   Chapter 10.2, Problem 31E

Chapter
Section
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 = t y = 2 t + 1 (b) x = cos θ y = 2 cos θ + 1 (c) x = e − t y = 2 e − t + 1 (d) x = e t y = 2 e t + 1

To determine
Any difference between the curves of the given parametric equations. Also, check is the graphs, orientations and the smoothness of the curves are same or not.

Explanation

Given:

(a) The parametric equation, x=t,y=2t+1

(b) The parametric equation, x=cosθ,y=2cosθ+1

(c) The parametric equation, x=et,y=2et+1

(d) The parametric equation, x=et,y=2et+1

Explanation:

(a) Consider the parametric function,

x=t …...…... (1)

and

y=2t+1 …...…... (2)

Substitute the value of x=t in equation (2),

y=2(x)+1=2x+1

The above equation is equation of straight line, y=2x+1.

Assume different values of t and find the different values for x and y by the parametric equation x=t and y=2t+1,

 t −2 −1 0 1 2 x=t −2 −1 0 1 2 y=2t+1 −3 −1 1 3 5

Plot all the points and join. The graph obtained is,

The arrow on the curve indicate the orientation of the curve. From the above graph, the orientation of the curve is up for the interval (,). That means the curve has a definite orientation for all real number.

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

(b) Consider the parametric function,

x=cosθ …...…... (3)

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

y=2cosθ+1 …...…... (4)

Substitute the value of cosθ=x in equation (4),

y=2(x)+1=2x+1

The above equation is equation of straight line, y=2x+1. This function is defined for [1,1] because the value of x varies between [1,1].

Assume different values of t and find the different values for x and y by the parametric equation x=cosθ and y=2cosθ+1,

 t −π −π2 0 π2 π x=cosθ −1 0 1 0 −1 y=2cosθ+1 −1 1 3 1 −1

The graph is plotted with the help of table is,

The arrow on the curve indicate the orientation of the curve. From the above graph, the orientation of the curve is up for certain interval and down for certain interval. That means the orientation of the curve is oscillate.

Therefore, from the above graph, the orientation of the curve changes at x=1 and x=1 or x=±1. That means,

x=±1cosθ=±1θ=nπ

Where, n belongs to integer.

Differentiate the equation (3) with respect to θ,

dxdθ=dcosθdθ=sinθ

At θ=nπ,

dxdθ=sinnπ=0

Differentiate the equation (4) with respect to θ,

dxdθ=ddθ(2cosθ+1)=2sinθ

At θ=nπ,

dxdθ=2sinnπ=0

From the above graph, the curve of the parametric function is straight line and that curve is not smooth because of dxdθ=0 and dydθ=0 at θ=nπ

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