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Find by implicit differentiation.
x²y² = 1
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Ch. 3 - Derivatives
Hass15th EditionThomas' CalculusISBN: 9780137616077
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TOC.tableOfContents
- Ch. 1 - Functions155
- Ch. 2 - Limits and Continuity240
- Ch. 3 - Derivatives337
- Ch. 4 - Applications of Derivatives293
- Ch. 5 - Integrals41
- Ch. 6 - Applications of Definite Integrals25
- Ch. 7 - Transcendental Functions490
- Ch. 8 - Techniques of Integration399
- Ch. 9 - First-Order Differential Equations66
- Ch. 10 - Infinite Sequences and Series132
- Ch. 11 - Parametric Equations and Polar Coordinates58
Capitolo 3, Problema 42a
Slopes and Tangent Lines
a. Horizontal tangent lines Find equations for the horizontal tangent lines to the curve y = x³ − 3x − 2. Also find equations for the lines that are perpendicular to these tangent lines at the points of tangency.
Guida verificata passo dopo passo1
First, find the derivative of the function y = x³ − 3x − 2 to determine the slope of the tangent line. The derivative, y', represents the slope of the tangent line at any point x. Use the power rule to differentiate: y' = d/dx(x³) - d/dx(3x) - d/dx(2).
The power rule states that d/dx(x^n) = n*x^(n-1). Applying this rule, the derivative of x³ is 3x², the derivative of 3x is 3, and the derivative of a constant, -2, is 0. Therefore, y' = 3x² - 3.
To find horizontal tangent lines, set the derivative equal to zero because horizontal lines have a slope of 0. Solve the equation 3x² - 3 = 0 for x.
Once you have the x-values where the slope is zero, substitute these x-values back into the original equation y = x³ − 3x − 2 to find the corresponding y-values. These points (x, y) are where the horizontal tangent lines occur.
To find the equations of lines perpendicular to these tangent lines at the points of tangency, recall that perpendicular lines have slopes that are negative reciprocals. Since the slope of the horizontal tangent line is 0, the perpendicular slope is undefined, indicating vertical lines. The equations of these lines are x = [x-value] for each point of tangency.
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Derivative
The derivative of a function at a point measures the rate at which the function's value changes as its input changes. It is represented as f'(x) and is crucial for finding slopes of tangent lines. For horizontal tangent lines, we set the derivative equal to zero, indicating that the slope at that point is flat.
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05:44
Derivatives
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. For horizontal tangents, the equation of the line can be expressed as y = k, where k is the y-coordinate of the point of tangency.
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05:13Slopes of Tangent Lines
Perpendicular Lines
Two lines are perpendicular if the product of their slopes is -1. If a tangent line has a slope of 0 (horizontal), the slope of the line perpendicular to it is undefined, indicating a vertical line. The equation of a vertical line can be expressed as x = a, where a is the x-coordinate of the point of tangency.
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05:13Slopes of Tangent Lines
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Find by implicit differentiation.
x² + xy + y² - 5x = 2
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