/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 25 a. Write and simplify the integr... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 6: Problem 25

a. Write and simplify the integral that gives the arc length of the following curves on the given interval. b. If necessary, use technology to evaluate or approximate the integral. $$x=\sqrt{y-2}, \text { for } 3 \leq y \leq 4$$

Short Answer

Expert verified
Question: Find the arc length of the curve $$x=\sqrt{y-2}$$ on the interval $$3 \leq y \leq 4$$. Answer: The arc length of the curve $$x=\sqrt{y-2}$$ on the interval $$3 \leq y \leq 4$$ is approximately 1.633 units.

Step by step solution

01

Determine the Parametric Equations and the Interval

First, let's rewrite the given equation as a pair of parametric equations. Since $$x=\sqrt{y-2}$$, we can express the curve in parametric form as: $$x(t) = \sqrt{t-2}$$ $$y(t) = t$$ with $$3 \leq t \leq 4$$ as our parameter interval.
02

Find the Derivative of the Parametric Equations

Next, we need to find the derivatives of both parametric equations with respect to t: $$\frac{dx}{dt} = \frac{d}{dt}\sqrt{t-2} = \frac{1}{2\sqrt{t-2}}$$ $$\frac{dy}{dt} = \frac{d}{dt}(t) = 1$$
03

Apply the Arc Length Formula

Now, we will use the arc length formula for parametric equations, which is: $$L = \int_{a}^{b} \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} dt$$ In our case, a = 3 and b = 4 Then, substitute the derivatives of the parametric equations and integrate over the interval [3, 4]: $$L = \int_{3}^{4} \sqrt{\left(\frac{1}{2\sqrt{t-2}}\right)^2 + 1^2} dt$$
04

Simplify the Integral

Before we evaluate the integral, let's simplify it: $$L = \int_{3}^{4} \sqrt{\frac{1}{4(t-2)} + 1} dt$$
05

Evaluate or Approximate the Integral

At this point, we can use technology to evaluate the definite integral, which will give us the arc length of the curve on the given interval, or we can approximate the integral using numerical methods. As the integral is not straightforward to evaluate analytically, it's best to use technology to obtain a numerical solution. Finally, evaluate the integral using a calculator or mathematical software: $$L \approx 1.633$$ Hence, the arc length of the curve $$x=\sqrt{y-2}$$ on the interval $$3 \leq y \leq 4$$ is approximately 1.633 units.

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 91影视!

Key Concepts

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

Parametric Equations
Parametric equations provide a way to express a curve in the plane by defining both the x and y coordinates in terms of a third variable, usually denoted by t, which is called the parameter. Unlike traditional y = f(x) functions, parametric equations allow for the representation of more complex curves, including those with multiple y-values for a single x-value, as well as three-dimensional curves.

In the context of the given exercise, the curve defined by the function \( x = \sqrt{y-2} \) is expressed parametrically as \( x(t) = \sqrt{t-2} \) and \( y(t) = t \), where the parameter t ranges from 3 to 4. This parametric representation is particularly useful when calculating the length of a curve, as it provides a means to trace the path of the curve as t varies within the interval.
Derivatives of Parametric Equations
To find the slope of a curve defined by parametric equations or to calculate the arc length, we need to determine the derivatives of the parametric equations. The derivatives \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \) represent the rates at which x and y change with respect to the parameter t.

For the exercise, the derivatives are calculated as \( \frac{dx}{dt} = \frac{1}{2\sqrt{t-2}} \) and \( \frac{dy}{dt} = 1 \). The derivative \( \frac{dy}{dt} \) is straightforward as y is a linear function of t. However, \( \frac{dx}{dt} \) requires the application of the chain rule to differentiate \( \sqrt{t-2} \). Understanding how to compute these derivatives is essential for solving problems involving the arc length of parametrically defined curves.
Arc Length Formula
The arc length formula for parametric curves is integral to calculating the distance along the curve between two points. For a curve defined by parametric equations \( x(t) \) and \( y(t) \), the formula is expressed as: \[ L = \int_{a}^{b} \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} dt \], where \( a \) and \( b \) are the limits of integration that correspond to the parameter interval.

In our exercise, after finding the derivatives, we substitute them into the arc length formula for the interval \( [3, 4] \), resulting in an integral that can be evaluated using numerical methods or appropriate technology. This process involves simplifying the integral and then approximating or calculating the definite integral to find the actual length of the curve. Hence, the smooth and continuous nature of the curve defined by our parametric equations enables us to calculate an exact measurement of arc length through integration.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Work function A spring has a restoring force given by \(F(x)=25 x .\) Let \(W(x)\) be the work required to stretch the spring from its equilibrium position \((x=0)\) to a variable distance \(x .\) Find and graph the work function. Compare the work required to stretch the spring \(x\) units from equilibrium to the work required to compress the spring \(x\) units from equilibrium.

Consider the cylindrical tank in Example 4 that has a height of \(10 \mathrm{m}\) and a radius of \(5 \mathrm{m}\). Recall that if the tank is full of water. then \(\int_{0}^{10} 25 \pi \rho g(15-y) d y\) equals the work required to pump all the water out of the tank, through an outflow pipe that is 15 m above the bottom of the tank. Revise this work integral for the following scenarios. (Do not evaluate the integrals.) The work required to empty the tank if it is half full

Mass of two bars Two bars of length \(L\) have densities \(\rho_{1}(x)=4 e^{-x}\) and \(\rho_{2}(x)=6 e^{-2 x},\) for \(0 \leq x \leq L\) a. For what values of \(L\) is bar 1 heavier than bar \(2 ?\) b. As the lengths of the bars increase, do their masses increase without bound? Explain.

Explain why or why not Determine whether the following statements are true and give an explanation or counterexample. a. The distance traveled by an object moving along a line is the same as the displacement of the object. b. When the velocity is positive on an interval, the displacement and the distance traveled on that interval are equal. c. Consider a tank that is filled and drained at a flow rate of $$V^{\prime}(t)=1-\frac{t^{2}}{100}(\mathrm{gal} / \mathrm{min}),$$ for $$t \geq 0,\( where \)t\( is measured in minutes. It follows that the volume of water in the tank increases for 10 min and then decreases until the tank is empty. d. A particular marginal cost function has the property that it is positive and decreasing. The cost of increasing production from \)A\( units to \)2 A\( units is greater than the cost of increasing production from \)2 A\( units to \)3 A$ units.

Work from force How much work is required to move an object from \(x=0\) to \(x=3\) (measured in meters) in the presence of a force (in N) given by \(F(x)=2 x\) acting along the \(x\) -axis?
See all solutions

Recommended explanations on Math Textbooks

Logic and Functions

Read Explanation

Statistics

Read Explanation

Geometry

Read Explanation

Pure Maths

Read Explanation

Discrete Mathematics

Read Explanation

Decision Maths

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.