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Question: How do you find the arc length of the curve \(f\left( x \right)=\dfrac{{{x}^{3}}}{6}+\dfrac{1}{2x}\)...

How do you find the arc length of the curve f(x)=x36+12xf\left( x \right)=\dfrac{{{x}^{3}}}{6}+\dfrac{1}{2x} over the interval [1,3]\left[ 1,3 \right] ?

Explanation

Solution

We have been given the equation of a curve whose arc length we have to find in a fixed interval. We shall use the formula of arc length as L=∫ab1+f′(x)2.dxL=\int\limits_{a}^{b}{\sqrt{1+{f}'{{\left( x \right)}^{2}}}.dx}. Thus, to apply this formula, we will first find the derivative of the given function and then substitute it in the formula of arc length to integrate it within the provided interval.

Complete step by step solution:
Given that f(x)=x36+12xf\left( x \right)=\dfrac{{{x}^{3}}}{6}+\dfrac{1}{2x}.
We shall first differentiate the function f(x)=x36+12xf\left( x \right)=\dfrac{{{x}^{3}}}{6}+\dfrac{1}{2x} with respect to variable-x and find the derivative, f′(x){f}'\left( x \right).
f′(x)=ddx(x36+12x){f}'\left( x \right)=\dfrac{d}{dx}\left( \dfrac{{{x}^{3}}}{6}+\dfrac{1}{2x} \right)
⇒f′(x)=ddx.x36+ddx.12x\Rightarrow {f}'\left( x \right)=\dfrac{d}{dx}.\dfrac{{{x}^{3}}}{6}+\dfrac{d}{dx}.\dfrac{1}{2x}
⇒f′(x)=ddx.x36+ddx.x−12\Rightarrow {f}'\left( x \right)=\dfrac{d}{dx}.\dfrac{{{x}^{3}}}{6}+\dfrac{d}{dx}.\dfrac{{{x}^{-1}}}{2}
Here, we shall use a property of differentials ddxxn=nxn−1\dfrac{d}{dx}{{x}^{n}}=n{{x}^{n-1}}.
⇒f′(x)=3x3−16+(−1)x−1−12\Rightarrow {f}'\left( x \right)=\dfrac{3{{x}^{3-1}}}{6}+\dfrac{\left( -1 \right){{x}^{-1-1}}}{2}
⇒f′(x)=3x26−x−22\Rightarrow {f}'\left( x \right)=\dfrac{3{{x}^{2}}}{6}-\dfrac{{{x}^{-2}}}{2}
⇒f′(x)=x22−12x2\Rightarrow {f}'\left( x \right)=\dfrac{{{x}^{2}}}{2}-\dfrac{1}{2{{x}^{2}}}
Taking 2 common in the denominator, we get
⇒f′(x)=12(x2−1x2)\Rightarrow {f}'\left( x \right)=\dfrac{1}{2}\left( {{x}^{2}}-\dfrac{1}{{{x}^{2}}} \right) ………………. Equation (1)
The formula of arc length, LL is given as:
L=∫ab1+f′(x)2.dxL=\int\limits_{a}^{b}{\sqrt{1+{f}'{{\left( x \right)}^{2}}}.dx}
Where,
a=a= lower limit of integral
b=b= upper limit of integral
f′(x)={f}'\left( x \right)=derivative of the function
Substituting the value of derivative of function, we get
L=∫131+(12(x2−1x2))2.dxL=\int\limits_{1}^{3}{\sqrt{1+{{\left( \dfrac{1}{2}\left( {{x}^{2}}-\dfrac{1}{{{x}^{2}}} \right) \right)}^{2}}}.dx}
We know that (12)2=14{{\left( \dfrac{1}{2} \right)}^{2}}=\dfrac{1}{4}, thus taking 12\dfrac{1}{2}outside and squaring it, we get
