Let r(t) = (cos t, sin t, t). What is the speed |r'(t)|?

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Multiple Choice

Let r(t) = (cos t, sin t, t). What is the speed |r'(t)|?

Explanation:
Speed is the length of the velocity vector r′(t). Differentiate componentwise: r′(t) = (−sin t, cos t, 1). The speed is |r′(t)| = sqrt[ (−sin t)^2 + (cos t)^2 + 1^2 ] = sqrt[ sin^2 t + cos^2 t + 1 ] = sqrt(1 + 1) = sqrt(2). Since sin^2 t + cos^2 t = 1, the horizontal part has unit speed and the vertical part adds in quadrature, giving the constant speed √2.

Speed is the length of the velocity vector r′(t). Differentiate componentwise: r′(t) = (−sin t, cos t, 1). The speed is |r′(t)| = sqrt[ (−sin t)^2 + (cos t)^2 + 1^2 ] = sqrt[ sin^2 t + cos^2 t + 1 ] = sqrt(1 + 1) = sqrt(2). Since sin^2 t + cos^2 t = 1, the horizontal part has unit speed and the vertical part adds in quadrature, giving the constant speed √2.

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