# Calculus for the Ambitous: Example 1.2.6

Example 1.2.6: Heron's formula states that the area A of a triangle whose sides have lengths a, b and c is given by

$A = \sqrt{s(s-a)(s-b)(s-c)}$, where s = $\frac{a+b+c}{2}$.

Use the formula to estimate the difference in area between a triangle whose sides have lengths $a$, $b$ and $c$ and a triangle whose sides have lengths $a + \delta a$, $b + \delta b$ and $c + \delta c$, where as usual $\delta a$, $\delta b$ and $\delta c$ are small in magnitude compared with $a$, $b$ and $c$.

Solution. Let us write $u = s - a$, $v = s - b$, $w = s - c$ and $T = suvw$ and let us denote the corresponding quantities for the new triangle by $u + \delta u$, $v + \delta v$, $w + \delta w$ and $T + \delta T$. The quantities $A + \delta A$ and $s + \delta s$ will have the obvious meanings.

Our earlier result on square roots tells us that

$A + \delta A = \sqrt{T + \delta T} \approx \sqrt{T} + \frac{1}{2 \sqrt{T}} \delta T = A + \frac{1}{2A} \delta T$

so that

$\delta A \approx \frac{1}{2A} \delta T$

to the first order. A simpler modification of our rule concerning products of two and three objects gives

$T + \delta T = (s + \delta s)(u + \delta u)(v + \delta v)(w + \delta w) \approx suvw + uvw \delta s + svw \delta u + suw \delta v + suv \delta w = T + uvw \delta s + svw \delta u + suw \delta v + suv \delta w$

so that

$\delta T = uvw \delta s + svw \delta u + suw \delta v + suv \delta w$

or, more neatly,

$\delta T = T(\frac{\delta s}{s} + \frac{\delta u}{u} + \frac{\delta v}{v} + \frac{\delta w}{w})$

to the first order.

Combining our two results gives

$\delta A \approx \frac{1}{2A} \times T(\frac{\delta s}{s} + \frac{\delta u}{u} + \frac{\delta v}{v} + \frac{\delta w}{w}) = \frac{\sqrt{A}}{2}(\frac{\delta s}{s} + \frac{\delta u}{u} + \frac{\delta v}{v} + \frac{\delta w}{w})$

to the first order. This result is nice and symmetric, but the reader may object that, for example, $\delta s$, depends on $\delta a$, $\delta b$ and $\delta c$. She is invited to do the next exercise.

I am stuck on the last step (the last '='), I don't understand how $\frac{1}{2A} \times T$ = $\frac{\sqrt{A}}{2}$, when we have already shown that $T = A^2$, therefore $\frac{1}{2A} \times T = \frac{A}{2}$.