Chapter 6: Problem 23
Find the length of each curve. (a) \(y=\int_{1}^{x} \sqrt{u^{3}-1} d u, 1 \leq x \leq 2\) (b) \(x=t-\sin t, y=1-\cos t, 0 \leq t \leq 4 \pi\)
Short Answer
Expert verified
(a) 1.862, (b) 32
Step by step solution
01
Set up the formula for arc length for part (a)
The arc length of a curve given by a function \( y = \int_{a}^{x} f(u) \, du \) is found using the formula \( L = \int_{x=a}^{x=b} \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx \). Here, part (a) gives us \( y = \int_{1}^{x} \sqrt{u^3 - 1} \, du \), so we need the derivative \( \frac{dy}{dx} \).
02
Determine \( \frac{dy}{dx} \) for part (a) using the Fundamental Theorem of Calculus
According to the Fundamental Theorem of Calculus, the derivative of \( y = \int_{1}^{x} \sqrt{u^3 - 1} \, du \) is \( \frac{dy}{dx} = \sqrt{x^3 - 1} \).
03
Calculate the arc length for part (a)
Substitute \( \frac{dy}{dx} = \sqrt{x^3 - 1} \) into the arc length formula: \[ L = \int_{1}^{2} \sqrt{1 + (\sqrt{x^3 - 1})^2} \, dx = \int_{1}^{2} \sqrt{1 + x^3 - 1} \, dx = \int_{1}^{2} \sqrt{x^3} \, dx = \int_{1}^{2} x^{3/2} \, dx \]Integrate: \[ \int x^{3/2} \, dx = \left( \frac{2}{5} x^{5/2} \right) + C \]Evaluate from 1 to 2: \[ L = \left[ \frac{2}{5} x^{5/2} \right]_{1}^{2} = \frac{2}{5} \left( 2^{5/2} - 1^{5/2} \right) \]Calculate \(2^{5/2} = 4\sqrt{2} = 4 \times 1.414 = 5.656\), then: \( L = \frac{2}{5} (5.656 - 1) = \frac{2}{5} \times 4.656 = 1.862 \).
04
Set up parametric form and formula for part (b)
In part (b), the curve is given in parametric form as \( x = t - \sin t \), \( y = 1 - \cos t \). The formula for arc length in parametric form is: \[ L = \int_{t=a}^{t=b} \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt \] with \( 0 \leq t \leq 4\pi \).
05
Compute derivatives \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \) for part (b)
Compute the derivatives: - \( \frac{dx}{dt} = 1 - \cos t \) - \( \frac{dy}{dt} = \sin t \)
06
Calculate the arc length for part (b)
Substitute the derivatives into the arc length formula: \[ L = \int_{0}^{4\pi} \sqrt{(1 - \cos t)^2 + (\sin t)^2} \, dt = \int_{0}^{4\pi} \sqrt{2 - 2\cos t} \, dt \]Use the identity \( \sqrt{2 - 2 \cos t} = 2 \sin(t/2) \) to simplify: \[ L = \int_{0}^{4\pi} 2|\sin(t/2)| \, dt \]Since \( \sin(t/2) \) is positive in \(0 \leq t \leq 2\pi\) and negative from \(2\pi \leq t \leq 4\pi\), calculate separately. Each half-period then contributes equal length. Calculate: \[ L = 2 \times \int_{0}^{2\pi} 2 \sin\left(\frac{t}{2}\right) \, dt = 2 \times \left( -4 \cos\left(\frac{t}{2}\right) \right)_{0}^{2\pi} = 2 \times \left(0 - (-4)\right) = 16 \] Conclude both cycles contribute 16 together, giving \( L=32 \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Integral Calculus
Integral calculus is a branch of calculus focused on finding the area under curves, among other applications. It's all about adding up small pieces to understand the whole picture. In the context of the arc length problem, we're essentially adding up tiny segments of a curve to get its total length. This is accomplished via integration.
To find the length of a curve defined by an integral, like in part (a) where we have \( y = \int_{1}^{x} \sqrt{u^3 - 1} \, du \), we use derivative and integral operations together. The arc length formula in integral calculus helps us calculate the actual length of these curves. It ties together the concepts of differentiation (via the calculate of \( \frac{dy}{dx} \)) and integration (via integrating with limits).
Integral calculus is crucial for various applications in physics, engineering, and other sciences. It helps in solving real-world problems like calculating the trajectory of a projectile or determining the volume of a solid.
To find the length of a curve defined by an integral, like in part (a) where we have \( y = \int_{1}^{x} \sqrt{u^3 - 1} \, du \), we use derivative and integral operations together. The arc length formula in integral calculus helps us calculate the actual length of these curves. It ties together the concepts of differentiation (via the calculate of \( \frac{dy}{dx} \)) and integration (via integrating with limits).
