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 3.8.80b

Chapter 3, Problem 3.8.80b

79–82. {Use of Tech} Visualizing tangent and normal lines
b. Graph the tangent and normal lines on the given graph.
x⁴ = 2x²+2y²; (x0, y0)=(2, 2) (kampyle of Eudoxus)

Verified step by step guidance

First, understand the problem: We need to find the equations of the tangent and normal lines to the curve given by \(x^4 = 2x^2 + 2y^2\) at the point \((x_0, y_0) = (2, 2)\).

To find the tangent line, we need the derivative of the curve. Start by implicitly differentiating the equation \(x^4 = 2x^2 + 2y^2\) with respect to \(x\). Use the chain rule for \(y^2\) since \(y\) is a function of \(x\).

After differentiating, solve for \(\frac{dy}{dx}\), which represents the slope of the tangent line at any point \((x, y)\) on the curve.

Substitute \((x_0, y_0) = (2, 2)\) into the derivative to find the slope of the tangent line at this specific point.

Use the point-slope form of a line, \(y - y_0 = m(x - x_0)\), where \(m\) is the slope found in the previous step, to write the equation of the tangent line. The normal line is perpendicular to the tangent line, so its slope is the negative reciprocal of the tangent line's slope. Use this to write 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. It represents the instantaneous rate of change of the function at that point, which can be found using the derivative. For the curve defined by the equation, the slope of the tangent line can be calculated by differentiating the equation implicitly.

Normal Line

The normal line at a point on a curve is perpendicular to the tangent line at that point. Its slope is the negative reciprocal of the slope of the tangent line. Understanding the normal line is essential for visualizing how the curve behaves at that point, and it can also be derived from the tangent line's slope.

Implicit Differentiation

Implicit differentiation is a technique used to differentiate equations where the dependent and independent variables are not isolated. In this case, the equation x⁴ = 2x² + 2y² involves both x and y. By differentiating both sides with respect to x, we can find dy/dx, which is necessary to determine the slopes of the tangent and normal lines.

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Related Practice

Textbook Question

21–30. Derivatives

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

f(x) = 4x²+1; a= 2,4

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Textbook Question

{Use of Tech} Hours of daylight The number of hours of daylight at any point on Earth fluctuates throughout the year. In the Northern Hemisphere, the shortest day is on the winter solstice and the longest day is on the summer solstice. At 40° north latitude, the length of a day is approximated by D(t) = 12−3 cos (2π(t+10) / 365), where D is measured in hours and 0≤t≤365 is measured in days, with t=0 corresponding to January 1.

b. Find the rate at which the daylight function changes.

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Textbook Question

Use a graphing utility to plot the curve and the tangent line.

y = cos x / 1−cos x; x = π/3

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Textbook Question

Surface area of a cone The lateral surface area of a cone of radius r and height h (the surface area excluding the base) is A = πr√r²+h².

b. Evaluate this derivative when r=30 and h=40.

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