Textbook Question
Length in Polar Coordinates
Find the lengths of the curves given by the polar coordinate equations in Exercises 51–54.
r = √(1 + cos 2θ), −π/2 ≤ θ ≤ π/2
1
views
Ch. 11 - Parametric Equations and Polar Coordinates
Hass15th EditionThomas' CalculusISBN: 9780137616077
Not the one you use?Change textbookSelect a different textbook
Table of contents
- Ch. 1 - Functions155
- Ch. 2 - Limits and Continuity240
- Ch. 3 - Derivatives337
- Ch. 4 - Applications of Derivatives293
- Ch. 5 - Integrals41
- Ch. 6 - Applications of Definite Integrals25
- Ch. 7 - Transcendental Functions489
- Ch. 8 - Techniques of Integration398
- Ch. 9 - First-Order Differential Equations60
- Ch. 10 - Infinite Sequences and Series119
- Ch. 11 - Parametric Equations and Polar Coordinates58
Chapter 11, Problem 11.PE.58
Graphing Conic Sections
Sketch the parabolas in Exercises 55–58. Include the focus and directrix in each sketch.
y² = −(8/3)x
Verified step by step guidance1
Identify the type of conic section given by the equation \(y^{2} = -\frac{8}{3}x\). Since the equation is in the form \(y^{2} = 4px\), it represents a parabola that opens either left or right.
Rewrite the equation to match the standard form \(y^{2} = 4px\). Here, \(4p = -\frac{8}{3}\), so solve for \(p\) by dividing both sides by 4: \(p = -\frac{8}{3} \times \frac{1}{4} = -\frac{2}{3}\).
Determine the vertex, focus, and directrix of the parabola. The vertex is at the origin \((0,0)\) because the equation is not shifted. The focus lies at \((p, 0)\), so substitute \(p = -\frac{2}{3}\) to get the focus at \(\left(-\frac{2}{3}, 0\right)\).
Find the equation of the directrix. For a parabola \(y^{2} = 4px\), the directrix is the vertical line \(x = -p\). Substitute \(p = -\frac{2}{3}\) to get the directrix \(x = \frac{2}{3}\).
Sketch the parabola opening to the left (since \(p\) is negative), plot the vertex at the origin, mark the focus at \(\left(-\frac{2}{3}, 0\right)\), and draw the directrix line \(x = \frac{2}{3}\). This completes the graph with all key features.
Verified video answer for a similar problem:
This video solution was recommended by our tutors as helpful for the problem above.
Video duration:
3mWas this helpful?
Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Parabola Definition
A parabola is the set of all points equidistant from a fixed point called the focus and a fixed line called the directrix. Understanding this geometric definition helps in identifying the focus and directrix from the equation and in sketching the parabola accurately.
Recommended video:
05:43
Definition of the Definite Integral
Standard Form of a Parabola
Parabolas can be expressed in standard forms such as y² = 4px or x² = 4py, where p represents the distance from the vertex to the focus (and directrix). Recognizing and rewriting the given equation into this form allows determination of the parabola's orientation and key features.
Recommended video:
3:40Circles in Standard Form Example 1
Focus and Directrix Calculation
From the standard form y² = 4px, the focus is located at (p, 0) and the directrix is the line x = -p (or similarly for x² = 4py). Calculating p from the equation's coefficients enables plotting these elements, which are essential for accurately sketching the parabola.
Recommended video:
5:28Horizontal Parabolas
Related Practice
Textbook Question
Lengths of Curves
Find the lengths of the curves in Exercises 13–19.
x = 5 cos t − cos 5t, y = 5 sin t − sin 5t, 0 ≤ t ≤ π/2
1
views
Textbook Question
Finding Parametric Equations
Find parametric equations and a parameter interval for the motion of a particle that starts at (a, 0) and traces the circle x²+y²=a².
a. once clockwise.
(There are many ways to do these, so your answers may not be the same as the ones at the back of the text.)
Textbook Question
Cartesian to Polar Equations
Find polar equations for the circles in Exercises 33–36. Sketch each circle in the coordinate plane and label it with both its Cartesian and polar equations.
x² + y² + 5y = 0
1
views
Textbook Question
Shifting Conic Sections
You may wish to review Section 1.2 before solving Exercises 39-56.
The hyperbola (y²/4) − (x²/5) = 1 is shifted 2 units down to generate the hyperbola (y + 2)²/4 − x²/5 = 1.
a. Find the center, foci, vertices, and asymptotes of the new hyperbola.
Textbook Question
Cycloid
a. Find the length of one arch of the cycloid x = a(t − sin t), y = a(1 − cos t).
1
views