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Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270336

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BuyFindarrow_forward

Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270336
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).

txy
-31.20-3.60
-2.81.23-3.43
-2.61.26-3.27
-2.41.30-3.11
-2.21.34-2.95
-21.40-2.80
-1.81.47-2.65
-1.61.56-2.50
-1.41.68-2.35
-1.21.82-2.18
-12.00-2.00
-0.82.22-1.78
-0.62.47-1.48
-0.42.72-1.09
-0.22.92-0.58
03.000.00
0.22.920.58
0.42.721.09
0.62.471.48
0.82.221.78
12.002.00
1.21.822.18
1.41.682.35
1.61.562.50
1.81.472.65
21.402.80
2.21.342.95
2.41.303.11
2.61.263.27
2.81.233.43
31.203.60

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).

txy
-31.05-3.15
-2.81.06-2.96
-2.61.06-2.77
-2.41.07-2.58
-2.21.09-2.39
-21.10-2.20
-1.81.12-2.01
-1.61.14-1.82
-1.41.17-1.64
-1.21.20-1.45
-11.25-1.25
-0.81.30-1.04
-0.61.37-0.82
-0.41.43-0.57
-0.21.48-0.30
01.500.00
0.21.480.30
0.41.430.57
0.61.370.82
0.81.301.04
11.251.25
1.21.201.45
1.41.171.64
1.61.141.82
1.81.122.01
21.102.20
2.21.092.39
2.41.072.58
2.61.062.77
2.81.062.96

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).

txy
-31.00-3.00
-2.81.00-2.80
-2.61.00-2.60
-2.41.00-2.40
-2.21.00-2.20
-21.00-2.00
-1.81.00-1.80
-1.61.00-1.60
-1.41.00-1.40
-1.21.00-1.20
-11.00-1.00
-0.81.00-0.80
-0.61.00-0.60
-0.41.00-0.40
-0.21.00-0.20
01.000.00
0.21.000.20
0.41.000.40
0.61.000.60
0.81.000.80
11.001.00
1.21.001.20
1.41

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