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The magnitude of the velocity is v = ωr sin(π/2) = ωr. Explicitly i j k v = ω × r = 0 0 ω = − ωyi + ωxj. x y 0 The acceleration is given by a = dv d = (ω × r) = ω × r + ω × r. dt dt Since ω is a constant, ω = 0. Moreover, r= v = ω × r. Thus a = ω × (ω × r) = ω 2 k × (k × r) . Hence, a is in the −r direction and its magnitude is equal to ω 2 r. Method II. The position vector can be explicitly written as r = x(t)i + y(t)j = r cos ωti + r sin ωtj. 19) v2 r. 20) We see immediately that the acceleration is toward the center with a magnitude of ω 2 r.

5. Find the coordinates of the foot of the perpendicular from the point (1, 2, 3) to the plane of the last example. 5. Let the position vector from the origin to the foot of the perpendicular be rp . The vector r1 −rp is perpendicular to the plane, therefore it is parallel to the unit normal vector n of the plane, r1 − rp = kn. It follows that |r1 − rp | = k. Since |r1 − rp | = d, so k = d. Thus, 4 rp = r1 − dn = i + 2j + 3k − √ 38 3 2 5 √ i− √ j+ √ k . 9 + 4 + 25 38 38 Hence, the coordinates of the foot of the perpendicular are 26 84 94 38 , 38 , 38 .

Now we evaluate db/ds and dn/ds. Since b is perpendicular to t, so b · t = 0. Afer differentiating we have d db dt db (b · t) = ·t+b· = · t + b · κn = 0. 54) Hence, db/ds · t = − κb · n. Since b is perpendicular to n, b · n = 0. Thus, db/ds · t = 0, which means db/ds is perpendicular to t. On the other hand, since b is a unit vector, so db/ds is perpendicular b. Therefore db/ds must be in the direction of n. 55) where γ, by definition, is the magnitude of db/ds and is called the torsion of the curve.

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A Guided Tour of Mathematical Physics by R. Sneider


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