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Magazine Chapter 8: Problem 12
Find the exact arc length of the parametric curve without eliminating the parameter. $$x=\cos t+t \sin t, \quad y=\sin t-t \cos t \quad(0 \leq t \leq \pi)$$
Short Answer
Expert verified
The exact arc length is \(\frac{\pi^2}{2}\).
Step by step solution
01
Find the Derivative of x with Respect to t
To find the derivative of the x-component with respect to the parameter \(t\), we use differentiation rules. The function is \(x = \cos t + t \sin t\). Using the product rule and chain rule: \[ \frac{dx}{dt} = -\sin t + \sin t + t \cos t = t \cos t \]
02
Find the Derivative of y with Respect to t
Now, find the derivative of the y-component with respect to \(t\). The function is \(y = \sin t - t \cos t\). Again, using the product and chain rules: \[ \frac{dy}{dt} = \cos t - (\cos t - t \sin t) = t \sin t \]
03
Compute the Expression for Arc Length
The formula for the arc length of a parametric curve is: \[ L = \int_{a}^{b} \sqrt{\left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2} \, dt \] Substitute \( \frac{dx}{dt} = t\cos t \) and \( \frac{dy}{dt} = t\sin t \) into this formula.
04
Simplify the Expression Inside the Integral
Simplify the expression under the square root:\[ \left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2 = (t \cos t)^2 + (t \sin t)^2 \] This simplifies to: \[ t^2(\cos^2 t + \sin^2 t) = t^2 \] using the Pythagorean identity \(\cos^2 t + \sin^2 t = 1\).
05
Evaluate the Integral for Arc Length
Now, compute the integral:\[ L = \int_{0}^{\pi} \sqrt{t^2} \, dt = \int_{0}^{\pi} t \, dt \] The integral becomes: \[ \left[ \frac{t^2}{2} \right]_0^\pi = \frac{\pi^2}{2} - 0 = \frac{\pi^2}{2} \] This gives the arc length of the parametric curve.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Parametric Equations
Parametric equations involve expressing the coordinates of the points on a curve as functions of a parameter, often denoted as \( t \). In our exercise, the parameter \( t \) dictates the values of \( x \) and \( y \), with the equations given by \( x = \cos t + t \sin t \) and \( y = \sin t - t \cos t \). These equations describe a curve in the plane by allowing the parameter \( t \) to vary over a specified interval, in this case from \( 0 \) to \( \pi \).
- The advantage of parametric equations is their ability to represent curves that are difficult to describe using standard \( y = f(x) \) formats.
- They are particularly useful in cases where the curve loops, twists, or behaves erratically in relation to one of the axes.
Differentiation in Calculus
Differentiation is a fundamental concept in calculus concerning the rate at which a function changes at any given point. In our exercise, we differentiated parametric equations \( x(t) \) and \( y(t) \) with respect to \( t \) to find \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \). This step provides the slopes of the tangent to the curve at any point described by \( t \).
- The derivative \( \frac{dx}{dt} = t \cos t \) shows the x-component's rate of change with respect to the parameter.
- Similarly, \( \frac{dy}{dt} = t \sin t \) shows the y-component's rate of change with respect to \( t \).
Integral Calculus
Integral calculus, the inverse process of differentiation, allows us to compute quantities like area and arc length under a curve. In this context, we used the concept to find the total arc length of a parametric curve by integrating a function of derivatives. The arc length formula \( L = \int_{a}^{b} \sqrt{\left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2} \; dt \) helps calculate this length by examining how far the curve travels between two values of \( t \).
- Integrating functions of derivatives provides a way to accumulate quantities over a continuous range.
- In our case, the integral \( \int_{0}^{\pi} t \; dt \) resulted in an arc length of \( \frac{\pi^2}{2} \).
Pythagorean Identity
The Pythagorean identity is an essential trigonometric principle stating that \( \cos^2 t + \sin^2 t = 1 \). This identity was used in our solution to simplify the expression under the square root in the arc length formula. By recognizing the sum of squares \( (t \cos t)^2 + (t \sin t)^2 \), we reduced it to \( t^2(\cos^2 t + \sin^2 t) = t^2 \). This simplification significantly eased the integration process.
- Using the identity saves computation time and effort by transforming complex trigonometric expressions into simpler algebraic forms.
- This simplification step is crucial when dealing with integrals of squared trigonometric functions.
