Textbook Question
Vertical tangent lines
b. Does the curve have any horizontal tangent lines? Explain.
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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 3.8.62b
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 guidance1
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.
Recommended video:
05:44Derivatives
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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06:15Graphing The Derivative
Related Practice
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.
x³+y³=2xy; (1, 1)
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.
Textbook Question
Explain why or why not Determine whether the following statements are true and give an explanation or counterexample.
b. d/dx(tan^−1 x) =sec² x
Textbook Question
31–32. Velocity functions A projectile is fired vertically upward into the air, and its position (in feet) above the ground after t seconds is given by the function s(t).
b. Determine the instantaneous velocity of the projectile at t=1 and t = 2 seconds.
s(t)= −16t²+100t
Textbook Question
Explain why or why not. Determine whether the following statements are true and give an explanation or counterexample.
b. ln(x + 1) + ln(x − 1) = ln(x² − 1), for all x.
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