Introduction to Motion and Acceleration

# At solution 12 at 1 0 1 4 1 0 4 2 1 1 4 1 2 1 4 2 1 4

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Unformatted text preview: s in fact a unit vector. Example 9 Show that if perpendicular to . Solution We wish to show that is a vector-valued function and · is constant, then is 0. To this end, observe that · Since is constant, its derivative is 0. Therefore, · tells us that · 0 By the properties of the dot product, · · 0 or 2 Therefore, · 0 · 0 and the vectors This is a critical result for our next step: The Unit Normal Vector 11 and are perpendicular to each other. The vector is a unit vector. Therefore, by definition its norm is constant (namely 1). From the previous and are perpendicular. Thus, we define the unit normal vector example it follows that , as the unit vector perpendicular to and such that , denoted by at / / unit normal Graphically these two vectors look like this: z T (t ) N (t ) y x Remark: Although we place these vectors with their tails at the moving point, they are really mathematical vectors in standard form so their tails should be at the origin (as in the figure above). It is only for visual convenience that are placed at the point under consideration. At any general point Provided that , , , the unit tangent and unit normal vectors are given by and Example 10 Suppose an object moves along a path whose parametric equations are cos 2 , sin 2 , Find the unit tangent and unit normal vectors at ; 0 . Solution First we find : 2 sin 2 , 2 cos 2 , 1 4 sin2 2 Thus, 1 √5 The unit tangent at 4 cos2 2 2 sin 2 , 2 cos 2 , 1 1 2 sin 2 , 2 cos 2 , 1 is 12 √5 1 √5 0, 2,1 · Next, the unit normal is given by Computing : 1 4 cos 2 , 4 sin 2 , 0 √5 Thus, 1 4 cos 2 , 4 sin 2 , 0 √5 1 · 16 cos2 2 16 sin 2 √5 4 cos 2 , 4 sin 2 , 0 4 Therefore, cos 2 , sin 2 , 0 and 1, 0,0 The figure below illustrates the curve together with the unit tangent and unit normal vectors at : - 1.0 - 0.5 0.0 0.5 1.0 6 4 2 1.0 0.5 0 0.0 - 0.5 - 1.0 13...
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