(a) Find the arc length function for the curve y=ln(sinx), 0<x<π, with starting point ( π/ 2,0 )

(a) Find the arc length function for the curve y=ln(sinx),...

Key Concepts

Graphical Representation

Graphing involves plotting functions to visually compare their behaviors. In this context, both the original function and its arc length function are plotted on the same screen, which is useful for understanding how the accumulation of arc length relates to the position on the curve. This approach is common in the visual analysis of mathematical functions.

Integration Techniques

Evaluating the arc length integral often involves advanced integration techniques including substitution and the use of trigonometric identities to simplify the integrand. Mastery of these techniques is important for solving integrals that arise in the computation of arc lengths and other related problems in calculus.

Arc Length Function

The arc length function is a tool in calculus that expresses the distance along a curve as a function of a variable. It is defined by the integral L(x) = ?[x0 to x] ?(1 + (dy/dt)²) dt, which accumulates the lengths of infinitesimal segments of the curve. This concept is essential for finding the distance traveled along any continuous curve.

Differentiation of Composite Functions

Differentiation of composite functions, especially when involving a logarithmic function composed with a trigonometric function, requires the application of the chain rule. This allows us to compute the derivative dy/dx for functions such as y = ln(sin x), which is a critical step in forming the integrand for the arc length formula.

Transcript

00:03 In part a, we are given a curve with a starting point, and we are asked to find the arc length function for this curve.

00:16 So the curve we're given has the equation y equals the natural log of sine of x, and the curve is defined for x between 0 and pi, and the starting point is at pi over 2.

00:53 0.

01:07 Well, first of all, we have that y prime is going to be 1 over sine x times the derivative of sine x, which is cosine x.

01:18 So we have cosine x over sine x, which is cotangent of x.

01:27 And therefore, 1 plus y prime squared is 1 plus cotangent squared of x.

01:37 And using trig identities, this is the same as co -secon squared as x, and therefore the square root of 1 plus y prime squared is the square root of cosecant squared of x, which is the absolute value of co -secon of x.

02:07 So the arc length function s of x is the integral from x equals pi over 2, and then we c up to x of the absolute value of cosecant of t d t now notice that for x greater than pi over two sine of x will automatically be less than one co -secent of x is going to be that's one over sine of x that's going to be greater than zero and so this will be the integral of jx is going to be just coseccccccccc d t.

03:45 And taking antiderivatives, this is the natural log of the absolute value of cosecant t minus cosine of t evaluated from pi over two to x is equal to, or this should be cotangent of t, my mistake, not cosine of t.

04:25 And so i get the natural log of absolute the value of cosecant of pi over 2, which is 1 over the sine of pi over 2, which is 1 minus cotangent of, sorry, x.

04:46 It's natural log of absolute value of co -secon of x minus cotangent of x minus the natural log of 1 minus 0.

05:06 And we have this is the same as the natural log of.

05:10 And then because co -seken x minus cotangent x is greater than 0 for x greater than pi over 2 and less than pi, it follows this is simply natural log of co -secon of x minus cotangent of x...

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