Calculus II - Arc Length

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Section 8.1 : Arc Length

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

$\displaystyle ds = \sqrt {1 + {{\left[ {\frac{{dy}}{{dx}}} \right]}^2}} \,dx$
$\displaystyle ds = \sqrt {1 + {{\left[ {\frac{{dx}}{{dy}}} \right]}^2}} \,dy$

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$\displaystyle ds = \sqrt {1 + {{\left[ {\frac{{dy}}{{dx}}} \right]}^2}} \,dx$ 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.

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

$\displaystyle ds = \sqrt {1 + {{\left[ {\frac{{dx}}{{dy}}} \right]}^2}} \,dy$ 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.

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