I am teaching myself calculus and trying to understand derivatives from a physics point of view. I am faced with this question: the edge of a square is being enlarged at $
2 \frac{cm}{sec}$. What is the rate of change of the area when the edge is 6 cm long. The solution in the book (Serge Lang, A first course in Calculus, 5th ed) is as follows:
$A = x^2 \\
\frac{dA}{dx} = 2x$
Using chain rule:
$ \frac{dA} {dt} =\frac{dA} {dx} * \frac{dx} {dt} \\
= 2x * 2 = 4x $
When the edge is 6: $\frac{dA}{dt} = 4*6 = 24$.
My main issue is that intuitively I had assume I can do the following:
At time $t_1, A_1 = x^2$
$\\ $At time $t_1+1, A_2 = (x+2)(x+2) =
x^2 + 4x + 4$
Then $\frac {dA}{dt}= \frac{A_2 – A_1} {t_1 +1 – t_1} = 4x+4$
Which does not match the answer, also, calculating difference of area when edge goes from 4 to 6 is 36 – 16 = 20 and not 24.
I would like to understand what was wrong in my logic. Also I considered this might be due to the fact that I am taking a long interval ( 1 sec. and derivatives evaluate the rate of change at exactly the point of change). But again the rate which we are using is defined as 2 $\frac {cm}{sec}$,we don't have any information about how it grows in shorter intervals, so I don't know if it make sense to think of shorter intervals than 1 second.
When I googled about derivates from definition are different from formulas I came against something called backward and forward differences and centered differences, it seems to be related to discrete mathematics which I have no experience with yet. But using the idea to evaluate the derivate at the center yields:
$(x+1)(x+1) – (x-1)(x-1) = 4x$
This also is correct as difference between 49 and 25.
Maybe someone can shade light on this?
(I guess same result can be reached if we take two differences, one forward and one backward and average them on the time interval of 2 second).
Also one final remark that confused me, the fact I can evaluate the area when $x = 5$ and $x = 6$ is based on the initial condition of the length of the edge. That is, is not there a hidden assumption that the increase is uniform everywhere during this one second? What I mean to point out is that if the edge was being increased in pulses and not uniformly then I can not assume that $x-1$ and $x+1$ (despite having a distance of 2 separating them) are actually legal to evaluate, because the only information I was given that $x_{next } = x_{old} + 2$ for every second that passes
Thanks.
Best Answer
To illustrate the discussion in the comments, view difference quotients as an approximation to the derivative.
In this case:
$$\frac{dA}{dt} \approx \frac{\Delta A}{\Delta x} \frac{\Delta x}{\Delta t}$$
Since $\frac{\Delta x}{\Delta t} = \frac{dx}{dt} = 2$ with $\Delta x =2$ when $\Delta t =1$ you will get 28.
Now try $\Delta t = \frac12$ then $\frac14$ or smaller and you should see this expression heading to $24$.