Question

How do you find the lengths of the curve $x=\dfrac{{{y}^{4}}+3}{6y}$for $3\le y\le 8$?

Answer 1

In the given question, we have been asked to find the length of the given curve and the upper limit and the lower limit of the expression is also given. The length of the curve can be found by differentiating the given expression. Then we find the distance between the two points by integrating the length and putting the limit after simplification. In this way we get the actual length of the curve.

Complete step by step answer:
We have given that,
$x=\dfrac{{{y}^{4}}+3}{6y}$ And the upper limit and the lower limit is given i.e. $3\le y\le 8$.
We have,
$x=\dfrac{{{y}^{4}}+3}{6y}$
Simplified the above expression, we get
$x=\dfrac{{{y}^{4}}+3}{6y}=\dfrac{{{y}^{4}}}{6y}+\dfrac{3}{6y}$
Cancelling out the common terms, we get
$x=\dfrac{{{y}^{4}}+3}{6y}=\dfrac{{{y}^{4}}}{6y}+\dfrac{3}{6y}=\dfrac{{{y}^{3}}}{6}+\dfrac{1}{2y}$
Thus,
$x=\dfrac{{{y}^{3}}}{6}+\dfrac{1}{2y}$
Differentiate with respect to ‘y’, we get
$x'=\dfrac{dx}{dy}\left( \dfrac{{{y}^{3}}}{6}+\dfrac{1}{2y} \right)$
Applying the derivative rules, we obtained
$x'=\dfrac{1}{2}\left( {{y}^{2}}-\dfrac{1}{{{y}^{2}}} \right)$
Now,
Length of the arcs is given by a formula:
$\Rightarrow S=\int\limits_{a}^{b}{\vartriangle {{x}_{i}}}\sqrt{{{\left( 1+\left( f'x \right) \right)}^{2}}}$
Thus,
Arc length is given by,
$\Rightarrow \int\limits_{3}^{8}{\sqrt{1+\dfrac{1}{4}{{\left( {{y}^{2}}-\dfrac{1}{{{y}^{2}}} \right)}^{2}}}}dy$
Expanding the square by using the property, i.e. ${{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}$
$\Rightarrow \int\limits_{3}^{8}{\sqrt{1+\dfrac{1}{4}\left( {{y}^{4}}-2+\dfrac{1}{{{y}^{4}}} \right)}}dy$
Combining and simplified the terms, we get
$\Rightarrow \dfrac{1}{2}\int\limits_{3}^{8}{\sqrt{{{y}^{4}}+2+\dfrac{1}{{{y}^{4}}}}}dy$
We know that, ${{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$
Applying this property, we obtained
$\Rightarrow \dfrac{1}{2}\int\limits_{3}^{8}{\sqrt{{{\left( {{y}^{2}}+\dfrac{1}{{{y}^{2}}} \right)}^{2}}}}dy$
Cancelling out the square root and the power of 2 with each other, we get
$\Rightarrow \dfrac{1}{2}\int\limits_{3}^{8}{\left( {{y}^{2}}+\dfrac{1}{{{y}^{2}}} \right)}dy$
Now,
Integrating the above expression, we obtained
$\Rightarrow \dfrac{1}{2}\left( \dfrac{{{y}^{3}}}{3}-\dfrac{1}{y} \right)_{3}^{8}$
Taking the limits, we will obtained
$\Rightarrow \dfrac{1}{2}\left[ \dfrac{{{8}^{3}}}{3}-\dfrac{1}{8} \right]-\dfrac{1}{2}\left[ \dfrac{{{3}^{3}}}{3}-\dfrac{1}{3} \right]$
Simplified the above expression by taking out common factors, we get
$\Rightarrow \dfrac{1}{6}\left[ {{8}^{3}}-{{3}^{3}} \right]-\dfrac{1}{2}\left[ \dfrac{1}{8}-\dfrac{1}{3} \right]$
Solving the above expression, we obtained
$\Rightarrow \dfrac{485}{6}+\dfrac{5}{48}$
Solving the above expression by taking the LCM of denominator, we get
$\therefore \dfrac{485}{6}+\dfrac{5}{48}=\dfrac{1295}{6}$

Hence, the length of the curve is $\dfrac{1295}{6}$ or 215.83.

Note: We can also calculate the length of the any portion of the given equation or expression by using equation as we know that the coordinates of the starting point and the end point of the curve and replacing the point ‘a’ and point ‘b’ with those coordinates in the equation and then find out the length of the given curve.