Ch. 6 - Applications of Integration

Briggs - Calculus: Early Transcendentals 3rd Edition

Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook

Briggs 3rd Edition

Ch. 6 - Applications of Integration

Problem 6.5.8

Chapter 6, Problem 6.5.8

Find the arc length of the line y = 4−3x on [−3, 2] using calculus and verify your answer using geometry.

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Recall the formula for the arc length of a curve defined by a function \(y = f(x)\) on the interval \([a, b]\): \[L = \int_{a}^{b} \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx\]

Identify the function and interval: here, \(y = 4 - 3x\) and the interval is \([-3, 2]\).

Compute the derivative \(\frac{dy}{dx}\): since \(y = 4 - 3x\), then \[\frac{dy}{dx} = -3\]

Substitute the derivative into the arc length formula: \[L = \int_{-3}^{2} \sqrt{1 + (-3)^2} \, dx = \int_{-3}^{2} \sqrt{1 + 9} \, dx = \int_{-3}^{2} \sqrt{10} \, dx\]

Evaluate the integral: since \(\sqrt{10}\) is constant, the integral becomes \[L = \sqrt{10} \times (2 - (-3)) = \sqrt{10} \times 5\]. To verify using geometry, recognize that the graph is a straight line segment between points \((-3, 4 - 3(-3))\) and \((2, 4 - 3(2))\). Calculate the distance between these points using the distance formula \[d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\] and confirm it matches the arc length.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Arc Length Formula

The arc length of a curve y = f(x) from x = a to x = b is found using the integral formula L = ∫_a^b √(1 + (dy/dx)^2) dx. This formula calculates the length of the curve by summing infinitesimal line segments along the curve.

Derivative of a Function

The derivative dy/dx represents the slope of the function y = f(x) at any point x. For the arc length formula, the derivative is squared and added to 1 inside the square root to account for both horizontal and vertical changes along the curve.

Geometric Interpretation of a Line Segment

Since y = 4 - 3x is a straight line, its arc length over [−3, 2] can be found using the distance formula between two points: √((x2 - x1)^2 + (y2 - y1)^2). This provides a geometric verification of the calculus result.

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A solid has a circular base; cross sections perpendicular to the base are squares. What method should be used to find the volume of the solid?

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Let R be the region bounded by the following curves. Find the volume of the solid generated when R is revolved about the given axis.

y=sin xon [0,π] and y=0 ; about the x-axis (Hint: Recall that sin^2 x=1 − cos2x / 2.

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Textbook Question

Let R be the region bounded by the following curves. Find the volume of the solid generated when R is revolved about the given axis.

y=x,y=x+2,x=0, and x=4 ; about the x-axis

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Find the area of the region described in the following exercises.

The region bounded by y=4x+4, y=6x+6, and x=4

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Textbook Question

9-34. Shell method Let R be the region bounded by the following curves. Use the shell method to find the volume of the solid generated when R is revolved about indicated axis.

{Use of Tech} y = 1 / (x² + 1)²,y=0,x=1, and x=2; about the y-axis

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Let R be the region bounded by the following curves. Find the volume of the solid generated when R is revolved about the given line.

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