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In Exercises \(13-20,\) do the following. \begin{equation}\begin{array}{l}{\text { a. Set up an integral for the length of the curve. }} \\ {\text { b. Graph the curve to see what it looks like. }} \\\ {\text { c. Use your grapher's or computer's integral evaluator to find }} \\\ {\text { the curve's length numerically. }}\end{array}\end{equation} $$y=x^{2},-1 \leq x \leq 2$$

Short Answer

Expert verified
Approximate the length of the curve on \(-1 \leq x \leq 2\) as roughly the value found from the integral \(\int_{-1}^{2} \sqrt{1 + 4x^2} \, dx \).

Step by step solution

01

Set up the Integral for Curve Length

To find the length of a curve defined by the function \( y = f(x) \) from \( x = a \) to \( x = b \), use the formula: \[ L = \int_{a}^{b} \sqrt{1 + \left( \frac{dy}{dx} \right)^2} \, dx \].For \( y = x^2 \), the derivative is \( \frac{dy}{dx} = 2x \). Substitute into the formula:\[ L = \int_{-1}^{2} \sqrt{1 + (2x)^2} \, dx = \int_{-1}^{2} \sqrt{1 + 4x^2} \, dx \].
02

Graph the Curve

Graph the function \( y = x^2 \) over the interval \(-1 \leq x \leq 2\). The graph is a parabola opening upwards with its vertex at the origin and symmetrical about the y-axis from \( x = -1 \) to \( x = 2 \).
03

Evaluate the Integral Numerically

Use a graphing calculator or a computer program capable of numerical integration (like a spreadsheet or math software) to approximate the value of the integral \( \int_{-1}^{2} \sqrt{1 + 4x^2} \, dx \). The numerical approximation gives the length of the curve. Ensure the settings on your calculator or software are set to an appropriate level of precision (e.g., at least four decimal places).

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Key Concepts

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

Definite Integral
A definite integral is a core concept in calculus that represents the accumulation of quantities, such as areas under curves or, as in this exercise, the length of a curve. We denote a definite integral as \( \int_{a}^{b} f(x) \, dx \), where \( a \) and \( b \) are the limits of integration, and \( f(x) \) is the function being integrated.

In the context of finding the length of a curve like \( y = x^2 \), the process involves setting up a definite integral that incorporates the derivative of the function. This method ensures that we're accurately calculating the whole of the curve's length over the specified interval \(-1 \leq x \le 2\).

To break it down:
  • The formula for the curve's length is \( L = \int_{a}^{b} \sqrt{1 + \left( \frac{dy}{dx} \right)^2} \, dx \).
  • For this exercise: \( a = -1 \) and \( b = 2 \).
  • This integral captures the continuously changing slope along the curve, contributed by the derivative inside the square root.
By using this specific form of the definite integral, we solidify the relationship between integration and geometry to measure lengths beyond straight lines.
Numerical Integration
Numerical integration is a technique used to find an approximate solution to definite integrals, especially when the integral is too complex for analytical methods. This approach is practical when calculating the integral of the function \( \int_{-1}^{2} \sqrt{1 + 4x^2} \, dx \), as seen in the exercise.

Since not all integrals can be solved symbolically, numerical methods such as the trapezoidal rule or Simpson's rule offer solutions. These methods subdivide the interval into smaller sections and compute the area under the curve, iteratively improving the estimate.

For practical implementation:
  • Use a graphing calculator or software that supports numerical integration.
  • Set the device to a high precision level to capture finer details of the curve.
  • The result grants an approximate length of the curve, useful when precise symbolic calculation is elusive.
Thus, numerical integration becomes an essential tool for mathematicians and engineers, providing concrete estimates when exact answers are impractical.
Derivative
The derivative represents the rate of change of a function with respect to its variable. It's particularly useful when finding the slope of a tangent line at any given point on a curve. In our exercise, the derivative helps us capture the slope as it changes along the parabola \( y = x^2 \).

Calculating the derivative:
  • The function \( y = x^2 \) has a derivative \( \frac{dy}{dx} = 2x \).
  • This derivative quantifies how steep the curve is at each point along the interval \(-1 \leq x \leq 2\).
  • Incorporating the derivative into the curve length formula provides the necessary component to measure the continuously varying slope.
Understanding derivatives is crucial for grasping how functions behave, especially when dealing with motion, optimization, and in this case, calculating curve lengths.
Parabola
A parabola is a symmetric, U-shaped curve. The standard form is \( y = x^2 \), which is the function we're examining in this exercise. Parabolas are fundamental in algebra and calculus due to their predictable properties and elegant symmetry.

Visual characteristics of the parabola \( y = x^2 \) over the interval \(-1 \leq x \leq 2\):
  • The vertex is at the origin (0,0), signifying the lowest point.
  • The parabola opens upwards, extending indefinitely.
  • It is symmetrical around the y-axis, a property that simplifies various calculations.
When we graph \( y = x^2 \) for \(-1 \leq x \leq 2\), we note this segment remains symmetrical and continuous, which aids in visualizing and setting the stage for calculating the curve's length accurately.

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Most popular questions from this chapter

Designing a wok You are designing a wok frying pan that will be shaped like a spherical bowl with handles. A bit of experimentation at home persuades you that you can get one that holds about 3L if you make it 9 \(\mathrm{cm}\) deep and give the sphere a radius of 16 \(\mathrm{cm}\) . To be sure, you picture the wok as a solid of revolution, as shown here, and calculate its volume with an integral. To the nearest cubic centimeter, what volume do you really get? (1 \(=1000 \mathrm{cm}^{3}\) )

In Exercises \(18,\) find the center of mass of a thin plate of constant density \(\delta\) covering the given region. The region bounded by the parabola \(x=y^{2}-y\) and the line \(y=x\)

The region in the first quadrant that is bounded above by the curve \(y=1 / \sqrt{x},\) on the left by the line \(x=1 / 4,\) and below by the line \(y=1\) is revolved about the \(y\) -axis to generate a solid. Find the volume of the solid by $$\text{a. the washer method}. \quad \text{b. the shell method}.$$

Find the volume of the torus generated by revolving the circle \((x-2)^{2}+y^{2}=1\) about the \(y\) -axis.

The region bounded by the curves \(y=\pm 4 / \sqrt{x}\) and the lines \(x=1\) and \(x=4\) is revolved about the \(y\) -axis to generate a solid. \begin{equation}\begin{array}{l}{\text { a. Find the volume of the solid. }} \\\ {\text { b. Find the center of mass of a thin plate covering the region if }} \\ {\text { the plate's density at the point }(x, y) \text { is } \delta(x)=1 / x \text { . }} \\ {\text { c. Sketch the plate and show the center of mass in your sketch. }}\end{array}\end{equation}

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