Make It New

Last year’s posts about Arc Length On The Cardiod, Vector Products, and integral-secant-cubed are relevant again. The cardiod was a recent in-class example; the “vector” note corrects a mistake I didn’t make until yesterday; the “surprisingly useful integral” is one I had students to look up in a table on the recent exam (and work with… nobody had this one all the way right but me though in principle it’s an “easy” problem when the table is available…).

What the heck. Here’s the problem: The polar equation $r = \theta$, for $\theta \in [0, \infty)$, determines an “Archimedean Spiral”, S. Determine the arc length along S for $\theta$ between 0 and $\pi\over6$ (exactly; a messy “formula”… check the work numerically). And the answer: ${{\pi\sqrt{36+ \pi^2}}\over{72}} + {1\over2} ln({{\pi +\sqrt{36+\pi^2}}\over6})$. This evaluates to about .5466… as does $\int_0^{\pi\over6} \sqrt{\theta^2 + 1}d\theta$… so we can be reasonably sure this is right.

In my own work messy expressions like this are almost never right the first time… on the copies of the handwritten solution I distributed yesterday, quite a bit of erased work is clearly visible. Rooting out every last little mistake until things are just right is of course a vital part of the process for a lot of problems. One student was real close.