Chapter 11: Q. 16 (page 879)

Given a twice-differentiable vector-valued function $r (t)$ and a point $t_{0}$ in its domain, what is the osculating plane at $r (t_{0})$ ?

Short Answer

Expert verified

The osculating plane at $B (t_{0}) = T (t_{0}) \times N (t_{0})$ .

Step by step solution

Step 1. Given information.

Consider the given question,

A twice-differentiable vector-valued function is $r (t_{0})$ is any point in the domain of $r (t)$ .

Step 2. Definition of osculating plane at rt0.

Assume $r (t) = <x (t), y (t), z (t)>$ be a differentiable vector function on the interval $I \subseteq R$ such that,

$T' (t_{0}) \neq 0$ where, $t_{0} \in I$ . Then the osculating plane at $r (t_{0})$ is defined as,

$B (t_{0}) . <x - x (t_{0}), y - y (t_{0}), z - z (t_{0})> = 0$

Where, $B (t_{0}) = T (t_{0}) \times N (t_{0})$ .

$T (t_{0}), N (t_{0})$ are the orthogonal unit vectors. so their cross product $B (t_{0})$ is another unit vector and is orthogonal to both $T (t_{0}), N (t_{0})$ .

The osculating plane contains the point $r (t_{0})$ and has $B (t_{0})$ as its normal vector.

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Given a twice-differentiable vector-valued function $r (t)$ and a point $t_{0}$ in its domain, what is the osculating plane at $r (t_{0})$ ?

Short Answer

Step by step solution

Step 1. Given information.

Step 2. Definition of osculating plane at rt0.

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Given a twice-differentiable vector-valued function r(t)and a point t0in its domain, what is the osculating plane at rt0?

Short Answer

Step by step solution

Step 1. Given information.

Step 2. Definition of osculating plane at rt0.

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Theoretical and Mathematical Physics

Discrete Mathematics

Calculus

Pure Maths

Decision Maths

Probability and Statistics

Study anywhere. Anytime. Across all devices.

Given a twice-differentiable vector-valued function $r (t)$ and a point $t_{0}$ in its domain, what is the osculating plane at $r (t_{0})$ ?