Graphs for which a calculus student can reasonably compute the arclengthm i12s,n89A123Vv
Given a differentiable real-valued function $f$, the arclength of its graph on $[a,b]$ is given by
$$\\int_a^b\\sqrt{1+\\left(f'(x)\\right)^2}\\,\\mathrm{d}x$$
For many choices of $f$ this can be a tricky integral to evaluate, especially for calculus students first learning integration. I've found a few choices of $f$ that make the computation pretty easy:
- Letting $f$ be linear is super easy, but then you don't even need the formula.
- Taking $f$ of the form $(\\text{stuff})^{\\frac{3}{2}}$ might work out nicely if $\\text{stuff}$ is chosen carefully.
- Calculating it for $f(x) = \\sqrt{1-x^2}$ is alright if you remember that $\\int\\frac{1}{x^2+1}\\,\\mathrm{d}x$ is $\\arctan(x)+C$.
- Letting $f(x) = \\ln(\\sec(x))$ results in $\\int\\sec(x)\\,\\mathrm{d}x$, which classically sucks.
But it looks like most choices of $f$ suggest at least a trig substitution $f'(x) \\mapsto \\tan(\\theta)$, and will be computationally intensive, and unreasonable to ask a student to do. Are there other examples of a function $f$ such that computing the arclength of the graph of $f$ won't be too arduous to ask a calculus student to do?
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4$\\begingroup$ Not an answer, but you can always combine this question with a Riemann sum approximation problem. Then the long integration computations are unnecessary and brings 2 concepts together. $\\endgroup$ – Dayton 15 hours ago
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9$\\begingroup$ This question is a bit infamous. All the calculus texts have the same two or three exercises. The reason is: no other cases are easy to compute. $\\endgroup$ – GEdgar 15 hours ago
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5$\\begingroup$ Perhaps more relevant on mathematics educators $\\endgroup$ – gen-z ready to perish 15 hours ago
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$\\begingroup$ May I ask: what is the "value added" to having students calculate the value? It might be interesting to have them compare the value based on your formula to the base definition value $ \\int_{a}^{b} \\left| f^{'}(t) \\right| dt $ . And then have them observe how quickly the series form converges. $\\endgroup$ – Carl Witthoft 5 mins ago
3 Answers
Another example: you can get $$ \\sqrt{1 + [f'(x)]^2} = ax + \\frac 1{2ax} $$ by taking $f(x) = \\frac 12 a x^2 - \\frac 1{4a} \\ln(x)$ for any constant $a$.
A possibly helpful way of reframing the question: we would like to know for which "nicely integrable" functions $g(x)$ is there a "reasonable" $f(x)$ satisfying $\\sqrt{1 + [f'(x)]^2} = g(x)$. In other words, for which nicely integrable $g(x)$ does the function $\\sqrt{[g(x)]^2 - 1}$ have a closed-form integral?
This example $$ y = a\\cosh \\frac{x}{a} $$ is quite simple for computations.
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3$\\begingroup$ Kids only just now learning integration have rarely heard of the hyperbolic functions $\\endgroup$ – gen-z ready to perish 15 hours ago
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9$\\begingroup$ @let'shaveabreakdown if you express the hyperbolic tangent in terms of exponentials, then you have a challenging (but arguably straightforward) integration problem $\\endgroup$ – Omnomnomnom 15 hours ago
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2$\\begingroup$ @Omnomnomnom Very very true! $\\endgroup$ – gen-z ready to perish 15 hours ago
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1$\\begingroup$ @Omnomnomnom Yes I've seen this sort of arclength question before: "The shape of a catenary is modelled by $f(x) = \\frac{1}{2a}(e^{ax}+e^{-ax})$ for some $a$ ... find the arclength." Taking the integral in this form is straightforward, but I think that it would be tough for a student to realize that $1 + (f'(x))^2$ is a square. $\\endgroup$ – Mike Pierce 15 hours ago
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3$\\begingroup$ @Mike It is notable that $1 + (f'(x))^2$ looks a lot like $(f'(x))^2$, except that the sign on the "cross-term" has flipped. I think that makes things a bit easier. $\\endgroup$ – Omnomnomnom 15 hours ago
You can try $f(x)=\\frac{\\sqrt{a^2e^{2ax}-1}-\\tan^{-1}\\sqrt{a^2e^{2ax}-1}}{a}$, which has arc-length $e^{ax}-1$ and isn't too hard to work with as long as you remember $\\frac{\\mathrm{d}}{\\mathrm{d}x}\\tan^{-1}x$.
But other than that, you could always define your function as an unsolved integral, $f(x)=\\int\\sqrt{L'(x)^2-1}\\ \\mathrm{d}x$. Then even when the function itself has no closed form, you can define a closed form for the arc-length, $L(x)$. Students can then use their knowledge of integration rules and the fundamental theorem of calculus to compute the arc-length.
For example, take $f(x)=\\int \\sqrt{\\sec^4x - 1} \\ \\mathrm{d}x$, which has a horribly unwieldy closed form when the integral is solved. Students could compute the arc length as
$$\\begin{aligned}L(x)&=\\int\\sqrt{1+{\\left({\\int \\sqrt{\\sec^4x - 1}\\ \\mathrm{d}x}'\\right)}^2}\\ \\mathrm{d}x\\\\ &=\\int\\sqrt{1+\\left[\\sqrt{\\sec^4x - 1}\\right]^2}\\ \\mathrm{d}x\\\\ &=\\int{\\sec^2x}\\ \\mathrm{d}x\\\\ &=\\tan x+C \\end{aligned}$$
Which has the added gratification of reducing the integral into a satisfyingly neat conclusion. Behind the scenes, this works because we chose $L(x)=\\tan(x)$, when we defined $f(x)=\\int \\sqrt{\\left(\\frac{\\mathrm{d}}{\\mathrm{d}x}\\tan x\\right)^2-1}\\ \\mathrm{d}x$.
The problem with that tactic is that you could only do it for a couple of problems since the students would soon see that your choice of $L(x)$ is the arc length. You'd probably also want to put a note in the question that students don't need to evaluate the integral form of $f(x)$, otherwise they'd get lost in it.
