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Ch.12 - Parametric and Polar Curves
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
All textbooksBriggs 3rd EditionCh.12 - Parametric and Polar CurvesProblem 12.4.69
Chapter 12, Problem 12.4.69

69–72. Tangent lines Find an equation of the line tangent to the following curves at the given point.
x² = -6y; (-6, -6)

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1
Rewrite the given curve equation \(x^{2} = -6y\) to express \(y\) explicitly in terms of \(x\). This will help in differentiating \(y\) with respect to \(x\).
Differentiate both sides of the equation \(x^{2} = -6y\) implicitly with respect to \(x\). Remember to apply the chain rule when differentiating \(y\) since \(y\) is a function of \(x\).
After differentiating, solve for \(\frac{dy}{dx}\), which represents the slope of the tangent line at any point on the curve.
Substitute the given point \((-6, -6)\) into the expression for \(\frac{dy}{dx}\) to find the slope of the tangent line at that specific point.
Use the point-slope form of a line, \(y - y_1 = m(x - x_1)\), where \(m\) is the slope found in the previous step and \((x_1, y_1)\) is the given point, 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.

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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Finding The Implicit Derivative

Slope of the Tangent Line

The slope of the tangent line at a given point on a curve is the value of the derivative dy/dx evaluated at that point. It represents the instantaneous rate of change of y with respect to x and determines the steepness and direction of the tangent line.
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Slopes of Tangent Lines

Equation of a Tangent Line

Once the slope of the tangent line and the point of tangency are known, the equation of the tangent line can be found using the point-slope form: y - y₁ = m(x - x₁), where m is the slope and (x₁, y₁) is the point on the curve.
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Equations of Tangent Lines
Related Practice
Textbook Question

73–76. Tangent lines Find an equation of the line tangent to the curve at the point corresponding to the given value of t.


x=cos t+t sin t,y=sin t−t cos t; t=π/4

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Textbook Question

39–50. Equations of ellipses and hyperbolas Find an equation of the following ellipses and hyperbolas, assuming the center is at the origin. 

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Textbook Question

77–80. Slopes of tangent lines Find all points at which the following curves have the given slope.


x = 4 cos t, y = 4 sin t; slope = 1/2

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Textbook Question

Multiple descriptions Which of the following parametric equations describe the same curve?

a. x = 2t², y = 4 + t; -4 ≤ t ≤ 4

b. x = 2t⁴, y = 4 + t²; -2 ≤ t ≤ 2

c. x = 2t^(2/3), y = 4 + t^(1/3); -64 ≤ t ≤ 64

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Textbook Question

15–30. Working with parametric equations Consider the following parametric equations.

a. Eliminate the parameter to obtain an equation in x and y.

b. Describe the curve and indicate the positive orientation.


x = cos t, y = 1 + sin t; 0 ≤ t ≤ 2π

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Textbook Question

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 segment of the parabola y=2x ²−4, where −1≤x≤5

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