The question asks, "Use the trapezoid rule (when $n=8$) to approximate the arc length of the graph of $y=2x^3-2x+1$ from $A (0,1)$ to $B(2,13)$"

I first graphed this out and found the points to have $(0,1)$, $(0.25, 0.53125)$, $(0.5, 0.25), (0.75, 0.3437), (1,1), (1.25, 2.40625), (1.5, 4.75), (1.75, 8.21875), (2,13)$ as the points of each trapezoid, and found the distance between each point using the distance formula and got an answer something like $13.92$. This is not the correct answer to this problem. Please help!

  • $\begingroup$ Do you know how to compute the arc length of a curve? Can you then use the trapezoidal rule to approximate said arc length? $\endgroup$ – mattos Apr 17 '16 at 3:02
  • $\begingroup$ Are you considering the arc length function, that is, $\sqrt{1+y'^2}$, or just $y$ itself? $\endgroup$ – Ian Apr 17 '16 at 3:08
  • $\begingroup$ BTW: this is multiple choice and there are such answer choices:A) 13.688, B)13.896 C)14.09, D)13.697, E)6.900 $\endgroup$ – SloAnderson24 Apr 17 '16 at 3:15
  • $\begingroup$ Because my answer isn't included and it doesn't seem to be a rounding issue I'm not sure what it is I'm doing wrong. I found the points of each fourth (0, .25, .5, .75, etc) then found the distance in between each point with the distance formula and added them together. I'm not sure what I did wrong. I checked and rechecked my numbers 4 times $\endgroup$ – SloAnderson24 Apr 17 '16 at 3:18
  • $\begingroup$ @SloAnderson24 You didn't answer mine nor Ian's questions, which makes it really hard to help you. $\endgroup$ – mattos Apr 17 '16 at 3:24

There are two reasonable interpretations of this problem. More experienced students would likely say to do $$s=\int ds=\int_0^2\sqrt{1+y^{\prime 2}(x)}dx\approx\frac h2\sum_{i=0}^8w_i\sqrt{1+(y^{\prime}(x_i))^2}\approx14.09254$$ Where $w_0=w_8=1$ and all other $w_i=2$. But I betcha that the questioner is doing something like $$s\approx\sum_{i=0}^7\sqrt{(x_{i+1}-x_i)^2+(y_{i+1}-y_i)^2}\approx13.92768$$ EDIT: That last sum could be rewritten as $$s\approx\sum_{i=0}^7\sqrt{1+\left(\frac{y_{i+1}-y_i}{x_{i+1}-x_i}\right)^2}(x_{i+1}-x_i)$$ And as such could be considered to be the midpoint rule using central differences to approximate the derivative of the function. Normally the midpoint rule is a little more accurate than the trapezoidal rule with error in the opposite direction. The is indeed the case here as the exact value seems to be close to $13.964791$, perhaps validating the original questioner's intuition for the problem.

  • $\begingroup$ That would indicate to me that the composer of the problem anticipated the first method, as in answer (C). $\endgroup$ – user5713492 Apr 17 '16 at 3:30
  • $\begingroup$ Thank you!!! Would the formula and work you've done for the first part still count as approximation? $\endgroup$ – SloAnderson24 Apr 17 '16 at 3:31
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    $\begingroup$ Sorry, I didn't make up the rules of this game :) For me, either method would be reasonable, but I'm not the referee. $\endgroup$ – user5713492 Apr 17 '16 at 3:33

I agree with user5713492.

Since the formula for determining the arc length, $L$, can be given as: $L= \int_a^b \! \sqrt{1+(\frac{dy}{dx})^2} \, \mathrm{d}x$

, where $a=0$ and $b=2$. The derivative of $y$, that is $ \frac{dy}{dx}$ can first be determined as:

$ \frac{dy}{dx} = 6x^2 -2$

Now, by letting the square-rooted term in the arc length formula be the function $g$ as follows and substituting for $\frac{dy}{dx}$ we have that,

$g(x) =\sqrt{1+(6x^2 -2)^2} $

and therefore, $L= \int_a^b \! g (x) \, \mathrm{d}x$

or put differently, $L= \int_a^b \! \sqrt{1+(6x^2 -2)^2} \, \mathrm{d}x$

We can now apply the trapezoidal rule to integrate numerically on the interval $[a, b]$.

$L \approx \frac{b-a}{2} \sum_{k=1}^n (g(x_{k+1}) -g(x_k))$ Where $n=8$. The trapezoidal rule formula can be re-written as: $L \approx \frac{b-a}{2n}[ g(x_1) +2g(x_2) + 2g(x_3) +\dots + 2g(x_7)+ g(x_8) ]$


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