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Consider the vector-valued function r (t) = < 2 + 2 t, 1 + 4 t, 3 + 4 t >. (a) Compute T(t),...

Question:

Consider the vector-valued function {eq}\displaystyle r (t) = \langle 2 + 2 t,\ 1 + 4 t,\ 3 + 4 t \rangle {/eq}.

(a) Compute {eq}T(t),\ N(t), {/eq} and {eq}B(t) {/eq}.

(b) Find the curvature {eq}\kappa (t) {/eq}.

Curvature:

Curvature {eq}\kappa {/eq} is a measure of the rate of change of direction of a curve. The curvature of a straight line is predictably zero. The curvature of circle or a sphere is uniform. The curvature is given as {eq}\kappa =\frac{\left\| T'\left( t \right) \right\|}{\left\| r'\left( t \right) \right\|} {/eq}, where {eq}r\left( t \right) {/eq} is the curve and {eq}T\left( t \right) {/eq} is the unit tangent vector given by {eq}T\left( t \right)=\frac{r'\left( t \right)}{\left\| r'\left( t \right) \right\|} {/eq}

Answer and Explanation: 1

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The given curve is {eq}r\left( t \right)=\left\langle 2+2t,1+4t,3+4t \right\rangle {/eq}

We will need the following:

{eq}\begin{align} & r'\left(...

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