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Ch. 3 - Derivatives
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
All textbooksBriggs 3rd EditionCh. 3 - DerivativesProblem 3.8.62b
Chapter 3, Problem 3.8.62b

60–62. {Use of Tech} Multiple tangent lines Complete the following steps. <IMAGE>
b. Graph the tangent lines on the given graph.
4x³ =y²(4−x); x=2 (cissoid of Diocles)

Verified step by step guidance
1
Identify the given equation of the curve: \(4x^3 = y^2(4-x)\). This is known as the cissoid of Diocles.
To find the tangent line at a specific point, we need to determine the derivative of the curve with respect to \(x\). Start by differentiating both sides of the equation implicitly with respect to \(x\).
Apply implicit differentiation: Differentiate \(4x^3\) to get \(12x^2\) and differentiate \(y^2(4-x)\) using the product rule, which gives \(2y \frac{dy}{dx} (4-x) - y^2\).
Set the derivatives equal: \(12x^2 = 2y \frac{dy}{dx} (4-x) - y^2\). Solve for \(\frac{dy}{dx}\) to find the slope of the tangent line.
Substitute \(x = 2\) into the derivative to find the slope at this point. Then, use the point-slope form of a line, \(y - y_1 = m(x - x_1)\), where \(m\) is the slope and \((x_1, y_1)\) is the point on the curve, to write the equation of the tangent line.

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Key Concepts

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

Tangent Lines

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. Understanding how to calculate and graph tangent lines is essential for analyzing the behavior of functions.
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Slopes of Tangent Lines

Derivatives

The derivative of a function measures how the function's output value changes as its input value changes. It is a fundamental concept in calculus that provides the slope of the tangent line at any point on the curve. To find the derivative, various rules such as the power rule, product rule, and chain rule can be applied, depending on the function's form.
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Derivatives

Graphing Techniques

Graphing techniques involve plotting points, lines, and curves on a coordinate plane to visually represent mathematical functions and their properties. When graphing tangent lines, it is important to accurately determine the point of tangency and the slope derived from the derivative. This visual representation aids in understanding the relationship between the function and its tangent lines.
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Graphing The Derivative
Related Practice
Textbook Question

Vertical tangent lines

b. Does the curve have any horizontal tangent lines? Explain.

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Textbook Question

A race Jean and Juan run a one-lap race on a circular track. Their angular positions on the track during the race are given by the functions θ(t) and ϕ(t), respectively, where 0≤t≤4 and t is measured in minutes (see figure). These angles are measured in radians, where θ=ϕ=0 represent the starting position and θ=ϕ=2π represent the finish position. The angular velocities of the runners are θ′(t) and ϕ′(t). <IMAGE>

b. Which runner has the greater average angular velocity?

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Textbook Question

Witch of Agnesi Let y(x²+4)=8 (see figure). <IMAGE>

b. Find equations of all lines tangent to the curve y(x²+4)=8 when y=1.

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Textbook Question

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)

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Textbook Question

City urbanization City planners model the size of their city using the function A(t) = - 1/50t² + 2t +20, for 0 ≤ t ≤ 50, where A is measured in square miles and t is the number of years after 2010.

b. How fast will the city be growing when it reaches a size of 38 mi²?

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Textbook Question

Derivatives and tangent lines

b. Determine an equation of the line tangent to the graph of f at the point (a,f(a)) for the given value of a.

f(x) = 1/3x-1; a= 2

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