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Section 8.1 : Arc Length

2. Set up, but do not evaluate, an integral for the length of \(x = \cos \left( y \right)\) , \(0 \le x \le \frac{1}{2}\) 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

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

\[x = \cos \left( y \right)\hspace{0.5in} \to \hspace{0.5in}\,\,\,y = {\cos ^{ - 1}}\left( x \right) = \arccos \left( x \right)\]

Which notation you use for the inverse tangent is not important since it will be “disappearing” once we take the derivative.

Speaking of which, here is the derivative of the function.

\[\frac{{dy}}{{dx}} = - \frac{1}{{\sqrt {1 - {x^2}} }}\] Show Step 2

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

\[ds = \sqrt {1 + {{\left[ {\frac{{dy}}{{dx}}} \right]}^2}} \,dx = \sqrt {1 + {{\left[ { - \frac{1}{{\sqrt {1 - {x^2}} }}} \right]}^2}} \,dx = \sqrt {1 + \frac{1}{{1 - {x^2}}}} \,dx = \sqrt {\frac{{2 - {x^2}}}{{1 - {x^2}}}} \,dx\] 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 \( ds\) and so we know that we need \(x\) limits of integration which we’ve been given in the problem statement.

\[L = \int_{{}}^{{}}{{ds}} = \require{bbox} \bbox[2pt,border:1px solid black]{{\int_{0}^{{\frac{1}{2}}}{{\sqrt {\frac{{2 - {x^2}}}{{1 - {x^2}}}} \,dx}}}}\]

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

We’ll need the derivative of the function first.

\[\frac{{dx}}{{dy}} = - \sin \left( y \right)\] Show Step 2

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

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

Next, note that the \(ds\) 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, in part (a) we solved the function for \(y\) to get,

\[y = {\cos ^{ - 1}}\left( x \right) = \arccos \left( x \right)\]

and all we need to do is plug \(x\) limits we were given into this to convert them to \(y\) limits. Doing this gives,

\[\begin{align*}x = & 0: \hspace{0.25in}y = {\cos ^{ - 1}}\left( 0 \right) = \arccos \left( 0 \right) = \frac{\pi }{2}\\ x = & \frac{1}{2}: \hspace{0.25in} y = {\cos ^{ - 1}}\left( {\frac{1}{2}} \right) = \arccos \left( {\frac{1}{2}} \right) = \frac{\pi }{3}\end{align*}\]

So, the corresponding \(y\) limits are : \(\frac{\pi }{3} \le y \le \frac{\pi }{2}\).

Note that we used both notations for the inverse cosine here but you only need to use the one you are comfortable with. Also, recall that we know that the range of the inverse cosine function is,

\[0 \le {\cos ^{ - 1}}\left( x \right) \le \pi \]

Therefore, there is only one possible value of \(y\) that we can get out of each value of \(x\).

Show Step 4

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

\[L = \int_{{}}^{{}}{{ds}} = \require{bbox} \bbox[2pt,border:1px solid black]{{\int\limits_{{\frac{\pi }{3}}}^{{\frac{\pi }{2}}}{{\sqrt {1 + {{\sin }^2}\left( y \right)} \,dy}}}}\]