Chapter 8: Problem 22
Find the length of the arc of the parabola \(y=x^{2}\) that lies between \((-1,1)\) and \((1,1)\).
Short Answer
Expert verified
The arc length is approximately 2.9578.
Step by step solution
01
Understand the Arc Length Formula
The formula for the arc length of a curve given by the equation \(y=f(x)\) from \(x=a\) to \(x=b\) is given by \[L = \int_{a}^{b} \sqrt{1 + \left(\frac{dy}{dx}\right)^{2}}\, dx.\]In this problem, \(f(x) = x^2\), and we are asked to find the arc length from \(x = -1\) to \(x = 1\).
02
Differentiate the Function
To use the arc length formula, we first need to compute \(\frac{dy}{dx}\). Given \(y = x^2\), the derivative is \(\frac{dy}{dx} = 2x\).
03
Integrate to Find the Arc Length
Substitute \(\frac{dy}{dx} = 2x\) into the arc length formula:\[L = \int_{-1}^{1} \sqrt{1 + (2x)^2} \, dx = \int_{-1}^{1} \sqrt{1 + 4x^2} \, dx.\]This is the integral we need to evaluate to find the arc length.
04
Solve the Integral
The integral \(\int \sqrt{1 + 4x^2} \, dx\) is a standard integral that can be solved using the trigonometric substitution \(x = \frac{1}{2}\tan(\theta)\), leading to:\[L = \frac{1}{2}\int \sec^3(\theta) \, d\theta.\]After evaluating, converting back to terms of \(x\), and computing the definite integral from \(x = -1\) to \(x = 1\), approximately, the arc length is \(2.9578\).
05
Final Answer
After evaluating the integral, the calculated arc length between ((-1,1)) and ((1,1)) on the parabola \(y=x^2\) is approximately \(2.9578\).
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Parabola
A parabola is a unique type of curve on a graph that is symmetric and U-shaped. The simplest form of a parabola is given by the equation \(y = x^2\). This curve opens upwards and its vertex, or lowest point, is at the origin (0,0).
- Parabolas can open upwards or downwards depending on the coefficient of \(x^2\). If positive, like in \(y = x^2\), the parabola opens up. If negative, it opens down.
- The axis of symmetry for a parabola defined by \(y = ax^2\) is the line \(x = 0\).
- The range of a parabola opening upwards is \([0, \infty)\).
Trigonometric Substitution
Trigonometric substitution is a technique used in calculus to simplify integrals that involve square roots. It involves substituting a trigonometric function for a variable to eliminate square roots.
- This method is particularly helpful in solving integrals of the form \(\int \sqrt{a^2 + b^2x^2} \, dx\).
- Common substitutions include \(x = a \sin(\theta)\), \(x = a \tan(\theta)\), or \(x = a \sec(\theta)\), depending on the integral.
Definite Integral
A definite integral calculates the exact area under a curve between two points along the x-axis. Unlike indefinite integrals, which yield a general formula, definite integrals provide a numeric value that represents this "area."
- The notation \(\int_{a}^{b} f(x) \, dx\) indicates integration of function \(f(x)\) from \(x = a\) to \(x = b\).
- For solving, compute an antiderivative of \(f(x)\) and evaluate it from \(x = a\) to \(x = b\).
Derivative Calculation
Derivatives measure the rate at which a function changes. They are fundamental in calculus because they provide a way to understand the shape of graphs and solve various real-world problems.
- The derivative of a function \(f(x)\) is denoted as \(f'(x)\) or \(\frac{df}{dx}\).
- For the function \(y = x^2\), the derivative is computed as \(\frac{dy}{dx} = 2x\).
- This derivative tells us the slope of the tangent to the curve at any point \(x\).