Chapter 9: Problem 42
Graph the curve and find its length. $$x=\cos t+\ln \left(\tan \frac{1}{2} t\right), \quad y=\sin t, \quad \pi / 4 \leqslant t \leqslant 3 \pi / 4$$
Short Answer
Expert verified
Integrate \( \int_{\pi/4}^{3\pi/4} \sqrt{(-\sin t + \frac{1}{2}\sec^2(\frac{t}{2}))^2 + (\cos t)^2} \, dt \).
Step by step solution
01
Parametric Equations Definition
The given parametric equations are \( x = \cos t + \ln \left( \tan \frac{1}{2} t \right) \) and \( y = \sin t \), with the parameter \( t \) ranging from \( \pi/4 \) to \( 3\pi/4 \). These equations represent the position of a point on the curve as \( t \) changes.
02
Derivative Calculation
To find the length of the curve, we start by calculating the derivatives of \( x \) and \( y \) with respect to \( t \). The derivative \( \frac{dx}{dt} = -\sin t + \frac{1}{2}\sec^2\left(\frac{t}{2}\right) \). The derivative \( \frac{dy}{dt} = \cos t \).
03
Arc Length Formula
The formula for the arc length of a parametric curve is \( L = \int_{a}^{b} \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt \). Here, \( a = \pi/4 \) and \( b = 3\pi/4 \).
04
Inside the Integral
Substitute the derivatives into the arc length formula: \( \left(\frac{dx}{dt}\right)^2 = (-\sin t + \frac{1}{2}\sec^2\left(\frac{t}{2}\right))^2 \) and \( \left(\frac{dy}{dt}\right)^2 = (\cos t)^2 \).
05
Simplification of Square Root Term
Combine and simplify the expression under the square root: \[ \sqrt{(-\sin t + \frac{1}{2}\sec^2\left(\frac{t}{2}\right))^2 + (\cos t)^2} \]. This needs further algebraic manipulation ensuring simplification to ease the integration.
06
Evaluate the Integral
Solve the integral \( \int_{\pi/4}^{3\pi/4} \sqrt{(-\sin t + \frac{1}{2}\sec^2\left(\frac{t}{2}\right))^2 + (\cos t)^2} \, dt \) using appropriate techniques such as substitution if needed, or numerical methods as this integral might be complex without a straightforward antiderivative.
07
Graph the Curve
Plot the parametric equations for \( x(t) \) and \( y(t) \) over the interval \( \pi/4 \leq t \leq 3\pi/4 \) using graphing software or a graphing calculator to visualize the shape of the curve.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Arc Length in Parametric Equations
Calculating the arc length of a parametric curve involves integrating the speed of a moving point as it travels along the curve. For a curve defined by parametric equations \( x(t) = f(t) \) and \( y(t) = g(t) \), the arc length \( L \) from \( t = a \) to \( t = b \) can be found using the formula: \[ L = \int_{a}^{b} \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt \] This formula comes from the Pythagorean theorem applied at an infinitesimally small level, meaning we're summing up the hypotenuses of right triangles formed by the horizontal \( \frac{dx}{dt} \) and vertical \( \frac{dy}{dt} \) changes, captured by the derivatives, along the curve. It’s crucial to correctly evaluate the derivatives \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \) for accuracy in length measurement. While this integral might sound complex, solutions often involve simplification techniques such as trigonometric identities or numerical methods if a direct integration proves challenging.
Role of Derivatives in Parametric Curves
Derivatives play a pivotal role in analyzing parametric curves, primarily by helping us understand the velocity vector of a moving point along the curve. In simple terms, when you have parametric equations like \( x = \cos t + \ln \left( \tan \frac{1}{2} t \right) \) and \( y = \sin t \), the derivatives \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \) indicate how fast these coordinates are changing with respect to the parameter \( t \).
- \( \frac{dx}{dt} = -\sin t + \frac{1}{2}\sec^2\left(\frac{t}{2}\right) \) measures the rate of change in the x-direction. It's influenced by the sine decrease and the secant increase parts of the equation.
- \( \frac{dy}{dt} = \cos t \) measures the rate of change in the y-direction, which varies with the cosine function.
Graphing Parametric Curves
Graphing parametric equations allows us to better visualize the path generated as the parameter \( t \) varies. Unlike traditional graphs, a parametric curve requires us to plot points using the pairs \((x(t), y(t))\) resulting from the equations. For the equations \( x = \cos t + \ln \left(\tan \frac{1}{2} t\right) \) and \( y = \sin t \), as \( t \) runs from \( \pi/4 \) to \( 3\pi/4 \), the shape of the path can be plotted for distinct points along these intervals using:
- Choose values of \( t \) within the given range. The given curve only spans from \( t = \pi/4 \) to \( t = 3\pi/4 \).
- Substitute \( t \) into the equations to get specific \( x \) and \( y \) values.
- Plot these \( (x, y) \) pairs on a coordinate plane. Repeat for several \( t \) values to see the curve's progression.