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### Section 8.1 : Arc Length

4. Determine the length of $$x = 4{\left( {3 + y} \right)^2}$$ , $$1 \le y \le 4$$.

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Since we are not told which $$ds$$ to use we will have to decide which one to use. In this case the function is set up to use the $$ds$$ in terms of $$y$$.

If we were to solve the function for $$y$$ (which we’d need to do in order to use the $$ds$$ that is in terms of $$x$$) we would put a square root into the function and those can be difficult to deal with in arc length problems.

So, let’s take the derivative of the given function and plug into the $$ds$$ formula.

$\frac{{dx}}{{dy}} = 8\left( {3 + y} \right)$ $ds = \sqrt {1 + {{\left[ {\frac{{dx}}{{dy}}} \right]}^2}} \,dy = \sqrt {1 + {{\left[ {8\left( {3 + y} \right)} \right]}^2}} \,dy = \sqrt {1 + 64{{\left( {3 + y} \right)}^2}} \,dy$

Note that we did not square out the term under the root. Doing that would greatly complicate the integration process so we’ll need to leave it as it is.

Show Step 2

In this case we don’t need to anything special to get the limits for the integral. Our choice of $$ds$$ contains a $$dy$$ which means we need $$y$$ limits for the integral and nicely enough that is what we were given in the problem statement.

So, the integral giving the arc length is,

$L = \int_{{}}^{{}}{{ds}} = \int_{1}^{4}{{\sqrt {1 + 64{{\left( {3 + y} \right)}^2}} \,dy}}$ Show Step 3

Finally, all we need to do is evaluate the integral. In this case all we need to do is use a trig substitution. We’ll not be putting a lot of explanation into the integration work so if you need a little refresher on trig substitutions you should go back to that section and work a few practice problems.

The substitution we’ll need is,

$3 + y = \frac{1}{8}\tan \theta \,\,\,\,\,\,\,\, \to \,\,\,\,\,\,\,\,\,\,dy = \frac{1}{8}{\sec ^2}\theta \,d\theta$

In order to properly deal with the square root we’ll need to convert the $$y$$ limits to $$\theta$$ limits. Here is that work.

\begin{align*}y = & 1:\,\,\,\,\,\,\,\,4 = \frac{1}{8}\tan \theta \,\,\,\,\, \to \,\,\,\,\,\tan \theta = 32\,\,\,\,\, \to \,\,\,\,\,\theta = {\tan ^{ - 1}}\left( {32} \right) = 1.5396\\ y = & 4:\,\,\,\,\,\,\,\,7 = \frac{1}{8}\tan \theta \,\,\,\,\, \to \,\,\,\,\,\tan \theta = 56\,\,\,\,\, \to \,\,\,\,\,\theta = {\tan ^{ - 1}}\left( {56} \right) = 1.5529\end{align*}

Now let’s deal with the square root.

$\sqrt {1 + 64{{\left( {3 + y} \right)}^2}} = \sqrt {1 + 64{{\left( {\frac{1}{8}\tan \theta } \right)}^2}} = \sqrt {1 + {{\tan }^2}\theta } = \sqrt {{{\sec }^2}\theta } = \left| {\sec \theta } \right|$

From the work above we know that $$\theta$$ is in the range $$1.5396 \le \theta \le 1.5529$$. This is in the first and fourth quadrants and cosine (and hence secant) is positive in this range. So,

$\sqrt {1 + 64{{\left( {3 + y} \right)}^2}} = \sec \theta$

Putting all of this together gives,

$L = \int_{1}^{4}{{\sqrt {1 + 64{{\left( {3 + y} \right)}^2}} \,dy}} = \frac{1}{8}\int_{{1.5396}}^{{1.5529}}{{{{\sec }^3}\theta \,d\theta }}$

Evaluating the integral gives,

$L = \int_{1}^{4}{{\sqrt {1 + 64{{\left( {3 + y} \right)}^2}} \,dy}} = \left. {\frac{1}{{16}}\left( {\tan \theta \sec \theta + \ln \left| {\tan \theta + \sec \theta } \right|} \right)} \right|_{1.5396}^{1.5529} = \require{bbox} \bbox[2pt,border:1px solid black]{{130.9570}}$

Note that if you used more decimal places than four here (the standard number of decimal places that we tend to use for these problems) you may have gotten a slightly different answer. Using a computer to get an “exact” answer gives 132.03497085.

These kinds of different answers can be a real issues with these kinds of problems and illustrates the potential problems if you round numbers too much.

Of course, there is also the problem of often not knowing just how many decimal places are needed to get an “accurate” answer. In many cases 4 decimal places is sufficient but there are cases (such as this one) in which that is not enough. Often the best bet is to simply use as many decimal places as you can to have the best chance of getting an “accurate” answer.