# Calculus 2 test question that even the professor could not solve (Parametric Arc length)

everyone got this question wrong from my calculus 3 midterm.

" find the arc length of the curve on the given interval: $$x=\sqrt(t), \space y=8t-6,\space on\space 0 \leq t \leq 3$$"

I set the problem up just fine, however at that point I got stuck.

$$\int_{0}^{3}\sqrt{\left ( \frac{1}{2\sqrt{t}} \right )^{2}+64} dt$$

we asked the teacher for a review, he got the problem set up then quit and said he would not count it against us. but it's killing me I need to know how to do it! thanks

• Note an equivalent problem would be to find the arc$-$length of $y=8x^2-6$ on $\big[0,\sqrt{3}\big]$. This gives you the integral $$\int_0^{\sqrt{3}}\sqrt{1+256x^2}dx$$ when you apply the Cartesian form of arc$-$length formula. Proceeding with your usual tangent substitution would give you a nasty $\sec^3$ integral, but it's not impossible. You just have to do by parts with a wrap around. Or partial fractions after a substitution if you're into that sort of thing. Apr 23, 2021 at 3:51
• Question upvoted for your curiosity! Apr 23, 2021 at 3:57
• Was your instructor the one who wrote the exam problem? This illustrates the hazard of putting something on an exam without working it out first: this is a reasonable homework problem, but takes too much writing for a mid-term exam without giving students either a useful anti-derivative or at least the reduction formula...
– user882145
Apr 23, 2021 at 4:04
• FYI the solution is $\frac{\operatorname{asinh}{\left(16 \sqrt{3} \right)}}{32} + \frac{\sqrt{2307}}{2}$ (computed via sympy) Apr 23, 2021 at 4:25

There is a way to perform the evaluation without resorting to trigonometric substitution, although it is hardly ideal. Note that with the substitution $$u^2 = \frac{1}{4t} + 64, \quad t = \frac{1}{4(u^2 - 64)}, \quad dt = - \frac{u}{2(u^2 - 64)^2} \, du,$$ and letting $$c = \sqrt{769/12}$$, \begin{align} \int_{t=0}^3 \sqrt{\frac{1}{4t} + 64} \, dt &= \int_{u=\infty}^c -\frac{u^2}{2(u^2 - 64)^2} \, du \\ &= \frac{1}{64} \int_{u=c}^\infty \frac{1}{u-8} - \frac{1}{u+8} + \frac{8}{(u-8)^2} + \frac{8}{(u-8)^2} \, du \\ &= \frac{1}{64} \left[\log \frac{u-8}{u+8} - \frac{8}{u-8} - \frac{8}{u+8} \right]_{u=c}^\infty \\ &= -\frac{1}{64} \log \frac{c-8}{c+8} + \frac{c}{4(c^2-64)} \\ &= \frac{\sqrt{2307}}{2} - \frac{1}{64} \log (1537 - 32 \sqrt{2307}). \end{align} While tractable, I would not expect such a computation to be performed by hand without a calculator within a short amount of time on a test. The partial fraction decomposition is not particularly difficult, but the evaluation of the antiderivative at $$c$$ involves a lot of tedious arithmetic that is not relevant for testing proficiency at calculus.
First simplify. $$\left( \frac{1}{2\sqrt{t}} \right) ^2 = \frac{1}{4t}$$. Next, factor out the $$4t$$: $$f(t) = \sqrt{\frac{1}{4t} + 64} = \frac{1}{2\sqrt{t}}\cdot \sqrt{1 + 256t}$$ Now let $$u = \sqrt{t}$$, so $$du = \frac{1}{2\sqrt{t}} dt$$ and $$f(t) dt = \sqrt{1+256u^2}du$$ From here you can use a trigonometric substitution where $$256u^2 = \tan^2 \theta$$, so $$16u = \tan \theta$$ and $$du = \frac{1}{16}\sec^2 \theta d\theta$$, resulting in $$f(t)dt = \sqrt{1 + \tan^2 \theta}\frac{1}{16}\sec^2 d\theta = \frac{1}{16}\sec^3 \theta d\theta$$ From there you can use a reduction of order formula to relate that integral to the integral of $$\sec \theta$$, which is doable.