M Example Tangent to a Curve

Determine the unit vector tangent to the curve: x = t2 + 1, y = 4t — 3, z = 2t2 — 6t for t = 2. Solution:

~dt dp dt dt [(t2 + 1)i + (4t — 3)j + (2t2 — 6t)k] = 2ti + 4j + (4t — 6)k v/(2t)2 + (4)2 + (4t — 6)2

2ti + 4j + (4t — 6)k \J (2t)2 + (4)2 + (4t — 6)2

Mathematica solution is shown in Fig. 3.4

■ Parametric Plot in 3D

ParametricPlot3D[{t^2 + 1, 4 t - 3, 2 t*2 - 6 t}, {t, 0, 4}]

ParametricPlot3D[{t^2 + 1, 4 t - 3, 2 t*2 - 6 t}, {t, 0, 4}]

- Graphics3D -

Figure 3.4: Mathematica Solution for the Tangent to a Curve in 3D

0 0

Post a comment