Ch. 3 - Derivatives

Briggs - Calculus: Early Transcendentals 3rd Edition

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

Briggs 3rd Edition

Ch. 3 - Derivatives

Problem 3.8.82b

Chapter 3, Problem 3.8.82b

79–82. {Use of Tech} Visualizing tangent and normal lines
b. Graph the tangent and normal lines on the given graph.
(x²+y²)² = 25/3 (x²-y²); (x0,y0) = (2,-1) (lemniscate of Bernoulli)

Verified step by step guidance

First, understand the given equation: \((x^2 + y^2)^2 = \frac{25}{3} (x^2 - y^2)\). This is the equation of a lemniscate of Bernoulli, a type of figure-eight curve.

To find the tangent line at the point \((x_0, y_0) = (2, -1)\), we need to differentiate the given equation implicitly with respect to \(x\). This involves using the chain rule and product rule.

After differentiating, substitute \(x = 2\) and \(y = -1\) into the derivative to find the slope of the tangent line at this point.

Use the point-slope form of a line, \(y - y_1 = m(x - x_1)\), where \(m\) is the slope found in the previous step, and \((x_1, y_1) = (2, -1)\), to write the equation of the tangent line.

The normal line is perpendicular to the tangent line. Therefore, its slope is the negative reciprocal of the tangent line's slope. Use the point-slope form again with this new slope to find the equation of the normal line.

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Key Concepts

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

Tangent Line

A tangent line to a curve at a given point is a straight line that touches the curve at that point without crossing it. The slope of the tangent line represents the instantaneous rate of change of the function at that point, which can be found using the derivative. For the given curve, the tangent line can be determined by calculating the derivative of the implicit function at the specified point (2, -1).

Normal Line

The normal line is perpendicular to the tangent line at a given point on a curve. Its slope is the negative reciprocal of the slope of the tangent line. To graph the normal line, one must first find the slope of the tangent line at the point of interest and then use this slope to determine the equation of the normal line, which will intersect the curve at the same point.

Implicit Differentiation

Implicit differentiation is a technique used to find the derivative of a function defined implicitly by an equation involving both x and y. In this case, the equation (x²+y²)² = 25/3 (x²-y²) defines y as a function of x. By differentiating both sides with respect to x and applying the chain rule, one can find dy/dx, which is essential for determining the slopes of the tangent and normal lines.

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Related Practice

Textbook Question

Derivatives and tangent lines

a. For the following functions and values of a, find f′(a).

f(x) = 8x; a = −3

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Textbook Question

Use definition (1) (p. 133) to find the slope of the line tangent to the graph of f at P.

f(x) = -3x² - 5x + 1; P(1,-7)

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Textbook Question

Use a graphing utility to graph the curve and the tangent line on the same set of axes.

y = (x + 5) / (x - 1); a = 3

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Textbook Question

{Use of Tech} Computing limits with angles in degrees Suppose your graphing calculator has two functions, one called sin x, which calculates the sine of x when x is in radians, and the other called s(x), which calculates the sine of x when x is in degrees.

b. Evaluate lim x→0 s(x) / x. Verify your answer by estimating the limit on your calculator.

318

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Textbook Question

21–30. Derivatives

b. Evaluate f'(a) for the given values of a.

f(t) = 3t⁴; a= -2, 2

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Textbook Question

45–50. Tangent lines Carry out the following steps. <IMAGE>

b. Determine an equation of the line tangent to the curve at the given point.