Question

Find the length of the curve $y=\ln (\csc x),\frac{\pi }{4}\le x\le \frac{\pi }{2}$

Short answer

The length of the curve is $\ln (\sqrt{2}+1)$.

Step-by-step solution

Recall the formula for curve length

To find the length of a curve given by a function $y=f(x)$, from $x=a$ to $x=b$, use the formula:$$L={\int }_a^b\sqrt{1+{(f^{\prime }(x))}^2}dx$$Here, $f^{\prime }(x)$ is the derivative of $f(x)$.

Find the derivative $f^{\prime }(x)$

The curve is given by $y=\ln (\csc x)$.First, find the derivative of $y$:$$\frac{d}{dx}(\ln (\csc x))=\csc x\cdot \cot x$$This results from using the derivative of $\ln u$ which is $\frac{1}{u}\cdot \frac{du}{dx}$, where $u=\csc x$. So $\frac{du}{dx}=-\csc x\cdot \cot x$.

Substitute into the curve length formula

Now substitute $f^{\prime }(x)=-\csc x\cot x$ into the formula for $L$.$$L={\int }_{\pi /4}^{\pi /2}\sqrt{1+(-\csc x\cot x{)}^2}dx$$Simplify inside the integral to find:$$\sqrt{1+{\csc }^{2}x\cdot {\cot }^{2}x}=\sqrt{1+\frac{{\cos }^{2}x}{{\sin }^{4}x}}=\sqrt{\frac{{\sin }^{4}x+{\cos }^{2}x}{{\sin }^{4}x}}$$

Simplify the integrand

Continue simplifying the expression for the integrand:$${\sin }^{4}x+{\cos }^{2}x={\sin }^{2}x({\sin }^{2}x+{\cos }^{2}x)$$Since ${\sin }^{2}x+{\cos }^{2}x=1$, this becomes ${\sin }^{2}x$.Thus, the expression simplifies to:$$\sqrt{\frac{{\sin }^{4}x+{\cos }^{2}x}{{\sin }^{4}x}}=\sqrt{\frac{1}{{\sin }^{2}x}}=\csc x$$

Evaluate the integral

Now, the integral becomes:$$L={\int }_{\pi /4}^{\pi /2}\csc xdx$$The antiderivative of $\csc x$ is $-\ln |\csc x+\cot x|$.Evaluate this from $\pi /4$ to $\pi /2$:$$L={[-\ln |\csc x+\cot x|]}_{\pi /4}^{\pi /2}$$Calculate:At $x=\pi /2$, $\csc x=1$ and $\cot x=0$. Thus, $\ln (1)=0$.At $x=\pi /4$, $\csc (\pi /4)=\sqrt{2}$ and $\cot (\pi /4)=1$. Thus, $\ln (\sqrt{2}+1)$.So,$$L=-[0-(-\ln (\sqrt{2}+1))]=\ln (\sqrt{2}+1)$$The length of the curve is $\ln (\sqrt{2}+1)$.

Key concepts

Curve Length

Calculating the length of a curve is an important concept in calculus, especially when dealing with functions between specific points. For a function expressed as a curve, the length from a starting point to an endpoint can be determined using a specific formula. Given the curve function $y=f(x)$, the curve length $L$ from $x=a$ to $x=b$ is calculated using the integral:

• $$L={\int }_a^b\sqrt{1+(f^{\prime }(x){)}^2}dx$$

This formula considers the derivative of the function, $f^{\prime }(x)$, which represents the slope of the function at any point.
The integral sums up these infinitesimal slopes along the curve to find the curve length. It's crucial to ensure all the values within the integral are computed correctly, using derivatives and simplifying the expression before evaluating the integral.

Integration Techniques

Integrating functions is a core skill in calculus that allows us to calculate areas, volumes, and in this case, curve lengths. In the context of our problem, we used an integral to determine the arc length of the given curve. The key was simplifying the integrand, $\sqrt{1+(\csc x\cot x{)}^2}$, which resulted from the differentiation process.
The special technique involved here is simplifying trigonometric expressions:

• Start by expressing all components of the integrand in terms of basic trigonometric functions.
• Use trigonometric identities, such as ${\sin }^{2}x+{\cos }^{2}x=1$, to simplify expressions.
• Simplification led to the expression $\csc x$, easing the calculation of the integral.

These techniques help in converting a complex integrand into a simpler form, thus making the integration process more manageable. Recognizing patterns and identities in trigonometric expressions is particularly valuable in solving such problems.

Trigonometric Derivatives

The process of finding the derivative of a trigonometric function, such as $y=\ln (\csc x)$, plays a crucial role in solving curve length problems. Derivatives give us the slope or rate of change of a function, which is pivotal for calculating arc lengths. Here is how trigonometric derivatives are applied in our solution:

• The function $y=\ln (\csc x)$ involves the natural logarithm of the cosecant function.
• To differentiate this function, apply the chain rule: the derivative of $\ln (u)$ is $\frac{1}{u}\cdot \frac{du}{dx}$.
• Since $u=\csc x$, the derivative $\frac{du}{dx}$ becomes $-\csc x\cdot \cot x$.

Recognizing and applying these trigonometric derivatives correctly is key to progressing from the function to its derivative.
Once the proper derivative is derived, it is then incorporated into the formula for curve length to continue with the calculation.