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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 12.1.71a
Chapter 12, Problem 12.1.71a

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 = t + 1/t, y = t − 1/t; t = 1

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1
Identify the parametric equations given: \(x = t + \frac{1}{t}\) and \(y = t - \frac{1}{t}\).
Find the derivatives of \(x\) and \(y\) with respect to the parameter \(t\). Compute \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\) by differentiating each expression:
\[\frac{dx}{dt} = 1 - \frac{1}{t^2}\]
\[\frac{dy}{dt} = 1 + \frac{1}{t^2}\]
Use the chain rule for parametric curves to find \(\frac{dy}{dx}\) by dividing \(\frac{dy}{dt}\) by \(\frac{dx}{dt}\):
\[\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{1 + \frac{1}{t^2}}{1 - \frac{1}{t^2}}\]
Evaluate \(\frac{dy}{dx}\) at the given value \(t = 1\) by substituting \(t = 1\) into the expression.

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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 t. Instead of y as a function of x, both x and y depend on t, allowing the description of more complex curves. Understanding how to work with these equations is essential for analyzing the curve's behavior.
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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 method allows us to find the slope of the tangent line to the curve at any parameter value t.
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Differentiation of Parametric Curves

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 us to find the slope of the curve at that specific point. This step is crucial for understanding the curve's instantaneous rate of change or tangent line slope at that parameter.
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Related Practice
Textbook Question

51–52. {Use of Tech} Arc length of polar curves Find the approximate length of the following curves.

The limaçon r=3−6cosθ

Textbook Question

53–57. Conic sections

a. Determine whether the following equations describe a parabola, an ellipse, or a hyperbola.

x = 16y²

Textbook Question

Volume of a hyperbolic cap 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 volume of the solid that is generated when R is revolved about the x-axis?

Textbook Question

Explain why or why not Determine whether the following statements are true and give an explanation or counterexample.  


a. The point with Cartesian coordinates (−2, 2) has polar coordinates (2√2, 3π/4), (2√2, 11π/4), (2√2, −5π/4), and (−2√2,−π/4).  

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Textbook Question

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 = cos t, y = 8 sin t; t = π/2

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Textbook Question

(Use of Tech) Finger curves: r = f(θ) = cos(aᶿ) - 1.5, where a = (1 + 12π)^(1/(2π)) ≈ 1.78933

a. Show that f(0) = f(2π) and find the point on the curve that corresponds to θ = 0 and θ = 2π.

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