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Find the length of the curve x = y^ 3 / 3 + 1 / 4 y from y = 1 to y = 2.

Question:

Find the length of the curve {eq}x=\frac{y^{3}}{3}+\frac{1}{4y} {/eq} from y = 1 to y = 2.

Arc Length of a Function:

When given a function in terms of {eq}x {/eq}, the length of the function from {eq}x=a {/eq} to {eq}x=b {/eq} is {eq}L=\int_{a}^{b}\sqrt{1+\left ( \frac{dy}{dx} \right )^{2}}dx {/eq}.

If the function is given in terms of {eq}y {/eq}, the length of the function from {eq}y=c {/eq} to {eq}y=d {/eq} is {eq}L=\int_{c}^{d}\sqrt{1+\left ( \frac{dx}{dy} \right )^{2}}dy {/eq}.

Answer and Explanation: 1

To find the length of the curve {eq}x=\frac{y^{3}}{3}+\frac{1}{4y} {/eq} from {eq}y = 1 {/eq} to {eq}y = 2 {/eq}, we will use {eq}L=\int_{c}^{d}\sqrt{1+\left ( \frac{dx}{dy} \right )^{2}}dy {/eq}.

{eq}\frac{dx}{dy}=y^{2}-\frac{1}{4y^{2}} {/eq}


{eq}\begin{align*} L&=\int_{1}^{2}\sqrt{1+\left ( y^{2}-\frac{1}{4y^{2}}\right )^{2}}dy\\ &=\int_{1}^{2}\sqrt{1+\left ( y^{4}-\frac{1}{2}+\frac{1}{16y^{4}}\right )}dy\\ &=\int_{1}^{2}\sqrt{\left ( y^{4}+\frac{1}{2}+\frac{1}{16y^{4}}\right )}dy\\ &=\int_{1}^{2}\sqrt{\left ( y^{2}+\frac{1}{4y^{2}}\right )^{2}}dy\\ &=\int_{1}^{2}\left ( y^{2}+\frac{1}{4y^{2}}\right )dy\\ &=\left ( \frac{y^{3}}{3}-\frac{1}{4y} \right )\bigg|_{1}^{2}\\ &=\left ( \frac{2^{3}}{3}-\frac{1}{4\left ( 2 \right )} \right )-\left ( \frac{1^{3}}{3}-\frac{1}{4\left ( 1 \right )} \right )\\ &=\left ( \frac{8}{3}-\frac{1}{8} \right )-\left ( \frac{1}{3}-\frac{1}{4} \right )\\ &=\frac{61}{24}-\frac{1}{12}\\ &=\boxed{\frac{59}{24}} \end{align*} {/eq}


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How to Find the Arc Length of a Function

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Chapter 12 / Lesson 12
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You don't always walk in a straight line. Sometimes, you want to know the distance between two points when the path is curved. In this lesson, you'll learn about finding the length of a curve.


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