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Math 4A CalcPlot3D Lab 03

Topic: Vector-valued Functions and Graphing a Space Curve

Go to the website to access the CalcPlot3D app.

CalcPlot3D – Monroe Community College

https://www.monroecc.edu/faculty/paulseeburger/calcnsf/CalcPlot3D/

Objective: The goal of Lab 03 show how CalcPlot3D constructs space curves and demonstrate how to

trace and display its path.

Here is a video that will demonstrate how to plot space curves using CalcPlot3D:

Apply what you have learned in the first two labs on plotting

surfaces along with the video for plotting space curves and

reproduce the image in section 12.1, example 4 on page 822.

Lab 03 DUE DATE is set in the Assignment

1

Math 4A CalcPlot3D Lab 03 (A Week 02Topic Assignment)

Topic: Vector-valued Functions, Unit Tangent, Unit Normal, Unit Binormal

Go to the website to access the CalcPlot3D app.

CalcPlot3D – Monroe Community College

https://www.monroecc.edu/faculty/paulseeburger/calcnsf/CalcPlot3D/

Objective: The goal of Lab 03 show how CalcPlot3D constructs space curves, trace the path of the curve,

and display the Frenet frame using its vector-valued function plotting capabilities

Here is a video that will demonstrate how to plot space curves using CalcPlot3D:

https://www.youtube.com/watch?v=YYftS8Kyvnk and discover how to display the Frenet Frame (or TNB-frame).

Use the space curve capabilities of CalcPlot3D to display

the space curve of the vector-valued function defined by

𝑟(𝑡) = (sin 𝑡 − 𝑡 cos 𝑡) 𝒊⃗ + (cos 𝑡 + 𝑡 sin 𝑡) 𝒋⃗ + 𝑡 ⃗𝒌⃗

and display at 𝑡 = 5 , the unit tangent vector, the unit

normal vector, and the unit binormal vector.

Lab 04 DUE DATE is set in the Assignment

4

= 1-

EXAMPLE 4 Representing a Graph: Vector-Valued Function

Sketch the space curve C represented by the intersection of the semiellipsoid

x2 2 2

1, 220

12 24

and the parabolic cylinder y = x2. Then find a vector-valued function to represent the

graph.

Solution The intersection of the two surfaces is shown in Figure 12.5. As in

Example 3, a natural choice of parameter is x = t. For this choice, you can use the

given equation y = xto obtain y = 12. Then it follows that

222

– 1 / 름

14_24 – 2r- r* _6 + 1)(4 – 1)

12

24

Because the curve lies above the xy-plane, you should choose the positive square root

for z and obtain the parametric equations

(6 + 12)(4 – 12)

*= 1, y = 1, and z =

The resulting vector-valued function is

(6 + 2)(4 – 12)

r(t) = ti + t?j +

-k, -2 sis 2.

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Vector-valued function

(Note that the k-component of r(t) implies -2 51 2.) From the points (-2,4, 0)

and (2, 4, 0) shown in Figure 12.5, you can see that the curve is traced as t increases

from -2 to 2.

21

24

24

24

6

Parabolic cylinder

(0, 0, 2)

C: x=1

y=r

(6 +124-1)

6

Ellipsoid

Curve in

space

(-2,4,0)

(2, 4,0)

The curve C is the intersection of the semiellipsoid and the parabolic cylinder.

Figure 12.5

Purchase answer to see full

attachment

Tags:

ellipsoid

parabola

parabolic cylinder

Space curve

Trace vector

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