Textbook Question
Find an equation of the line tangent to the following curves at the given value of x.
y = 4 sin x cos x; x = π/3
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Ch. 3 - Derivatives
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243
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Table of contents
- Ch. 1 - Functions210
- Ch. 2 - Limits371
- Ch. 3 - Derivatives633
- Ch. 4 - Applications of the Derivative412
- Ch. 5 - Integration328
- Ch. 6 - Applications of Integration331
- Ch. 7 - Logarithmic, Exponential Functions, and Hyperbolic Functions177
- Ch. 8 - Integration Techniques394
- Ch. 9 - Differential Equations179
- Ch. 10 - Sequences and Infinite Series339
- Ch. 11 - Power Series218
- Ch.12 - Parametric and Polar Curves215
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
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 guidance1
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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05:14
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.
Recommended video:
05:13Slopes of Tangent Lines
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.
Recommended video:
04:50Critical Points
Related Practice
Textbook Question
Highway travel A state patrol station is located on a straight north-south freeway. A patrol car leaves the station at 9:00 A.M. heading north with position function s = f(t) that gives its location in miles t hours after 9:00 A.M. (see figure). Assume s is positive when the car is north of the patrol station. <IMAGE>
a. Determine the average velocity of the car during the first 45 minutes of the trip.
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Textbook Question
21–30. Derivatives
a. Use limits to find the derivative function f' for the following functions f.
f(x) = 4x²+1; a= 2,4
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Textbook Question
The line tangent to the curve y=h(x) at x=4 is y = −3x+14. Find an equation of the line tangent to the following curves at x=4.
y = (x²-3x)h(x)
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Textbook Question
Derivatives and tangent lines
a. For the following functions and values of a, find f′(a).
f(x) = 1/ √x; a= 1/4
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Textbook Question
79–82. {Use of Tech} Visualizing tangent and normal lines <IMAGE>
a. Determine an equation of the tangent line and the normal line at the given point (x0, y0) on the following curves. (See instructions for Exercises 73–78.)
(x²+y²)² = 25/3 (x²-y²); (x0,y0) = (2,-1) (lemniscate of Bernoulli)
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