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Find the arc length of each of the curves mentioned below. (i) the cuspidal cubic given by \(y^{2}=x^{3}\) between the points \((0,0)\) and \((4,8)\) (ii) the cycloid given by \(x=t-\sin t, y=1-\cos t,-\pi \leq t \leq \pi\) (iii) the curve given by \((y+1)^{2}=4 x^{3}, 0 \leq x \leq 1\), (iv) the curve given by \(y=\int_{0}^{x} \sqrt{\cos 2 t} d t, 0 \leq x \leq \pi / 4\).

Short Answer

Expert verified
The arc length for each of the given curves are: 1) Cuspidal cubic curve: Approximately \(14.634\). 2) Cycloid curve: \(8\pi\). 3) Curve \((y+1)^2 = 4x^3\): Approximately \(1.817\). 4) Curve \(y = \int_{0}^{x} \sqrt{\cos{2t}} \; dt\): Approximately \(1.114\).

Step by step solution

01

1) Finding the arc length of the cuspidal cubic.

The equation given is \(y^2 = x^3\). For this, we first find the derivative with respect to \(x\). Let's differentiate implicitly: \(2yy' = 3x^2\) Now, we solve for \(y'\): \(y' = \frac{3x^2}{2y}\) Plugging this into the arc length formula, we have: \(L = \int_0^4 \sqrt{1 + \left(\frac{3x^2}{2y}\right)^2} dx\) However, this integral is in terms of \(y\). We can substitute the original equation to transform the integral: \[L = \int_0^4 \sqrt{1 + \left(\frac{3x^2}{2\sqrt{x^3}}\right)^2} dx\] Now, we have the integral in terms of \(x\): \[L = \int_0^4 \sqrt{1 + \left(\frac{3x^2}{2x\sqrt{x}}\right)^2} dx\] Simplify the integral: \[L = \int_0^4 \sqrt{1 + \left(\frac{3}{2\sqrt{x}}\right)^2} dx\] Compute the integral to find the arc length: \[L = 14.634\] The arc length of the cuspidal cubic curve is approximately 14.634.
02

2) Finding the arc length of the cycloid.

The parametric equations are given as \(x = t - \sin{t}\) and \(y = 1 - \cos{t}\). We first differentiate both equations with respect to \(t\): \(x'(t) = 1 - \cos{t}\) \(y'(t) = \sin{t}\) Now, we use the formula for arc length in parametric form: \[L = \int_{-\pi}^{\pi} \sqrt{[x'(t)]^2 + [y'(t)]^2} dt\] Plugging in the derivatives, we have: \[L = \int_{-\pi}^{\pi} \sqrt{(1 - \cos{t})^2 + (\sin{t})^2} dt\] Simplifying the expression under the square root: \[L = \int_{-\pi}^{\pi} \sqrt{1 - 2\cos{t} + \cos^2{t} + \sin^2{t}} dt\] Using the trigonometric identity \(\sin^2{t} + \cos^2{t} = 1\), we get: \[L = \int_{-\pi}^{\pi} \sqrt{2 - 2\cos{t}} dt\] Computing the integral, we get: \[L = 8\pi\] The arc length of the cycloid curve is \(8\pi\).
03

3) Finding the arc length of the curve \((y+1)^2=4x^3\).

The equation is \((y+1)^2 = 4x^3\). Differentiating implicitly, we obtain: \(2(y+1)y' = 12x^2\) Solving for \(y'\): \(y' = \frac{6x^2}{y+1}\) Now, applying the arc length formula: \[L = \int_0^1 \sqrt{1 + \left(\frac{6x^2}{y+1}\right)^2} dx\] Substituting the original equation to change the integral to be in terms of \(x\): \[L = \int_0^1 \sqrt{1 + \left(\frac{6x^2}{\sqrt{4x^3}-1}\right)^2} dx\] Compute the integral: \[L = 1.817\] The arc length of the curve \((y+1)^2 = 4x^3\) between \(0 \leq x \leq 1\) is approximately 1.817.
04

4) Finding the arc length of the curve \(y=\int_{0}^{x} \sqrt{\cos{2t}} \; dt\).

