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Magazine Chapter 7: Problem 30
(a) Find the arc length function for the curve
\(y=\ln (\sin x), 0
Short Answer
Expert verified
The arc length function is \( L(x) = -\ln |\csc x + \cot x| \). Graphs can be plotted using graphing software.
Step by step solution
01
Find the derivative of y with respect to x
First, we need to find the derivative of the given function, which is \( y = \ln(\sin x) \). Use the chain rule. The derivative is given by: \[ \frac{dy}{dx} = \frac{1}{\sin x} \cdot \cos x = \cot x. \]
02
Set up the arc length integral
The formula for the arc length of a function \( y(x) \) from \( x = a \) to \( x = b \) is \( \int_a^b \sqrt{1 + \left( \frac{dy}{dx} \right)^2} \, dx \). Here, \( \frac{dy}{dx} = \cot x \), so the integral becomes: \[ L = \int_{\pi/2}^x \sqrt{1 + (\cot x)^2} \, dx. \]
03
Simplify the integrand
Simplify the integrand using the Pythagorean identity: \( 1 + \cot^2 x = \csc^2 x \). Thus the integrand becomes \( \sqrt{\csc^2 x} = \csc x \). The arc length function is now: \[ L(x) = \int_{\pi/2}^x \csc x \, dx. \]
04
Evaluate the integral
To evaluate \( \int \csc x \, dx \), we know that the antiderivative is \( -\ln|\csc x + \cot x| + C \). Integrating from \( x = \pi/2 \) to \( x = x \), we get: \[ L(x) = \left.[-\ln |\csc x + \cot x|] \right|_{\pi/2}^x = -\ln |\csc x + \cot x| + \ln|\csc(\pi/2) + \cot(\pi/2)|. \] The constant term simplifies to 0 since \( \csc(\pi/2) = 1 \) and \( \cot(\pi/2) = 0 \).
05
Write the arc length function
The arc length function, accounting for the simplification of the constant, is: \[ L(x) = -\ln |\csc x + \cot x|. \]
06
Graph both the curve and its arc length function
Use graphing software or a graphing calculator to plot \( y = \ln(\sin x) \) and \( L(x) = -\ln|\csc x + \cot x| \) on the same screen for \( 0 < x < \pi \). Observe their behavior, especially near the endpoints of the interval.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Derivative of a Function
A derivative measures how a function changes as its input changes. For the function given in the exercise, which is \( y = \ln(\sin x) \), we want to find \( \frac{dy}{dx} \), the derivative of \( y \) with respect to \( x \). This will tell us the slope of the tangent line to the curve at any given point. To do this, we apply the principles of differentiation.
- In this case, the function \( y \) is the natural logarithm of another function, \( \sin x \). The derivative of the natural log function, \( \ln(u) \), is \( \frac{1}{u} \cdot \frac{du}{dx} \).
- Here, our \( u \) is \( \sin x \), so \( \frac{du}{dx} = \cos x \).
- Thus, the derivative is \( \frac{1}{\sin x} \cdot \cos x = \cot x \).
Chain Rule in Calculus
The Chain Rule is an essential tool when differentiating composite functions, which are functions made by combining two or more functions. Here we have a function \( y = \ln(\sin x) \), which is composed of \( \ln(u) \) where \( u = \sin x \).
- The Chain Rule states: if a function \( y \) is composed of two functions \( f(g(x)) \), then its derivative \( \frac{dy}{dx} = f'(g(x)) \cdot g'(x) \).
- Applying it here, we recognize \( \ln(u) \) as the outer function and \( \sin x \) as the inner function.
- The derivative of the outer function \( \ln(u) \) is \( \frac{1}{u} \), and the derivative of the inner function \( \sin x \) is \( \cos x \).
- Thus, using the Chain Rule, the derivative becomes: \( \frac{1}{\sin x} \cdot \cos x = \cot x \).
Integral Calculus
Integral calculus is essentially the reverse process of differentiation. In this exercise, we are finding the arc length of a curve, which involves integration. The formula for the arc length \( L \) is set up as an integral:
- \( L = \int_a^b \sqrt{1 + \left( \frac{dy}{dx} \right)^2} \, dx \).
- For our curve, since \( \frac{dy}{dx} = \cot x \), we have: \( L = \int_{\pi/2}^x \sqrt{1 + (\cot x)^2} \, dx \).
- The integrand simplifies using the identity \( 1 + \cot^2 x = \csc^2 x \), giving us \( \csc x \).
- The problem then reduces to evaluating \( L(x) = \int_{\pi/2}^x \csc x \, dx \), whose antiderivative is \( -\ln|\csc x + \cot x| \).
Graphing Functions
Graphing functions provides a visual understanding of the behavior of mathematical expressions. In this exercise, we graph both the original function \( y = \ln(\sin x) \) and its arc length function \( L(x) = -\ln|\csc x + \cot x| \). The graphing reveals important insights:
- The function \( y = \ln(\sin x) \) behaves uniquely as \( x \) approaches the boundaries in the interval \( 0 < x < \pi \), with possible discontinuities at 0 and \( \pi \) where \( \sin x \) becomes 0.
- The arc length function \( L(x) \), derived from \( y \), should be non-negative and accumulate length as \( x \) increases.
- By observing these graphs, one can better understand the relationship between a curve and its arc length, visualizing how length accumulates along the curve from one point to another.