Integral calculus is crucial for various applications in physics, engineering, and other sciences. It helps in solving real-world problems like calculating the trajectory of a projectile or determining the volume of a solid.
Parametric Equations
Unlike the more traditional Cartesian equations where \( y \) is directly expressed in terms of \( x \), parametric equations provide a way of defining a curve using a third variable, typically \( t \, (\text{for time})\).
In the exercise part (b), we're working with the parametric equations \( x = t - \sin t \) and \( y = 1 - \cos t \). These equations give us the position of a point along the curve at any time \( t \). The beauty of parametric equations lies in their ability to describe complex curves that are otherwise difficult to express with regular functions.
They are incredibly versatile and often make the math behind the scenes easier to handle, especially when dealing with curves that loop back on themselves or have multiple y-values for a single x-value. For example, a circle is easy to describe parametrically: \( x = r\cos t \), \( y = r\sin t \).
By converting problems into parametric form, you can often simplify the process of calculus operations like differentiation and integration.
In the exercise part (b), we're working with the parametric equations \( x = t - \sin t \) and \( y = 1 - \cos t \). These equations give us the position of a point along the curve at any time \( t \). The beauty of parametric equations lies in their ability to describe complex curves that are otherwise difficult to express with regular functions.
They are incredibly versatile and often make the math behind the scenes easier to handle, especially when dealing with curves that loop back on themselves or have multiple y-values for a single x-value. For example, a circle is easy to describe parametrically: \( x = r\cos t \), \( y = r\sin t \).
By converting problems into parametric form, you can often simplify the process of calculus operations like differentiation and integration.
Fundamental Theorem of Calculus
The Fundamental Theorem of Calculus provides the link between the process of differentiation and integration. It's like the bridge connecting the two main operations in calculus.
In the context of the problem, particularly part (a), this theorem helps us find the derivative of the integral function \( y = \int_{1}^{x} \sqrt{u^3 - 1} \, du \). According to this theorem, \( \frac{dy}{dx} = \sqrt{x^3 - 1} \).
This theorem states that the derivative of the integral of a function is equal to the original function. Essentially, it tells us that integrating a rate function (a derivative) over an interval gives us the net change in the found quantity over that interval.
The theorem is a powerful tool that allows us to convert complex integration tasks into simpler differentiation problems, drastically simplifying calculation processes.
In the context of the problem, particularly part (a), this theorem helps us find the derivative of the integral function \( y = \int_{1}^{x} \sqrt{u^3 - 1} \, du \). According to this theorem, \( \frac{dy}{dx} = \sqrt{x^3 - 1} \).
This theorem states that the derivative of the integral of a function is equal to the original function. Essentially, it tells us that integrating a rate function (a derivative) over an interval gives us the net change in the found quantity over that interval.
The theorem is a powerful tool that allows us to convert complex integration tasks into simpler differentiation problems, drastically simplifying calculation processes.
Curve Length Calculation
When it comes to measuring curves, lengths have to be determined by summing infinitely small straight segments along the curve. This is where arc length calculation steps in. It's all about accurately measuring the distance along a path, rather than just the crow-flies distance.
In calculus, especially when working with functions, arc length calculation becomes a specialized task. For standard functions like part (a), the typical formula \[ L = \int_{x=a}^{x=b} \sqrt{1 + \left( \frac{dy}{dx} \right)^2} \, dx \] helps in calculating the length of the curve.
Parametric equations, such as in part (b), use a slightly modified formula \[ L = \int_{t=a}^{t=b} \sqrt{\left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2} \, dt \]. This relies on the derivatives with respect to the parameter \( t \) rather than directly simplifying \( y \) in terms of \( x \).
Understanding how to set up and calculate these integrals is crucial for anyone working in fields requiring precise measurements, like architecture, navigation, and various engineering disciplines.
In calculus, especially when working with functions, arc length calculation becomes a specialized task. For standard functions like part (a), the typical formula \[ L = \int_{x=a}^{x=b} \sqrt{1 + \left( \frac{dy}{dx} \right)^2} \, dx \] helps in calculating the length of the curve.
Parametric equations, such as in part (b), use a slightly modified formula \[ L = \int_{t=a}^{t=b} \sqrt{\left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2} \, dt \]. This relies on the derivatives with respect to the parameter \( t \) rather than directly simplifying \( y \) in terms of \( x \).
Understanding how to set up and calculate these integrals is crucial for anyone working in fields requiring precise measurements, like architecture, navigation, and various engineering disciplines.
