   Chapter 10, Problem 43RE

Chapter
Section
Textbook Problem

The curves defined by the parametric equations x = t 2 − c t 2 + 1     y = t ( t 2 − c ) t 2 + 1 are called strophoids (from a Greek word meaning “to turn or twist”). Investigate how these curves vary as c varies.

To determine

To find: The variations of curve with the variation in c value from the given equation.

Explanation

Given:

The given parametric equations are x=t2ct2+1 and y=t(t2c)t2+1.

Calculation:

Take the value of c=3.

Calculate the value of x.

x=t2ct2+1

Substitute the values of 3 for t and 3 for c in the above equation.

x=(32)(3)(32)+1x=1.2

Calculate the value of y.

y=t(t2c)t2+1

Substitute the values of 3 for t and 3 for c in the above equation.

y=t(t2c)t2+1=3((32)(3))(32)+1=3.6

Similarly, calculate the remaining values.

Tabulate the values for c=3 as shown in below table (1).

 t x y -3 1.20 -3.60 -2.8 1.23 -3.43 -2.6 1.26 -3.27 -2.4 1.30 -3.11 -2.2 1.34 -2.95 -2 1.40 -2.80 -1.8 1.47 -2.65 -1.6 1.56 -2.50 -1.4 1.68 -2.35 -1.2 1.82 -2.18 -1 2.00 -2.00 -0.8 2.22 -1.78 -0.6 2.47 -1.48 -0.4 2.72 -1.09 -0.2 2.92 -0.58 0 3.00 0.00 0.2 2.92 0.58 0.4 2.72 1.09 0.6 2.47 1.48 0.8 2.22 1.78 1 2.00 2.00 1.2 1.82 2.18 1.4 1.68 2.35 1.6 1.56 2.50 1.8 1.47 2.65 2 1.40 2.80 2.2 1.34 2.95 2.4 1.30 3.11 2.6 1.26 3.27 2.8 1.23 3.43 3 1.20 3.60

Take the value of c=1.5.

Calculate the value of x.

x=t2ct2+1

Substitute the values of 3 for t and 1.5 for c in the above equation.

x=(32)(1.5)(32)+1x=1.05

Calculate the value of y.

y=t(t2c)t2+1

Substitute the values of 3 for t and 1.5 for c in the above equation.

y=t(t2c)t2+1=3((32)(1.5))(32)+1=3.15

Similarly, calculate the remaining values.

Tabulate the values for c=1.5 as shown in below table (2).

 t x y -3 1.05 -3.15 -2.8 1.06 -2.96 -2.6 1.06 -2.77 -2.4 1.07 -2.58 -2.2 1.09 -2.39 -2 1.10 -2.20 -1.8 1.12 -2.01 -1.6 1.14 -1.82 -1.4 1.17 -1.64 -1.2 1.20 -1.45 -1 1.25 -1.25 -0.8 1.30 -1.04 -0.6 1.37 -0.82 -0.4 1.43 -0.57 -0.2 1.48 -0.30 0 1.50 0.00 0.2 1.48 0.30 0.4 1.43 0.57 0.6 1.37 0.82 0.8 1.30 1.04 1 1.25 1.25 1.2 1.20 1.45 1.4 1.17 1.64 1.6 1.14 1.82 1.8 1.12 2.01 2 1.10 2.20 2.2 1.09 2.39 2.4 1.07 2.58 2.6 1.06 2.77 2.8 1.06 2.96

Take the value of c=1.

Calculate the value of x.

x=t2ct2+1

Substitute the values of 3 for t and 1 for c in the above equation.

x=(32)(1)(32)+1x=1.0

Calculate the value of y.

y=t(t2c)t2+1

Substitute the values of 3 for t and 1 for c in the above equation.

y=t(t2c)t2+1=3((32)(1))(32)+1=3.0

Similarly, calculate the remaining values.

Tabulate the values for c=1 as shown in below table (3).

 t x y -3 1.00 -3.00 -2.8 1.00 -2.80 -2.6 1.00 -2.60 -2.4 1.00 -2.40 -2.2 1.00 -2.20 -2 1.00 -2.00 -1.8 1.00 -1.80 -1.6 1.00 -1.60 -1.4 1.00 -1.40 -1.2 1.00 -1.20 -1 1.00 -1.00 -0.8 1.00 -0.80 -0.6 1.00 -0.60 -0.4 1.00 -0.40 -0.2 1.00 -0.20 0 1.00 0.00 0.2 1.00 0.20 0.4 1.00 0.40 0.6 1.00 0.60 0.8 1.00 0.80 1 1.00 1.00 1.2 1.00 1.20 1.4 1

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