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Q. 9

Expert-verifiedFound in: Page 879

Book edition
1st

Author(s)
Peter Kohn, Laura Taalman

Pages
1155 pages

ISBN
9781429241861

Given a twice-differentiable vector-valued function $r\left(t\right)$, what is the definition of the principal unit normal vector $N\left(t\right)$?

The definition of the principal unit normal vector $N\left(t\right)$ is given by,

$N\left(t\right)=\frac{T\text{'}\left(t\right)}{||T\text{'}\left(t\right)||}$

Consider the given question,

A twice-differentiable vector-valued function is $r\left(t\right)$.

From the question,

Both $r\left(t\right),r\text{'}\text{'}\left(t\right)$ exists.

It is clear that the principal unit normal vectors a unit vector with magnitude $1$ and is orthogonal to the tangent vectors.

$T\left(t\right)$ always points towards the direction in which the curve bends.

From the definition of the principal unit normal vector, if $r\left(t\right)$ is a differentiable vector function $I\subseteq R$, then the principal unit normal vector at $r\left(t\right)$ is denoted by $N\left(t\right)$ is defined as,

$N\left(t\right)=\frac{T\text{'}\left(t\right)}{||T\text{'}\left(t\right)||}$

Where, $T\text{'}\left(t\right)\ne 0,\phantom{\rule{0ex}{0ex}}T\left(t\right)=\frac{r\text{'}\left(t\right)}{||r\text{'}\left(t\right)||}$

Assume $r\left(t\right)=>">\mathrm{cos}\alpha ,\mathrm{sin}at$ where $\alpha >0$.

$r\text{\'}\left(t\right)=>">-\alpha \mathrm{sin}\alpha t,\alpha \mathrm{cos}\alpha t$

From the given question,

$T\text{\'}\left(t\right)=>">-\mathrm{cos}\alpha t,-\alpha \mathrm{sin}\alpha t$

Now the principal unit normal vector to $r\left(t\right)=>">\mathrm{cos}\alpha t,\mathrm{sin}\alpha t$ is given below,

$N\left(t\right)=\frac{T\text{'}\left(t\right)}{||T\text{'}\left(t\right)||}\phantom{\rule{0ex}{0ex}}=\frac{>}{-\alpha \mathrm{cos}\alpha t,-\alpha \mathrm{sin}\alpha t}\alpha $

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