Calculus II - Arc Length

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Section 8.1 : Arc Length

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3. Determine the length of $y = 7{\left( {6 + x} \right)^{\frac{3}{2}}}$ , $189 \le y \le 875$ .

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Start Solution

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 $x$ . Note as well that if we solve the function for $x$ (which we’d need to do in order to use the $ds$ that is in terms of $y$ ) we would still have a fractional exponent and the derivative will not work out as nice once we plug it into the $ds$ formula.

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

$\frac{{dy}}{{dx}} = \frac{{21}}{2}{\left( {6 + x} \right)^{\frac{1}{2}}}$

$\begin{align*}ds & = \sqrt {1 + {{\left[ {\frac{{dy}}{{dx}}} \right]}^2}} \,dx = \sqrt {1 + {{\left[ {\frac{{21}}{2}{{\left( {6 + x} \right)}^{\frac{1}{2}}}} \right]}^2}} \,dx = \sqrt {1 + \frac{{441}}{4}\left( {6 + x} \right)} \,dx\\ & = \sqrt {\frac{{2650}}{4} + \frac{{441}}{4}x} \,dx = \frac{1}{2}\sqrt {2650 + 441x} \,dx\end{align*}$

We did a little simplification that may or may not make the integration easier. That will probably depend upon the person doing the integration and just what they find the easiest to deal with. The point is there are several forms of the $ds$ that we could use here. All will give the same answer.

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Next, we need to deal with the limits for the integral. The $ds$ that we choose to use in the first step has a $dx$ in it and that means that we’ll need $x$ limits for our integral. We, however, were given $y$ limits in the problem statement. This means we’ll need to convert those to $x$ ’s before proceeding with the integral.

To do convert these all we need to do is plug them into the function we were given in the problem statement and solve for the corresponding $x$ . Doing this gives,

$\begin{align*}y = & 189:\,\,\,\,\,\,\,\,\,189 = 7{\left( {6 + x} \right)^{\frac{3}{2}}}\,\,\,\,\, \to \,\,\,\,\,\,6 + x = {27^{\frac{2}{3}}} = 9\,\,\,\,\,\,\, \to \,\,\,\,\,\,\,x = 3\\ y = & 875:\,\,\,\,\,\,\,\,\,875 = 7{\left( {6 + x} \right)^{\frac{3}{2}}}\,\,\,\,\, \to \,\,\,\,\,\,6 + x = {125^{\frac{2}{3}}} = 25\,\,\,\,\,\,\, \to \,\,\,\,\,\,\,x = 19\end{align*}$

So, the corresponding ranges of $x$ ’s is : $3 \le x \le 19$ .

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The integral giving the arc length is then,

$L = \int_{{}}^{{}}{{ds}} = \int_{3}^{{19}}{{\frac{1}{2}\sqrt {2650 + 441x} \,dx}}$ Show Step 4

Finally, all we need to do is evaluate the integral. In this case all we need to do is use a quick Calc I substitution. We’ll leave most of the integration details to you to verify.

The arc length of the curve is,

$L = \int_{3}^{{19}}{{\frac{1}{2}\sqrt {2650 + 441x} \,dx}} = \left. {\frac{1}{{1323}}{{\left( {2650 + 441x} \right)}^{\frac{3}{2}}}} \right|_3^{19} = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{1}{{1323}}\left( {{{11029}^{\frac{3}{2}}} - {{3973}^{\frac{3}{2}}}} \right) = 686.1904}}$