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Ch.12 - Parametric and Polar Curves
Briggs - Calculus: Early Transcendentals 3rd Edition
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Tutti i libri di testoBriggs 3rd EditionCh.12 - Parametric and Polar CurvesProblema 12.4.71
Capitolo 12, Problema 12.4.71

69–72. Tangent lines Find an equation of the line tangent to the following curves at the given point.
y² - x²/64 = 1; (6, -5/4)

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Recognize that the given curve is an implicit function defined by the equation \(y^{2} - \frac{x^{2}}{64} = 1\). Since \(y\) is not explicitly solved for \(x\), we will use implicit differentiation to find \(\frac{dy}{dx}\), the slope of the tangent line.
Differentiate both sides of the equation with respect to \(x\). Remember to apply the chain rule when differentiating \(y^{2}\), treating \(y\) as a function of \(x\): \(\frac{d}{dx}\left(y^{2}\right) - \frac{d}{dx}\left(\frac{x^{2}}{64}\right) = \frac{d}{dx}(1)\) which becomes \(2y \cdot \frac{dy}{dx} - \frac{2x}{64} = 0\).
Solve the resulting equation for \(\frac{dy}{dx}\) to express the slope of the tangent line in terms of \(x\) and \(y\): \(2y \cdot \frac{dy}{dx} = \frac{2x}{64}\) so \(\frac{dy}{dx} = \frac{2x}{64 \cdot 2y} = \frac{x}{64y}\).
Substitute the given point \((6, -\frac{5}{4})\) into the expression for \(\frac{dy}{dx}\) to find the slope of the tangent line at that point: \(m = \frac{6}{64 \cdot (-\frac{5}{4})}\).
Use the point-slope form of the equation of a line to write the equation of the tangent line: \(y - y_1 = m(x - x_1)\), where \((x_1, y_1) = (6, -\frac{5}{4})\) and \(m\) is the slope found in the previous step.

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Implicit Differentiation

Implicit differentiation is used when a function is given in an implicit form, such as an equation involving both x and y. Instead of solving for y explicitly, we differentiate both sides with respect to x, treating y as a function of x, and apply the chain rule to find dy/dx.
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05:14
Finding The Implicit Derivative

Equation of a Tangent Line

The equation of a tangent line at a point on a curve is given by y - y₁ = m(x - x₁), where m is the slope of the tangent line at that point. Once the derivative dy/dx is found, substituting the point's coordinates gives the slope needed to write the tangent line equation.
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Percorso guidato
05:14
Equations of Tangent Lines

Hyperbola and Its Properties

The given curve y² - x²/64 = 1 represents a hyperbola. Understanding its shape and properties helps in visualizing the curve and verifying the correctness of the tangent line. The hyperbola's implicit form requires implicit differentiation to find slopes at specific points.
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06:21
Properties of Functions
Pratica correlata
Domanda del libro di testo

37–52. Curves to parametric equations Find parametric equations for the following curves. Include an interval for the parameter values. Answers are not unique.


The lower half of the circle centered at (−2, 2) with radius 6, oriented in the counterclockwise direction

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31–38. Equations of parabolas Find an equation of the following parabolas. Unless otherwise specified, assume the vertex is at the origin.

A parabola symmetric about the y-axis that passes through the point (2, -6)

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25–30. Converting coordinates Express the following polar coordinates in Cartesian coordinates.


(1, 2π/3)

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45–60. Areas of regions Find the area of the following regions.


The region inside the limaçon r = 4 - 2 cos θ

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31–36. Eliminating the parameter Eliminate the parameter to express the following parametric equations as a single equation in x and y.


x=tan t, y=sec ² t−1 

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25–28. Horizontal and vertical tangents Find the points at which the following polar curves have horizontal or vertical tangent lines.

r = 4 cos θ

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