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Section 8.1 : Arc Length

1. Set up, but do not evaluate, an integral for the length of \(y = \sqrt {x + 2} \) , \(1 \le x \le 7\) using,

  1. \(\displaystyle ds = \sqrt {1 + {{\left[ {\frac{{dy}}{{dx}}} \right]}^2}} \,dx\)
  2. \(\displaystyle ds = \sqrt {1 + {{\left[ {\frac{{dx}}{{dy}}} \right]}^2}} \,dy\)
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a \(\displaystyle ds = \sqrt {1 + {{\left[ {\frac{{dy}}{{dx}}} \right]}^2}} \,dx\) Show All Steps Hide All Steps
Start Solution

We’ll need the derivative of the function first.

\[\frac{{dy}}{{dx}} = \frac{1}{2}{\left( {x + 2} \right)^{ - \,\frac{1}{2}}} = \frac{1}{{2{{\left( {x + 2} \right)}^{\,\frac{1}{2}}}}}\] Show Step 2

Plugging this into the formula for \(\displaystyle ds\) gives,

\[ds = \sqrt {1 + {{\left[ {\frac{{dy}}{{dx}}} \right]}^2}} \,dx = \sqrt {1 + {{\left[ {\frac{1}{{2{{\left( {x + 2} \right)}^{\,\frac{1}{2}}}}}} \right]}^2}} \,dx = \sqrt {1 + \frac{1}{{4\left( {x + 2} \right)}}} \,dx = \sqrt {\frac{{4x + 9}}{{4\left( {x + 2} \right)}}} \,dx\] Show Step 3

All we need to do now is set up the integral for the arc length. Also note that we have a \(dx\) in the formula for \(\displaystyle ds\) and so we know that we need \(x\) limits of integration which we’ve been given in the problem statement.

\[L = \int_{{}}^{{}}{{ds}} = \require{bbox} \bbox[2pt,border:1px solid black]{{\int_{1}^{7}{{\sqrt {\frac{{4x + 9}}{{4x + 8}}} \,dx}}}}\]

b \(\displaystyle ds = \sqrt {1 + {{\left[ {\frac{{dx}}{{dy}}} \right]}^2}} \,dy\) Show All Steps Hide All Steps
Start Solution

In this case we first need to solve the function for \(x\) so we can compute the derivative in the \(\displaystyle ds\).

\[y = \sqrt {x + 2} \hspace{0.5in} \to \hspace{0.5in}x = {y^2} - 2\]

The derivative of this is,

\[\frac{{dx}}{{dy}} = 2y\] Show Step 2

Plugging this into the formula for \(\displaystyle ds\) gives,

\[ds = \sqrt {1 + {{\left[ {\frac{{dx}}{{dy}}} \right]}^2}} \,dy = \sqrt {1 + {{\left[ {2y} \right]}^2}} \,dy = \sqrt {1 + 4{y^2}} \,dy\] Show Step 3

Next, note that the \(\displaystyle ds\) has a \(dy\) in it and so we’ll need \(y\) limits of integration.

We are only given \(x\) limits in the problem statement. However, we can plug these into the function we were given in the problem statement to convert them to \(y\) limits. Doing this gives,

\[x = 1:y = \sqrt 3 \hspace{0.25in}\hspace{0.25in}\hspace{0.25in}x = 7:y = \sqrt 9 = 3\]

So, the corresponding \(y\) limits are : \(\sqrt 3 \le y \le 3\).

Show Step 4

Finally, all we need to do is set up the integral.

\[L = \int_{{}}^{{}}{{ds}} = \require{bbox} \bbox[2pt,border:1px solid black]{{\int_{{\sqrt 3 }}^{3}{{\sqrt {1 + 4{y^2}} \,dy}}}}\]