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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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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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Related Practice
Textbook Question

81–88. Arc length Find the arc length of the following curves on the given interval.


x = 2t sin t - t² cos t, y = 2t cos t + t² sin t; 0 ≤ t ≤ π

Textbook Question

13–30. Graphing conic sections Determine whether the following equations describe a parabola, an ellipse, or a hyperbola, and then sketch a graph of the curve. For each parabola, specify the location of the focus and the equation of the directrix; for each ellipse, label the coordinates of the vertices and foci, and find the lengths of the major and minor axes; for each hyperbola, label the coordinates of the vertices and foci, and find the equations of the asymptotes.


4x² - y² = 16

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Textbook Question

41–44. Intersection points and area  Find all the intersection points of the following curves. Find the area of the entire region that lies within both curves


r = 1 + sin θ and r = 1 + cos θ

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Textbook Question

33–40. Areas of regions Make a sketch of the region and its bounding curves. Find the area of the region.


The region inside the circle r = 8 sin θ

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Textbook Question

15–22. Sets in polar coordinates Sketch the following sets of points.


1 < r < 2 and π/6 ≤ θ ≤ π/3

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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 line that passes through the points P(1, 1) and Q(3, 5), oriented in the direction of increasing x

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