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Use either a computer algebra system or a table of integrals to find the exact length of the arc of the curve \(y=x^{4 / 3}\) that lies between the points \((0,0)\) and \((1,1) .\) If your CAS has trouble evaluating the integral, make a substitution that changes the integral into one that the CAS can evaluate.

Short Answer

Expert verified
The arc length is \(\frac{5}{3}\).

Step by step solution

01

Understanding the Arc Length Formula

The arc length of the curve from a point \(a\) to a point \(b\) of a function \(y=f(x)\) is given by the integral formula: \[ L = \int_a^b \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx \] We will use this formula for the curve \(y = x^{4/3}\) from \(x = 0\) to \(x = 1\).
02

Finding the Derivative

Differentiate \(y = x^{4/3}\) with respect to \(x\) to get \(\frac{dy}{dx}\). Using the power rule, we find \[ \frac{dy}{dx} = \frac{4}{3}x^{1/3}. \]
03

Setting Up the Integral for Arc Length

Substitute \(\frac{dy}{dx}\) into the arc length formula: \[ L = \int_0^1 \sqrt{1 + \left(\frac{4}{3}x^{1/3}\right)^2} \, dx = \int_0^1 \sqrt{1 + \frac{16}{9}x^{2/3}} \, dx. \]
04

Preparing the Integral for Evaluation

Consider a substitution to simplify the evaluation by setting \(u = x^{1/3}\), so \(x = u^3\) and \(dx = 3u^2 \, du\). Then replace \(x\) and \(dx\) in the integral with \(u\) terms: \[ L = \int_0^1 \sqrt{1 + \frac{16}{9}u^2} \cdot 3u^2 \, du. \]
05

Evaluating the Integral

Now evaluate the integral using a computer algebra system:\[ L = 3 \int_0^1 u^2 \sqrt{1 + \frac{16}{9}u^2} \, du. \] After evaluating, the integral gives us the arc length as \(L = \frac{5}{3}.\)
06

Conclusion - Arc Length Result

The exact length of the arc for \(y = x^{4/3}\), from \(0,0\) to \(1,1\), is \(\frac{5}{3}\).

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Key Concepts

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

Integral Evaluation
Integral evaluation is the process of calculating the value of an integral, whether it be a definite or an indefinite integral. In the context of arc length calculations, this involves applying a specific formula to determine the length of a curve between two points. The integral formula for arc length is:\[ L = \int_a^b \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx \]In the given exercise, we evaluated the arc length from point \((0,0)\) to point \((1,1)\) of the curve \(y = x^{4/3}\). This requires setting up the integral within the boundaries of these points.Using an integral evaluation, we set our limits of integration from 0 to 1. This allowed us to compute the arc length by evaluating the integral, which represents the accumulated tiny lengths of segments along the curve.

To proceed, simplifying the integral is sometimes necessary, which can be done by substituting complex expressions. This brings us to the substitution method, which we explore next.
Substitution Method
The substitution method is a powerful technique used to simplify the evaluation of integrals. It involves changing variables to transform a complex integral into a simpler one. In our exercise, we encountered a formidable integral while calculating the arc length:\[ L = \int_0^1 \sqrt{1 + \frac{16}{9}x^{2/3}} \, dx \]To ease the evaluation, we used substitution. We set:- \( u = x^{1/3} \)
- Thus, \( x = u^3 \) and consequently, \( dx = 3u^2 \, du \)With this substitution, the integral becomes:\[ L = \int_0^1 \sqrt{1 + \frac{16}{9}u^2} \cdot 3u^2 \, du \]Substituting in terms of \( u \) simplified the integral by changing the form of the expression under the square root, enabling easier computation with a computer algebra system.

By transforming variables, the substitution method can effectively streamline and make tractable even the most complicated integrals.
Derivative of Function
The derivative of a function is a fundamental concept that measures how a function changes as its input changes. It is imperative in the calculation of arc lengths, as it is integral to the arc length formula.To find the derivative of the function \( y = x^{4/3} \), we use the power rule of differentiation. The power rule states that:\[ \frac{d}{dx}(x^n) = nx^{n-1} \]Applying this rule, the derivative \( \frac{dy}{dx} \) of \( y = x^{4/3} \) can be calculated as:\[ \frac{dy}{dx} = \frac{4}{3}x^{1/3} \]This derivative is then substituted into the arc length formula to determine how the curve stretches between the two points \((0,0)\) and \((1,1)\). The square of this derivative informs us about the instantaneous rate of change in the direction of the curve, directly affecting the total length measurement.Understanding derivatives is crucial, not only for solving problems involving arc length but also for exploring myriad calculus applications like optimization and studying motion dynamics.

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Most popular questions from this chapter

(a) A model for the shape of a bird's egg is obtained by rotating about the \(x\) -axis the region under the graph of $$ f(x)=\left(a x^{3}+b x^{2}+c x+d\right) \sqrt{1-x^{2}} $$ Use a CAS to find the volume of such an egg. (b) For a Red-throated Loon, \(a=-0.06, b=0.04\) \(c=0.1,\) and \(d=0.54 .\) Graph \(f\) and find the volume of an egg of this bird.

When gas expands in a cylinder with radius \(r,\) the pressure at any given time is a function of the volume: \(P=P(V)\) . The force exerted by the gas on the piston (see the figure) is the product of the pressure and the area: \(F=\pi r^{2} P .\) Show that the work done by the gas when the volume expands from volume \(V_{1}\) to volume \(V_{2}\) is $$W=\int_{V_{1}}^{V_{2}} P d V$$

\(29-32=\) Sketch the direction field of the differential equation. Then use it to sketch a solution curve that passes through the given point. $$y^{\prime}=x-x y, \quad(1,0)$$

A wedge is cut out of a circular cylinder of radius 4 by two planes. One plane is perpendicular to the axis of the cylinder. The other intersects the first at an angle of \(30^{\circ}\) along a diameter of the cylinder. Find the volume of the wedge.

(a) Newton's Law of Gravitation states that two bodies with masses \(m_{1}\) and \(m_{2}\) attract each other with a force $$F=G \frac{m_{1} m_{2}}{r^{2}}$$ where \(r\) is the distance between the bodies and \(G\) is the gravitational constant. If one of the bodies is fixed, find the work needed to move the other from \(r=a\) to \(r=b\). (b) Compute the work required to launch a 1000-kg satellite vertically to a height of 1000 \(\mathrm{km} .\) You may assume that the earth's mass is \(5.98 \times 10^{24} \mathrm{kg}\) and is concentrated at its center. Take the radius of the earth to be \(6.37 \times 10^{6} \mathrm{m}\) and \(G=6.67 \times 10^{-11} \mathrm{N} \cdot \mathrm{m}^{2} / \mathrm{kg}^{2}\).

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