Textbook Question
13-26 Implicit differentiation Carry out the following steps.
a. Use implicit differentiation to find dy/dx.
³√x+³√y⁴ = 2;(1,1)
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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 3.8.80a
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.)
x⁴ = 2x²+2y²; (x0, y0)=(2, 2) (kampyle of Eudoxus)
Verified step by step guidance1
First, understand that the problem involves finding the equations of the tangent and normal lines to the curve at a specific point. The curve given is x⁴ = 2x² + 2y², and the point is (x₀, y₀) = (2, 2).
To find the equation of the tangent line, we need to calculate the derivative of the curve with respect to x, which involves implicit differentiation since y is also a function of x. Start by differentiating both sides of the equation x⁴ = 2x² + 2y² with respect to x.
Apply implicit differentiation: For the left side, differentiate x⁴ to get 4x³. For the right side, differentiate 2x² to get 4x, and differentiate 2y² using the chain rule to get 4y(dy/dx). Set the derivatives equal: 4x³ = 4x + 4y(dy/dx).
Solve for dy/dx, which represents the slope of the tangent line at the point (x₀, y₀). Substitute x = 2 and y = 2 into the differentiated equation to find the specific slope at the given point.
Once you have the slope of the tangent line, use the point-slope form of a line equation, y - y₀ = m(x - x₀), where m is the slope you found, to write the equation of the tangent line. For the normal line, use the negative reciprocal of the tangent slope as the slope of the normal line and apply the point-slope form again.
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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. The slope of the tangent line is equal to the derivative of the function at that point, which represents the instantaneous rate of change of the function. To find the equation of the tangent line, one typically uses the point-slope form of a line, incorporating the slope derived from the derivative.
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Slopes of Tangent Lines
Normal Line
The normal line at a point on a curve is a line that is perpendicular to the tangent line at that point. Its slope is the negative reciprocal of the slope of the tangent line. To find the equation of the normal line, one can use the point-slope form as well, substituting the point coordinates and the normal slope, which is derived from the tangent slope.
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05:13Slopes of Tangent Lines
Implicit Differentiation
Implicit differentiation is a technique used to differentiate equations that define y implicitly in terms of x, rather than explicitly as y = f(x). This method involves differentiating both sides of the equation with respect to x and applying the chain rule when differentiating terms involving y. It is particularly useful for finding derivatives of curves defined by equations like x⁴ = 2x² + 2y², where y cannot be easily isolated.
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Related Practice
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.
g(s) = 4s³ - 8s² +4s / 4s
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Textbook Question
Comparing velocities Two stones are thrown vertically upward, each with an initial velocity of 48 ft/s at time t=0. One stone is thrown from the edge of a bridge that is 32 feet above the ground, and the other stone is thrown from ground level. The height above the ground of the stone thrown from the bridge after t seconds is f(t) = − 16t²+48t+32. and the height of the stone thrown from the ground after t seconds is g(t) = −16t²+48t.
a. Show that the stones reach their high points at the same time.
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Textbook Question
Use definition (2) (p. 135) to find the slope of the line tangent to the graph of f at P.
f(x) = 1/x; P (1,1)
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Textbook Question
Vertical tangent lines If a function f is continuous at a and lim x→a| f′(x)|=∞, then the curve y=f(x) has a vertical tangent line at a, and the equation of the tangent line is x=a. If a is an endpoint of a domain, then the appropriate one-sided derivative (Exercises 71–72) is used. Use this information to answer the following questions.
73. {Use of Tech} Graph the following functions and determine the location of the vertical tangent lines.
a. f(x) = (x-2)^1/3
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Textbook Question
Deriving trigonometric identities
a. Differentiate both sides of the identity cos 2t = cos² t−sin² t to prove that sin 2 t= 2 sin t cos t.
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