Textbook Question
Area of a sector of a hyperbola: Consider the region R bounded by the right branch of the hyperbola x²/a² - y²/b² = 1 and the vertical line through the right focus
a. What is the area of R?
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 12.1.67a
67–72. Derivatives Consider the following parametric curves.
a. Determine dy/dx in terms of t and evaluate it at the given value of t.
x = 2 + 4t, y = 4 − 8t; t = 2
Verified step by step guidance1
Identify the parametric equations given: \(x = 2 + 4t\) and \(y = 4 - 8t\).
Find the derivatives of \(x\) and \(y\) with respect to the parameter \(t\): compute \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\).
Calculate \(\frac{dx}{dt}\) by differentiating \(x = 2 + 4t\) with respect to \(t\), and calculate \(\frac{dy}{dt}\) by differentiating \(y = 4 - 8t\) with respect to \(t\).
Use the formula for the derivative of \(y\) with respect to \(x\) in parametric form: \(\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}\).
Evaluate \(\frac{dy}{dx}\) at the given value \(t = 2\) by substituting \(t = 2\) into the expressions for \(\frac{dy}{dt}\) and \(\frac{dx}{dt}\), then compute the ratio.
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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, usually denoted as t. Instead of y as a function of x, both x and y depend on t, allowing the description of more complex curves and motions.
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Parameterizing Equations
Derivative of Parametric Curves (dy/dx)
For parametric curves, the derivative 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 gives the slope of the tangent line to the curve at a given parameter value.
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Evaluating Derivatives at a Specific Parameter Value
After finding the general expression for dy/dx in terms of t, substituting the given value of t allows calculation of the slope of the curve at that specific point. This step is essential for understanding the behavior of the curve at particular parameter values.
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Related Practice
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a. Find the distance traveled during this 30-minute period.
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Textbook Question
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a. Express the region formed by the channel as a set in polar coordinates.
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Textbook Question
Intersecting lines Consider the following pairs of lines. Determine whether the lines are parallel or intersecting. If the lines intersect, then determine the point of intersection.
a. x = 1 + s, y = 2s and x = 1 + 2t, y = 3t
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Textbook Question
Explain why or why not Determine whether the following statements are true and give an explanation or counterexample.
b. On every ellipse, there are exactly two points at which the curve has slope s, where s is any real number.