How long should it take to work an arclength problem on a polynomial ofby hand?

Question

I was going over calculus homework with my son the other day, who was frustrated because this problem took him three hours to solve:

Find the arclength between 2 and 5 of the function:

$f(x)$ $=$ $x^5 \over 10$ $+$ $1 \over 6x^3$

...which we solved by taking the first derivative and integrating the "square root of d/dx-squared plus one" over the domain. But, he spun his wheels over the complexity of the indefinite integral and became frustrated.

Assuming he took this to someone near graduation who is effective as a math help TA, How much time would that student mentor need to solve this? What kinds of principles and techniques would he teach my son so he could get there in three minutes instead of 180?

Answer 1

tick tick tick....

$\int_2^5 \sqrt{1+\frac {dy}{dx}^2} \ dx\\ y = \frac {x^5}{10} + \frac 1{6x^3}\\ \frac {dy}{dx} = \frac 12 x^4 - \frac 12 x^{-4}\\ \sqrt{1+\frac {dy}{dx}^2} = \sqrt {1 + \frac 14x^8 - 2(\frac 12x^4)(\frac 12x^{-4}) + \frac 14 x^{-8}} \\ \sqrt {1 + \frac 12 x^8 - \frac 12 + \frac 14 x^{-8}}\\ \sqrt {\frac 12 x^8 + \frac 12 + \frac 14 x^{-8}}\\ \frac 12 x^4 + \frac 12 x^{-4}$

And now we see that we are going to get a figure that looks very similar to our starting figure.

$\int_2^5 \frac 12 x^4 + \frac 12 x^{-4} \ dx\\ \frac {x^5}{10} - \frac 1{6x^3}|_2^5\\ \frac {5^5}{10} - \frac {2^5}{10} - \frac 1{6(5^3)} + \frac 1{6(2^3)}$

That should take about 10-15 minutes.

If you want to plug it into your calculator, another minute to get. $309.3$

Or if you are anti-calculator

$5^5 - 2^5 = 3125 - 32 = 3093\\ \frac 1{5^3} - \frac 1{2^3} = \frac {2^3 - 5^3}{10^3} = -0.117\\ 309.3 + \frac {0.117}{6} = 309.3195$

It will take a minute or two longer.

Answer 2

I just gave it a shot. This is a contrived case which makes it easier to solve than a typical arc length problem. Still, it took me about five minutes to do (writing out each step), so a three minute time is quite unreasonable. Three hours is a lot longer than it should have taken him as well. I would think fifteen minutes might be a reasonable expectation.

The question you should ask him, is after he figured out how to solve it, how long would it take him to do it again from scratch on a fresh sheet of paper.

Ced

Answer 3

$f'(x) = \frac12(x^4-x^{-4})$ -->
$\int \sqrt{1+f'(x)^2} \ dx = \int \sqrt{1+\frac{x^8}4-\frac12+\frac1{4x^8}}$
$= \int \sqrt{\frac{x^8}4+\frac12+\frac1{4x^8}} = \int \sqrt{(\frac{x^4}2-\frac{x^{-4}}2)^2} = \int {x^4\over2} - {1\over(2x^4)} \ dx$

Easy peasy, and if you know this trick, then 3 minutes is a piece of cake!
(A favorite quote from my calculus class is "You should have learned that in pre-calc!" This should've been learned from Algebra 1)