Question

How do you find arc length of the curve $f\left( x \right)=2{{\left( x-1 \right)}^{\dfrac{3}{2}}}$ over the interval [1,5]?

Answer 1

The question asks us to find the length of the arc for the following function in the given interval. This question needs the concept of both integration and differentiation. The formula used to find the length of the arc of the curve is $s=\int\limits_{a}^{b}{\sqrt{1+{{\left( \dfrac{dy}{dx} \right)}^{2}}}}dx$. Here $y$is the curve, s is the length of the curve. This formula works for any kind of curve.

Complete step-by-step solution:
According to the question, the curve given is $y$ which is $f\left( x \right)=2{{\left( x-1 \right)}^{\dfrac{3}{2}}}$. On applying the formula $s=\int\limits_{a}^{b}{\sqrt{1+{{\left( \dfrac{dy}{dx} \right)}^{2}}}}dx$ to find the length of the curve. Our first step is to differentiate the given function, $f\left( x \right)=2{{\left( x-1 \right)}^{\dfrac{3}{2}}}$.
On differentiating the function we get:
$\dfrac{dy}{dx}=\dfrac{d\left[ 2{{\left( x-1 \right)}^{\dfrac{3}{2}}} \right]}{dx}$
The formula used here to differentiate the curve is, if a term ${{x}^{n}}$ is given then on differentiation it becomes $n\times {{x}^{n-1}}$ , so on applying same on the above function we get:
$\Rightarrow 2\times \dfrac{3}{2}{{\left( x-1 \right)}^{\dfrac{1}{2}}}$
$\Rightarrow 3\sqrt{x-1}$
Now, we will ahead towards the 2nd step which is to find the value of ${{\left( \dfrac{dy}{dx} \right)}^{2}}$
$\Rightarrow {{\left( \dfrac{dy}{dx} \right)}^{2}}={{[3\sqrt{x-1}]}^{2}}$
On squaring the above term we get:
$\Rightarrow 9(x-1)$
On finding the value of $\sqrt{1+{{\left( \dfrac{dy}{dx} \right)}^{2}}}$, we get
$\Rightarrow \sqrt{1+9(x-1)}$
$\Rightarrow \sqrt{9x-8}$
Further, we will integrate the equation $\sqrt{1+{{\left( \dfrac{dy}{dx} \right)}^{2}}}$with respect to $dx$ with the limit $[1,5]$.
Use the formula $\int\limits_{a}^{b}{\sqrt{1+{{\left( \dfrac{dy}{dx} \right)}^{2}}}}dx$, where $a$is $1$ and $b$ is$5$. Hence putting the value in the above formula, we get
$\Rightarrow \int\limits_{a}^{b}{\sqrt{9x-8}}dx$
On integrating the number, we use the formula ${{x}^{n}}$ we get ${{\dfrac{x}{n+1}}^{n+1}}$, so applying the same we get:
$\Rightarrow \int\limits_{1}^{5}{\sqrt{9x-8}dx}$
$\Rightarrow \int\limits_{1}^{5}{\dfrac{2}{3\times 9}{{(9x-8)}^{\dfrac{3}{2}}}dx}$
On multiplying the denominator we get
$\Rightarrow \left[ \dfrac{2}{27}{{(9x-8)}^{\dfrac{3}{2}}}dx \right]_{1}^{5}$
We will now apply the limits on the above expression
$\Rightarrow \dfrac{2}{27}\left[ {{\left( 9\times 5-8 \right)}^{\dfrac{3}{2}}}-{{(9\times 1-8)}^{\dfrac{3}{2}}} \right]$
On calculating we get
$\Rightarrow \dfrac{2}{27}\left[ {{\left( 45-8 \right)}^{\dfrac{3}{2}}}-{{(9-8)}^{\dfrac{3}{2}}} \right]$
$\Rightarrow \dfrac{2}{27}\left[ {{\left( 37 \right)}^{\dfrac{3}{2}}}-{{(1)}^{\dfrac{3}{2}}} \right]$
$\Rightarrow \dfrac{2}{27}\left[ {{\left( 37 \right)}^{\dfrac{3}{2}}}-1 \right]$
$\therefore$The length of the arc for the given curve in the interval $\left[ 1,5 \right]$ is $\dfrac{2}{27}\left[ {{\left( 37 \right)}^{\dfrac{3}{2}}}-1 \right]$.

Note: We need to keep in mind the formulas of differentiation and integration. The steps are lengthy, so we need to take care of the calculation in each step. Proper and step by step solutions will help you to solve the problem without any error.