   Chapter 10.3, Problem 75E

Chapter
Section
Textbook Problem

# Investigate the family of curves with polar equations r = 1 + c cos θ, where c is a real number. How does the shape change as c changes?

To determine

To investigate: The family of curves for the polar equation r=1+ccosθ by plotting graphs for different values of c .

Explanation

Given:

The polar equation is r=1+ccosθ .

Here, the real number is c .

Calculation:

Write the x and y coordinates for the polar equation.

x=rcosθ (1)

y=rsinθ (2)

Substitute (1+ccosθ) for r in equation (1),

x=(1+ccosθ)cosθ (3)

Substitute (1+ccosθ) for r in equation (2),

y=(1+ccosθ)sinθ (4)

i) When c=0

Substitute 0 for c and 0 for θ in equation (3),

x={1+[0×cos(0)]}cos(0)=[1+(0×1)]1=1

Substitute 0 for c and 0 for θ in equation (4),

y={1+[0×cos(0)]}sin(0)=[1+(0×1)]0=0

Similarly, calculate the values of x and y for each value of θ ranging from 0° to 40° and tabulate as shown in the below table.

 θ (degrees) θ (rad) x y 0 0 1 0 10 0.174611111 0.984794173 0.173725175 20 0.349222222 0.939639127 0.342167081 30 0.523833333 0.865908101 0.500203119 40 0.698444444 0.765843378 0.643027154

Graph:

Plot the curve using the points (x,y) obtained when c=0 as shown below in figure 1.

Thus, the curve for the polar equation r=1+ccosθ is plotted for c=0 as shown in figure 1. The curve is a circle with radius 1.

ii) When c=0.25 ,

Substitute 0.25 for c and 0 for θ in equation (3),

x={1+[0.25×cos(0)]}cos(0)=[1+(0.25×1)]1=1.25

Substitute 0.25 for c and 0 for θ in equation (4),

y={1+[0.25×cos(0)]}sin(0)=[1+(0.25×1)]0=0

Similarly, calculate the values of x and y for each value of θ ranging from 0° to 40° and tabulate as shown in the below table.

 θ (degrees) θ (rad) x y 0 0 1.25 0 10 0.174611111 1.227249064 0.21649606 20 0.349222222 1.160369549 0.422545475 30 0.523833333 1.053357311 0.608485602 40 0.698444444 0.912472398 0.766141676

Graph:

The curve was plotted with the points (x,y) obtained by (c=0.25) and different values of θ as shown in figure-2.

Thus, the curve for the polar equation r=1+ccosθ is plotted for c=0.25 as shown in figure 2. The curve moves towards right and gets enlarged.

iii) When c=0.75 ,

Substitute 0.75 for c and 0 for θ in equation (3),

x={1+[0.75×cos(0)]}cos(0)=[1+(0.75×1)]1=1.75

Substitute 0.25 for c and 0 for θ in equation (4),

y={1+[0

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