Q. 12 - Calculus

Question

The arc length of the curve is traced out by the graph of $f\left(x\right)=\ln (csx x)$ on the interval $\left[\frac{\pi }{4},\frac{\pi }{2}\right]$.

Step-by-Step Solution

Answer

The arc length of the function $f\left(x\right)=\ln (csx x)$ on the given interval is $\ln (\sqrt{2}+1)\approx 0.88.$

Step 1. Given information.

given,

    $f\left(x\right)=\ln (csc x)$

Step 2. Solution:

Recalling that the exact value of the arc length of a function $f\left(x\right)$, which is differentiable and has continuous derivate on an interval $[ a,b ]$, is computed by the definite integral.  

$L={\int }_0^b \sqrt{1+{\left(f^{\mathrm{\prime }}\left(x\right)\right)}^2}dx .....\left(1\right)$

Observe that the function $f\left(x\right)=\ln (csx x)$ is a differentiable function and has a continuous derivative on the interval $\left[\frac{\pi }{4},\frac{\pi }{2}\right]$ .Differentiate the function with respect to $x$ by using the chain rules of differentiation.

$\begin{array}{r}{f}^{\mathrm{\prime }}\left(x\right)=\frac{1}{\csc x}\cdot \frac{d}{dx}(\csc x)\\ =\frac{1}{\csc x}\cdot (-\csc x\cot x)\\ =-\cot x\end{array}$

Step 3. Substitute this value of f ' ( x ) in the integral on the right-hand side of the equation and evaluate the integral using the known method of integration

$\begin{array}{r}L={\int }_{\pi /4}^{\pi /2} \sqrt{1+(-\cot x{)}^{2}x}dx\\ ={\int }_{\pi /4}^{\pi /2} \sqrt{1+{\cot }^{2}x}dx \end{array}$

Using trigonometric identity to solve the integral 

$L={\int }_{\pi /4}^{\pi /2} \csc xdx$

Using the trigonometric integration 

$\int \csc xdx=-\ln (\csc x+\cot x)$

Using the above formula to calculate the arc length

$\begin{array}{r}L=-[\ln (\csc x+\cot x){]}_{\pi /4}^{\pi /2} \\ =-\left[\ln \left(\csc x\frac{\pi }{2}+\cot \frac{\pi }{2}\right)-\ln \left(\csc \frac{\pi }{4}+\cot \frac{\pi }{4}\right)\right]\\ =-[\ln (1+0)-\ln (\sqrt{2}+1\left)\right] \\ =\ln (\sqrt{2}+1) \end{array}$

Therefore the arc length of the given function is $\ln (\sqrt{2}+1)\approx 0.88$