Rate of change of the area of a triangle

Solution 1:

Trying to solve this with differentiation would be a bloody nightmare. Using Heron's formula $Area = \sqrt{s(s-a)(s-b)(s-c)}$ (where s is $\frac{a+b+c}{2}$ and $a, b, c$ all represent side lengths). Now sure, you COULD find the derivative of that, but you'd go through pages. Instead, since this is just a multiple choice question, let's try a "cheap" way to do this. I assume you know that the average rate of change over a very small area becomes very close to the instantaneous rate of change (the whole $\Delta x \to 0$ you find in the limit form of a derivative). So, using that let's try to find the change in area over $0.01$ seconds by finding the area at $t = 0$ and then at $t = 0.01$.

At $t=0$: $A = \sqrt{44(44-40)(44-32)(44-16)} = 243.178947$

At $t=0.01$:

$a = 40.02$, $b = 32.05$, $c = 16.03$, $s = 44.05$, so: $A = \sqrt{44.05(44.05-40.02)(44.05-32.05)(44.05-16.03)} = 244.315020$

Now for the last step: $$\frac{dA}{dt} \approx \frac{244.315020-243.178947}{0.01} = 113.607 m^2/sec$$ So clearly option d is the correct choice. As for why our answer is slightly different? That's the whole idea of this, it's an estimate! We don't need the exact answer, just one that is close enough we can confidently make our choice. If we reduced our step size even further like to $0.001$ seconds, our prediction would be even better (with $0.001$ seconds it's actually $113.685$, nearly perfect). This a perfect example of a problem where you should just do "enough" to answer it. Work smarter, not harder!!!