differential area I'm just trying to refresh my calculus a bit, I'm stuck on a question and I'd love some insight.
A square measures 0.9cm on each side when drawn with a pencil. When traced over with a marker, it measures 0.95cm on each side. Use the differential of area, $dA$, to estimate the increase in area of the square.
What I would do is this:
$$s_p = 0.9cm$$
$$s_m = 0.95cm$$
$$ds = s_m- s_p = 0.05cm$$
$$dA = s_m * ds = .95cm * .05cm = 0.475cm^2$$
 A: The formula for the area of a square is $A = \ell^2$, where $\ell$ is the length of each side. Hence
$$\frac{\mathrm{d}A}{\mathrm{d}\ell} = 2\ell$$
This can also be written as $\mathrm{d}A = 2\ell\,\mathrm{d}\ell$. Here $\mathrm{d}A$ is the infinitesimal difference in the area and $\mathrm{d}\ell$ is the infinitesimal difference in the length of each side. You have $\ell = 0.9$ and $\mathrm{d}\ell = 0.05$. Hence
$$\mathrm{d}A = 2\times 0.9 \times 0.05 = 0.09$$
A: A square of side length $x$ (units) has area $A(x)=x^2$. The derivative of $A$ with respect to the side length is $$A'(x)={dA\over dx}=2x.$$
But, the derivative gives a rate of change. If the side length changes by an amount $\Delta x$, starting from a point $x=x_0$, then the area changes by an amount $\Delta A$.  The relationship between these changes and the derivative is
$$
{dA\over dx}\biggl|_{x=x_0}\approx {\Delta A\over \Delta x}.
$$
This follows from the (limit) definition of derivative.
In your problem $x$ changes from $x=x_0=.9$ to $x=.95$, so $\Delta x=.05$. Then using the above, the corresponding change in area is
$$\Delta A\approx {dA\over dx}\biggl|_{x=.9} \cdot\,\Delta x=2(.9)\cdot(.05)=.09\,{\text {cm}}^2$$
