# Consider the curve x=\frac{y^{3}}{3}+\frac{1}{4y}, 1\leq y\leq3. (a) Find the length of the...

## Question:

Consider the curve {eq}x=\frac{y^{3}}{3}+\frac{1}{4y} {/eq}, {eq}1\leq y\leq3 {/eq}. Find the length of the curve.

## Length of a Function:

The path that the function traces is continuous and then we can find the length of that trace, using the definite integration calculus. Where the formula which is used is:

{eq}L=\int_{y_1}^{y_2}\sqrt{1+x'^2}dy\\ {/eq} x' is the derivative wrt y

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In the problem, we have the curve {eq}x=\frac{y^{3}}{3}+\frac{1}{4y} {/eq}, {eq}1\leq y\leq3 {/eq}

So now in order to find the length of the...

How to Find the Arc Length of a Function

from

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.