⇒L=∫131+14(x2−1x2)2.dx\Rightarrow L=\int\limits_{1}^{3}{\sqrt{1+\dfrac{1}{4}{{\left( {{x}^{2}}-\dfrac{1}{{{x}^{2}}} \right)}^{2}}}.dx}
We will expand the terms within the square, (x2−1x2)2{{\left( {{x}^{2}}-\dfrac{1}{{{x}^{2}}} \right)}^{2}}using the algebraic property (a−b)2=a2+b2−2ab{{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab where a=x2a={{x}^{2}} and b=1x2b=\dfrac{1}{{{x}^{2}}}.
⇒L=∫131+14(x4+1x4−2x2(1x2)).dx\Rightarrow L=\int\limits_{1}^{3}{\sqrt{1+\dfrac{1}{4}\left( {{x}^{4}}+\dfrac{1}{{{x}^{4}}}-2{{x}^{2}}\left( \dfrac{1}{{{x}^{2}}} \right) \right)}.dx}
⇒L=∫131+14x4+14x4−(14)2x2(1x2).dx\Rightarrow L=\int\limits_{1}^{3}{\sqrt{1+\dfrac{1}{4}{{x}^{4}}+\dfrac{1}{4{{x}^{4}}}-\left( \dfrac{1}{4} \right)2{{x}^{2}}\left( \dfrac{1}{{{x}^{2}}} \right)}.dx}
⇒L=∫131+x44+14x4−12.dx\Rightarrow L=\int\limits_{1}^{3}{\sqrt{1+\dfrac{{{x}^{4}}}{4}+\dfrac{1}{4{{x}^{4}}}-\dfrac{1}{2}}.dx}
⇒L=∫13x44+14x4+12.dx\Rightarrow L=\int\limits_{1}^{3}{\sqrt{\dfrac{{{x}^{4}}}{4}+\dfrac{1}{4{{x}^{4}}}+\dfrac{1}{2}}.dx}
We shall take LCM here to simplify these terms.
⇒L=∫13x8+2x4+14x4.dx\Rightarrow L=\int\limits_{1}^{3}{\sqrt{\dfrac{{{x}^{8}}+2{{x}^{4}}+1}{4{{x}^{4}}}}.dx}
⇒L=∫13(x4+12x2)2.dx\Rightarrow L=\int\limits_{1}^{3}{\sqrt{{{\left( \dfrac{{{x}^{4}}+1}{2{{x}^{2}}} \right)}^{2}}}.dx}
⇒L=∫13x4+12x2.dx\Rightarrow L=\int\limits_{1}^{3}{\dfrac{{{x}^{4}}+1}{2{{x}^{2}}}.dx}
⇒L=∫13x22.dx+∫1312x2.dx\Rightarrow L=\int\limits_{1}^{3}{\dfrac{{{x}^{2}}}{2}.dx+}\int\limits_{1}^{3}{\dfrac{1}{2{{x}^{2}}}.dx}
⇒L=[x36]13−[12x]13\Rightarrow L=\left[ \dfrac{{{x}^{3}}}{6} \right]_{1}^{3}-\left[ \dfrac{1}{2x} \right]_{1}^{3}
⇒L=[336−136]−[12(3)−12(1)]\Rightarrow L=\left[ \dfrac{{{3}^{3}}}{6}-\dfrac{{{1}^{3}}}{6} \right]-\left[ \dfrac{1}{2\left( 3 \right)}-\dfrac{1}{2\left( 1 \right)} \right]
⇒L=[276−16]−[16−12]\Rightarrow L=\left[ \dfrac{27}{6}-\dfrac{1}{6} \right]-\left[ \dfrac{1}{6}-\dfrac{1}{2} \right]
⇒L=[266]−[−26]\Rightarrow L=\left[ \dfrac{26}{6} \right]-\left[ -\dfrac{2}{6} \right]
⇒L=286\Rightarrow L=\dfrac{28}{6}
⇒L=143\Rightarrow L=\dfrac{14}{3}
Therefore, the arc length of the curve f(x)=x36+12xf\left( x \right)=\dfrac{{{x}^{3}}}{6}+\dfrac{1}{2x} over the interval [1,3]\left[ 1,3 \right]is 143\dfrac{14}{3} units.

Note: We must have prior knowledge of the various formulae of integration and in order to proceed with such problems. The property of integration used in this problem is ∫xn.dx=xn+1n+1+C\int{{{x}^{n}}.dx=\dfrac{{{x}^{n+1}}}{n+1}+C} but since the limits of integration were already provided, thus we removed the constant of integration, C and substituted the upper and lower limits further to reach the final solution.