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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 3.8.64.a
Chapter 3, Problem 3.8.64.a

Vertical tangent lines
a. Determine the points where the curve x+y³−y=1 has a vertical tangent line (see Exercise 60).

Verified step by step guidance
1
First, understand that a vertical tangent line occurs when the derivative of the curve with respect to x is undefined, which typically happens when the denominator of the derivative is zero.
To find the derivative, start by implicitly differentiating the given equation x + y³ - y = 1 with respect to x. Remember that y is a function of x, so use the chain rule for terms involving y.
Differentiate each term: the derivative of x with respect to x is 1, the derivative of y³ with respect to x is 3y²(dy/dx), and the derivative of -y with respect to x is -(dy/dx). Set the derivative equal to zero to find where the tangent is vertical.
Combine the differentiated terms to form the equation: 1 + 3y²(dy/dx) - (dy/dx) = 0. Rearrange this to solve for dy/dx: dy/dx = -1 / (3y² - 1).
Identify the points where the denominator 3y² - 1 equals zero, as these are the points where the derivative is undefined, indicating a vertical tangent. Solve 3y² - 1 = 0 to find the values of y, and substitute back into the original equation to find the corresponding x values.

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Key Concepts

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

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, we differentiate the equation x + y³ - y = 1 with respect to x, treating y as a function of x. This allows us to find dy/dx, which is essential for determining the slope of the tangent line at any point on the curve.
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Finding The Implicit Derivative

Vertical Tangent Lines

A vertical tangent line occurs when the slope of the tangent line approaches infinity, which mathematically corresponds to dy/dx being undefined. This situation typically arises when the denominator of the derivative expression equals zero while the numerator does not. Identifying points where this occurs is crucial for determining the locations of vertical tangents on the curve.
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Critical Points

Critical points are values of x where the derivative of a function is either zero or undefined. In the context of finding vertical tangents, we focus on points where dy/dx is undefined, as these indicate potential vertical tangents. Analyzing these points helps in understanding the behavior of the curve and identifying where the tangent lines are vertical.
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Related Practice
Textbook Question

{Use of Tech} A damped oscillator The displacement of a mass on a spring suspended from the ceiling is given by y=10e−t2cos(πt8)y=10e^{-\(\frac{t}{2}\)}\(\cos\[\left\)(\(\frac{\pi t}{8}\]\right\)).

a. Graph the displacement function. 

Textbook Question

Find an equation of the line tangent to the given curve at a.

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

Textbook Question

7–14. Find the derivative the following ways:

a. Using the Product Rule (Exercises 7–10) or the Quotient Rule (Exercises 11–14). Simplify your result.

f(w) = w³ -w / w

Textbook Question

Airline travel The following figure shows the position function of an airliner on an out-and-back trip from Seattle to Minneapolis, where s = f(t) is the number of ground miles from Seattle t hours after take-off at 6:00 A.M. The plane returns to Seattle 8.5 hours later at 2:30 P.M. <IMAGE>

a. Calculate the average velocity of the airliner during the first 1.5 hours of the trip (0 ≤ t ≤ 1.5).

Textbook Question

62–65. {Use of Tech} Graphing f and f'

a. Graph f with a graphing utility.

f(x) = (x−1) sin^−1 x on [−1,1]

Textbook Question

Use the graph of f in the figure to do the following. <IMAGE>

a. Find the values of x in (-2,2) at which f is not continuous.