   Chapter 10.6, Problem 19E

Chapter
Section
Textbook Problem

Graph the conics r = e/(1 − e cos θ) with e = 0.4, 0.6, 0.8, and 1.0 on a common screen. How does the value of e affect the shape of the curve?

To determine

To find: The variation of shape of curve with respect to eccentricity provided.

Explanation

Given:

The given polar equation is r=e1ecosθ with e=0.4,0.6,0.8and1.0.

Calculation:

Write the condition of conic:

• If eccentricity is less than one will make conic of Ellipse
• If eccentricity is equal to one will make conic of Parabola
• If eccentricity is greater than one will make conic of Hyperbola

Write the polar equation with different eccentricities.

r=0.410.4cosθEllipser=0.610.6cosθEllipser=0.810.8cosθEllipser=111cosθParabola

The value of x and y with respect to ‘r’ is,

x=rcosθy=rsinθ

Consider the eccentricity of 0.4.

Calculate the value of r.

r=0.410.4cosθ

Substitute 0° for θ in the above equation.

r=0.410.4cos(0°×π180)=0.67

Calculate the value of x.

x=rcosθ

Substitute 0.67 for r and 0° for θ in the above equation

x=0.67×cos(0°×π180)=0.67

Calculate the value of y.

y=rsinθ

Substitute 0.67 for r and 0° for θ in the above equation

y=0.67×sin(0°×π180)=0

Similarly calculate the remaining values.

Tabulate the calculated value as shown in the table (1).

 θ r=0.41−0.4cosθ x=rcosθ y=rsinθ 0 0.67 0.67 0.00 10 0.66 0.65 0.11 20 0.64 0.60 0.22 30 0.61 0.53 0.31 40 0.58 0.44 0.37 50 0.54 0.35 0.41 60 0.50 0.25 0.43 70 0.46 0.16 0.44 80 0.43 0.07 0.42 90 0.40 0.00 0.40 100 0.37 -0.06 0.37 110 0.35 -0.12 0.33 120 0.33 -0.17 0.29 130 0.32 -0.20 0.24 140 0.31 -0.23 0.20 150 0.30 -0.26 0.15 160 0.29 -0.27 0.10 170 0.29 -0.28 0.05 180 0.29 -0.29 0.00

Similarly calculate the remaining values till 360 degrees.

Consider the eccentricity of 0.6.

Calculate the value of r.

r=0.610.6cosθ

Substitute 0° for θ in the above equation.

r=0.610.6cos(0°×π180)=1.5

Calculate the value of x.

x=rcosθ

Substitute 1.5 for r and 0° for θ in the above equation

x=1.5×cos(0°×π180)x=1.5

Calculate the value of y.

y=rsinθ

Substitute 1.5 for r and 0° for θ in the above equation

y=1.5×sin(0°×π180)y=0

Similarly calculate the remaining values.

Tabulate the calculated value as shown in the table (2).

 θ r=0.61−0.6cosθ x=rcosθ y=rsinθ 0 1.50 1.50 0.00 10 1.47 1.44 0.25 20 1.38 1.29 0.47 30 1.25 1.08 0.62 40 1.11 0.85 0.71 50 0.98 0.63 0.75 60 0.86 0.43 0.74 70 0.75 0.26 0.71 80 0.67 0.12 0.66 90 0.60 0.00 0.60 100 0.54 -0.09 0.54 110 0.50 -0.17 0.47 120 0.46 -0.23 0.40 130 0.43 -0.28 0.33 140 0.41 -0.31 0

Still sussing out bartleby?

Check out a sample textbook solution.

See a sample solution

The Solution to Your Study Problems

Bartleby provides explanations to thousands of textbook problems written by our experts, many with advanced degrees!

Get Started

In Exercises 4756, solve the given equation for the indicated variable. 27=32x1

Finite Mathematics and Applied Calculus (MindTap Course List)

pH The pH of lime juice is 1.9. Find the hydrogen ion concentration.

Precalculus: Mathematics for Calculus (Standalone Book) 