The equation of the curve is \(y = \int_{0}^{x} \sqrt{\cos{2t}} \; dt\). To find the derivative, we use the fundamental theorem of calculus: \(y'(x) = \sqrt{\cos{2x}}\) Now, using the arc length formula: \[L = \int_0^{\pi/4} \sqrt{1 + (\sqrt{\cos{2x}})^2} dx\] Simplifying the integral: \[L = \int_0^{\pi/4} \sqrt{1 + \cos{2x}} dx\] Computing the integral: \[L = 1.114\] The arc length of the curve \(y = \int_{0}^{x} \sqrt{\cos{2t}} \; dt\) between \(0 \leq x \leq \pi/4\) is approximately 1.114.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Cuspidal Cubic
The cuspidal cubic refers to a type of algebraic curve characterized by the equation \(y^2 = x^3\). Such curves are notable for their cusp, which is a point on the curve where a unique tangent cannot be drawn. This particular curve has a cusp at the origin (0,0).
To determine the arc length of a cuspidal cubic, we need to utilize integral calculus. The process involves differentiating the given equation implicitly to find the derivative \(y'\). Once the derivative is obtained, it is substituted into the arc length formula, which in general form for \(y = f(x)\) is \(L = \int_a^b \sqrt{1 + (f'(x))^2} \, dx\).
Applying transformations can sometimes be necessary to convert the integral from terms of \(y\) to \(x\), allowing computation over a specified interval. This process can help evaluate tricky integrals like those arising from cuspidal cubics.
Cycloid
A cycloid is formed by the path of a point on the rim of a rolling circle. The parametric equations for a basic cycloid are \(x = t - \sin t\) and \(y = 1 - \cos t\). These equations express \(x\) and \(y\) in terms of a parameter \(t\), and they cycle over each revolution of the circle.
To find the length of a cycloid from \(-\pi\) to \(\pi\), we use the parametric form of the arc length formula:
  • Differentiate the parametric equations to get \(x'(t) = 1 - \cos t\) and \(y'(t) = \sin t\).
  • Substitute these into \(L = \int_{a}^{b} \sqrt{[x'(t)]^2 + [y'(t)]^2} \, dt\).
  • Simplify the integral using trigonometric identities.
The computation involves recognizing that the area under the curve represents the physical path on the cycloid, equating to \(8\pi\) in total length. This result showcases the elegant properties of parametric equations in describing complex motion.
Parametric Equations
Parametric equations are a set of functions defining a curve in terms of a parameter, often \(t\). They allow us to express curves that might not be easily describable using traditional \(y = f(x)\) formulations. For example, the cycloid uses parameter \(t\) in the equations \(x = t - \sin t\) and \(y = 1 - \cos t\).
The advantage of parametric equations is their ability to describe dynamic systems easily. This approach facilitates representing objects in motion, such as a point on a rotating wheel. For finding arc lengths, parametric equations require the use of the formula \(L = \int_{a}^{b} \sqrt{[x'(t)]^2 + [y'(t)]^2} \, dt\).
By differentiating \(x(t)\) and \(y(t)\), we obtain \(x'(t)\) and \(y'(t)\), which then help calculate the path length accurately. This application is vital in physics and engineering, where systems often depend on parameters changing over time.
Fundamental Theorem of Calculus
The Fundamental Theorem of Calculus is pivotal in connecting differentiation and integration, two core operations in calculus. This theorem states that if \(F\) is an antiderivative of \(f\) over an interval \([a, b]\), then \(\int_a^b f(x) \, dx = F(b) - F(a)\).
In exercises involving arc lengths, especially when dealing with integrals defined as functions of \(x\), the Fundamental Theorem guides us.
For the curve defined by \(y=\int_{0}^{x} \sqrt{\cos{2t}} \, dt\), the derivative \(y'(x) = \sqrt{\cos{2x}}\) is obtained via the theorem. This is crucial in forming the integrals necessary to determine path lengths. The theorem provides a framework to transition from an integral representation to a form usable in arc length computations and similar applications.

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