Let C be a smooth curve with parametrization r(t). We know from class that at time t its acceleration vector has the form

a(t) = aT(t)T(t) + aN(t)N(t),
where aT(t) and aN(t) are the tangential and normal components of the acceleration. Use this fact to show that, for each t, the vector r0(t)×r00(t) is parallel to B(t).

Using the given fact to show that, for each t, the vector r'(t)×r"(t) is parallel to B(t), below is proof.

B(t) = λ [r'(t) * r"(t)]   where, λ = 1 / [ aₙ(t) ||r'(t)|| ]

Step-by-step explanation:

a(t) = aT(t)T(t) + aN(t)N(t), where aT(t) and aN(t) are the tangential and normal components of the acceleration.

In the attached workings, the above fact is used to show that for every 't', the vector r'(t)×r"(t) is parallel to B(t).

the correct answer to your question is a. 4^4

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1 number

if we use a graphing calculator, there is only one such number that satisfies the conclusion of the mean value theorem, that is, a number such that

where the right-hand side is the average rate of change of f on the interval [0.5,0.75], which is the slope of the secant line.

in the attached graph, the black line is the secant line which has its slope as the average rate of change of f on [0.5,0.75], while the orange line is the only tangent line to the graph of f in the interval that satisfies the mvt conclusion.