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=t ²−1, y=t ³ +t; t=2
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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 12.1.78
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
Verified step by step guidance1
Recall that the slope of the tangent line to a parametric curve given by \(x = x(t)\) and \(y = y(t)\) is found by computing \(\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}\).
Differentiate both parametric equations with respect to \(t\): compute \(\frac{dx}{dt}\) for \(x = 2 \cos t\) and \(\frac{dy}{dt}\) for \(y = 8 \sin t\).
Write the expression for the slope \(\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}\) using the derivatives found in the previous step.
Set the slope equal to the given value \(-1\) and solve the resulting equation for \(t\).
Substitute the values of \(t\) found back into the parametric equations \(x = 2 \cos t\) and \(y = 8 \sin t\) to find the corresponding points on the curve.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Parametric Equations and Curves
Parametric equations express the coordinates of points on a curve as functions of a parameter, often denoted t. Understanding how x and y depend on t allows us to analyze the curve's behavior and find points corresponding to specific conditions, such as a given slope.
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Parameterizing Equations
Derivative of Parametric Curves (dy/dx)
For parametric curves, the slope of the tangent line dy/dx is found by dividing the derivative of y with respect to t by the derivative of x with respect to t, i.e., (dy/dt) / (dx/dt). This ratio gives the instantaneous rate of change of y with respect to x at any parameter value t.
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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 expression for dy/dx equal to the given slope and solve for the parameter t. Then, substitute t back into the parametric equations to find the corresponding (x, y) points on the curve.
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Related Practice
Textbook Question
45–60. Areas of regions Find the area of the following regions.
The region inside one leaf of the rose r = cos 5θ
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Textbook Question
45–60. Areas of regions Find the area of the following regions.
The region inside the limaçon r = 4 - 2 cos θ
Textbook Question
31–36. Eliminating the parameter Eliminate the parameter to express the following parametric equations as a single equation in x and y.
x=t,y= √(4−t²) a
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Textbook Question
11–20. Slopes of tangent lines Find the slope of the line tangent to the following polar curves at the given points.
r = 2θ; (π/2, π/4)
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Textbook Question
75–76. Graphs to polar equations Find a polar equation for each conic section. Assume one focus is at the origin.