Textbook Question
77–80. Slopes of tangent lines Find all points at which the following curves have the given slope.
x = 2 cos t, y = 8 sin t; slope = -1
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Ch.12 - Parametric and Polar Curves
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 12, Problem 12.1.80
77–80. Slopes of tangent lines Find all points at which the following curves have the given slope.
x = 2 + √t, y = 2 - 4t; slope = -8
Verified step by step guidance1
Identify the given parametric equations: \(x = 2 + \sqrt{t}\) and \(y = 2 - 4t\).
Recall that the slope of the tangent line to a parametric curve is given by \(\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}\).
Compute the derivatives with respect to \(t\): \(\frac{dx}{dt} = \frac{d}{dt}(2 + \sqrt{t})\) and \(\frac{dy}{dt} = \frac{d}{dt}(2 - 4t)\).
Set the slope equal to the given value: \(\frac{dy}{dx} = -8\), which means \(\frac{\frac{dy}{dt}}{\frac{dx}{dt}} = -8\).
Solve the resulting equation for \(t\), then substitute back into the original parametric equations to find the corresponding points \((x, y)\).
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Parametric Equations
Parametric equations express the coordinates of points on a curve as functions of a parameter, often denoted as t. Here, x and y are given in terms of t, allowing us to analyze the curve's behavior by studying these functions.
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Guided course
08:02
Parameterizing Equations
Derivative of Parametric Curves
The slope of the tangent line to a parametric curve is found by computing dy/dx = (dy/dt) / (dx/dt). This requires differentiating both x(t) and y(t) with respect to t and then dividing the results to find the instantaneous rate of change.
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06:49Differentiation of Parametric Curves
Finding Points with a Given Slope
To find points where the curve has a specific slope, set the derivative dy/dx equal to the given slope and solve for the parameter t. Substituting t back into x(t) and y(t) gives the coordinates of the points with that slope.
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05:45Understanding Slope Fields
Related Practice
Textbook Question
Write the equations that are used to express a point with polar coordinates (r, θ) in Cartesian coordinates.
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57–64. Graphing polar curves Graph the following equations. Use a graphing utility to check your work and produce a final graph.
r = sin 3θ
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33–40. Areas of regions Make a sketch of the region and its bounding curves. Find the area of the region.
The region inside one leaf of r = cos 3θ
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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.
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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 segment starting at P(0, 0) and ending at Q(2, 8)
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