What is the principal unit normal vector to the curve at the specified value of the parameter?

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What is the principal unit normal vector to the curve at the specified value of the parameter?

$r (t) = \sqrt{2} t ˆ i + e^{t} ˆ j + e^{- t} ˆ k; t = 0$ $r(t)=sqrt(2)thati+e^thatj+e^-thatk; t=0$

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Answer 1 · 2018-01-02T01:37:41

$ˆ n = \frac{\sqrt{2}}{2} ˆ i + \frac{1}{2} ˆ j - \frac{1}{2} ˆ k$ $hatn = sqrt2/2hati+ 1/2hatj-1/2hatk$

Explanation:

The normal vector at any point on the curve is the gradient of the curve evaluated at the specified parametric value.

Compute the gradient:

$\nabla r (t) = \frac{\partial (\sqrt{2} t)}{\partial t} ˆ i + \frac{\partial (e^{t})}{\partial t} ˆ j + \frac{\partial (e^{- t})}{\partial t} ˆ k$ $gradr(t) = (del(sqrt2t))/(del t)hati+ (del(e^t))/(del t)hatj+(del(e^-t))/(del t)hatk$

$\nabla r (t) = \sqrt{2} ˆ i + e^{t} ˆ j - e^{- t} ˆ k$ $gradr(t) = sqrt2hati+ e^thatj-e^-thatk$

A normal vector, $\to n$ $vecn$ , is the gradient evaluated at $t = 0$ $t=0$

$\to n = \sqrt{2} ˆ i + ˆ j - ˆ k$ $vecn = sqrt2hati+ hatj-hatk$

We make it the unit vector by dividing by its magnitude:

$ˆ n = \frac{\sqrt{2} ˆ i + ˆ j - ˆ k}{\sqrt{{(\sqrt{2})}^{2} + 1^{2} + {(- 1)}^{2}}}$ $hatn = (sqrt2hati+ hatj-hatk)/sqrt((sqrt2)^2+ 1^2+ (-1)^2)$

$ˆ n = \frac{\sqrt{2} ˆ i + ˆ j - ˆ k}{\sqrt{4}}$ $hatn = (sqrt2hati+ hatj-hatk)/sqrt4$

$ˆ n = \frac{\sqrt{2}}{2} ˆ i + \frac{1}{2} ˆ j - \frac{1}{2} ˆ k$ $hatn = sqrt2/2hati+ 1/2hatj-1/2hatk$

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What is the principal unit normal vector to the curve at the specified value of the parameter?

$r (t) = \sqrt{2} t ˆ i + e^{t} ˆ j + e^{- t} ˆ k; t = 0$ $r(t)=sqrt(2)thati+e^thatj+e^-thatk; t=0$

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What is the principal unit normal vector to the curve at the specified value of the parameter?

r(t)=√2tˆi+etˆj+e−tˆk;t=0r(t)=sqrt(2)thati+e^thatj+e^-thatk; t=0

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$r (t) = \sqrt{2} t ˆ i + e^{t} ˆ j + e^{- t} ˆ k; t = 0$ $r(t)=sqrt(2)thati+e^thatj+e^-thatk; t=0$