# Vector-valued Functions and Graphing

1 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: https://www.youtube.com/watch?v=YYftS8Kyvnk&t=53s 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 π(π‘) = (sin π‘ β π‘ cos π‘) πβ + (cos π‘ + π‘ sin π‘) πβ + π‘ βπβ and display at π‘ = 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 / λ¦ 14_24 β 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 β 12) r(t) = ti + t?j + -k, -2 sis 2. 6 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