\begingroup$Even though $\frac{ds}{dt}$ looks like a fraction, it is not one. Explicitly, it means $$\frac {ds}{dt} = \lim_{h\to 0} \frac {s(t+h)-s(t)}{h}$$ and this is what a derivative is. Doing this out for $s (t)=16t^2$ provides
$$\begin {align*} \frac {ds}{dt} &=& \lim_{h\to 0} \frac {s(t+h)-s(t)}{h} \\ &=& \lim_{h\to 0} \frac {16 (t+h)^2-16t^2}{h} \\ &=& \lim_{h\to 0} \frac {32th+16h^2}{h} \\ &=& \lim_{h\to 0} 32t+16h \\ &=& 32t. \end{align*}$$
$\endgroup$ $\begingroup$s=16t^2.
but,dv/dt=32t
But,the velocity is the rate of change of displacement it means that is their having something is displacement &we take the difference between initial and final velocity and then divide by time
$\endgroup$ 1 $\begingroup$First, we need to define what a derivative is.
A derivative is a function that gives us the rate in change odd another function at any given time,. The rate of change is often referred to as the slope of the line.
That settled.
In order to calculate the slope of a function at a given point without use derivatives, is complicated unless the function of a straight line, in which case we use: m = (y2 - y1)/(x2 - x1).
For other functions, we know that the slope is not constant, so we need to use something a little bit more complicated, than the previous function: m = (f (dx + x1) - f (x1)) / ( dx + x - x).
Where m will give us the slope of the line at x when dx is equal to zero.
The above function when expanded and simplified using algebra and division gives us the slope of the line at x.
(This is how we prove that we can take a derivative).
Admittedly this is not an answer to your question? Howdy I think it is an answer to what you were really asking.
$\endgroup$ 1
