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Q. 9

Expert-verified

Calculus

Found in: Page 879

Book edition 1st

Author(s) Peter Kohn, Laura Taalman

Pages 1155 pages

ISBN 9781429241861

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Short Answer

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

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

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

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1. Given information.

Consider the given question,

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

Step 2. Write the definition.

From the question,

Both $r (t), r'' (t)$ exists.

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

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

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

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

Where, $T' (t) \neq 0, T (t) = \frac{r' (t)}{||r' (t)||}$

Step 3. Consider an example.

Assume $r (t) = <\cos α, \sin a t>$ where $α > 0$ .

$r' (t) = <- α \sin α t, α \cos α t> ||r' (t)|| = \sqrt{{(- α \sin α t)}^{2} + {(α c o s α t)}^{2}} = α T (t) = \frac{r' (t)}{||r' (t)||} = \frac{<- α \sin α t, α \cos α t>}{α} = <- \sin α t, \cos α t>$

Step 4. Write the principal unit normal vector to rt.

From the given question,

$T' (t) = <- \cos α t, - α \sin α t> ||T' (t)|| = \sqrt{{(- α \cos α t)}^{2} + {(- α \sin α t)}^{2}} = α$

Now the principal unit normal vector to $r (t) = <\cos α t, \sin α t>$ is given below,

$N (t) = \frac{T' (t)}{||T' (t)||} = \frac{<- α \cos α t, - α \sin α t>}{α} = <- \cos α t, - \sin α t>$

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What is the dot product of the vector functions $r_{1} (t) = (x_{1} (t), y_{1} (t)) and r_{2} (t) = (x_{2} (t), y_{2} (t)) ?$

Given a twice-differentiable vector-valued function $r (t)$ and a point $t_{0}$ in its domain, what are the geometric relationships between the unit tangent vector $T (t_{0})$ , the principal unit normal vector $N (t_{0})$ , and ${\frac{d N}{d t}}_{t_{0}}$ ?

Let $r_{1} (f)$ and $r_{2} (t)$ be differentiable vector functions with three components each. Prove that

\frac{d}{d t} (r_{1} (t) - r_{2} (t)) = r_{1}^{'} (t) - r_{2} (t) + r_{1} (t) - r_{2}^{'} (t) ∣

(This is Theorem 11.11 (c).)

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