Chapter 10.2, Problem 38E

### Calculus (MindTap Course List)

11th Edition
Ron Larson + 1 other
ISBN: 9781337275347

Chapter
Section

### Calculus (MindTap Course List)

11th Edition
Ron Larson + 1 other
ISBN: 9781337275347
Textbook Problem

# Comparing Plane Curves In Exercises 35–38, determine any differences between the curves of the parametric equations. Are the graphs the same? Are the orientations the same? Are the curves smooth? Explain. ( a )   x = 2   cos   θ     y = 2   sin   θ ( b ) x = 4 t 2 − 1 / | t |     y = 1 / t ( c ) x = t y = 4 − t ( d ) x = − 4 − e 2 t     y = e t

To determine
Any difference between the curves of the following parametric equations. The graphs, orientations and the smoothness of the curves are same or not.

Explanation

Given:

(a) The parametric equation, x=2cosÎ¸,y=2sinÎ¸

(b) The parametric equation, x=4t2âˆ’1/|t|,y=1/t

(c) The parametric equation, x=t,y=4âˆ’t

(d) The parametric equation, x=âˆ’4âˆ’e2t,y=et

Explanation:

(a) Consider the parametric function,

x=2cosÎ¸ â€¦â€¦ (1)

Since, the rage of the cosine function is, [âˆ’1,1]. So, the value of x varies between [âˆ’2,2].

y=2sinÎ¸ â€¦â€¦ (2)

Sum of square of equation (1)and (2),

x2+y2=(2cosÎ¸)2+(2sinÎ¸)2=4(cos2Î¸+sin2Î¸)=4

The above equation is equation of straight line, x2+y2=4.

For different value of t the different value of xÂ andÂ y are obtained,

 t Ï€ 2Ï€3 Ï€2 Ï€3 0 x âˆ’2 âˆ’1 0 1 2 y 0 3 2 3 0

The graph is plotted with the help of table is,

ss

The arrow on the curve indicate the orientation of the curve. From the above calculation and graph, the orientation of the curve anticlockwise for the interval [âˆ’2,2]. That means the curve has a definite orientation for [âˆ’2,2].

From the above graph, the curve of the parametric function is circle and that curve is smooth for all real number.

(b) Consider the parametric function,

x=4t2âˆ’1|t| â€¦â€¦ (3)

The above function is defined for,

4t2âˆ’1â‰¥04t2â‰¥1t2â‰¥14tâ‰¤âˆ’12Â orÂ tâ‰¥12

So, the domain of the function is tâ‰¤âˆ’12Â orÂ tâ‰¥12. The value of x varies from [0,2].

Consider the equation another equation,

y=1t â€¦â€¦ (4)

Sum of square of equation (3) and (4),

x2+y2=(4t2âˆ’1|t|)2+(1t)2=4t2âˆ’1t2+1t2=4t4t2=4

The above equation is equation of straight line, x2+y2=4. This function is defined for [0,2] because the value of x varies between [0,2]. So, the domain of the function is 0â‰¤xâ‰¤2.

For different value of t the different value of xÂ andÂ y are obtained,

 t âˆ’12 âˆ’1 12 1 x 0 3 0 3 y âˆ’2 âˆ’1 2 1

The graph is plotted with the help of table is,

From the above graph the value of x varies from [0,2]. So, the domain of the function is [0,2]

The arrow on the curve indicate the orientation of the curve. From the above calculation and graph, the orientation of the curve anticlockwise. That means the orientation has definite orientation for [0,2].

From the above graph, the curve of the parametric function is circle and that curve is smooth for all real number.

(c) Consider the parametric function,

x=t â€¦â€¦ (5)

The above function is defined for,

tâ‰¥0

So, the domain of the function is tâ‰¥0.

Consider the equation another equation,

y=4âˆ’t â€¦â€¦ (6)

The above function is defined for,

4âˆ’tâ‰¥0tâ‰¤4

That means the function is defined for 0â‰¤tâ‰¤4

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