Chapter 11: Q. 53 (page 890)

Use Exercise 52 to show that $\frac{d N}{d s} = - k T + ζ B,$ where κ is the curvature. (Hint: Differentiate N = B × T with respect to arc length.)

Short Answer

Expert verified

The given statement is proved.

Step by step solution

Step 1. Given Information.

The Relationship between $\frac{d T}{d s}, κ, and N$ can be defined as letting C be the graph of a vector function r(s) defined on an intervalI and parametrized by

arc length s. If the unit tangent vector T has a nonzero derivative, then $\frac{d T}{d s} = κ N .$

Step 2. Showing that dNds=-kT+ζB.

To show that $\frac{d N}{d s} = - k T + ζ B,$ where κ is the curvature, we will use the relationship between $\frac{d T}{d s}, κ, and N .$

By using the relationship,

role="math" localid="1649676899612" $\frac{d T}{d s} = \frac{d}{d s} (N) = \frac{d}{d s} (B \times T) [N = B \times T] = \frac{d B}{d s} \times T + B \times \frac{d T}{d s} = - ζ N \times T + B \times k N [\frac{d B}{d s} = - ζ N and \frac{d T}{d s} = k N] = ζ B - k T [B \times N = - T a n d - N \times T = B]$

Hence proved.

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Use Exercise 52 to show that $\frac{d N}{d s} = - k T + ζ B,$ where κ is the curvature. (Hint: Differentiate N = B × T with respect to arc length.)

Short Answer

Step by step solution

Step 1. Given Information.

Step 2. Showing that dNds=-kT+ζB.

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Use Exercise 52 to show that dNds=-kT+ζB, where κ is the curvature. (Hint: Differentiate N = B × T with respect to arc length.)

Short Answer

Step by step solution

Step 1. Given Information.

Step 2. Showing that dNds=-kT+ζB.

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Logic and Functions

Pure Maths

Decision Maths

Statistics

Calculus

Mechanics Maths

Study anywhere. Anytime. Across all devices.

Use Exercise 52 to show that $\frac{d N}{d s} = - k T + ζ B,$ where κ is the curvature. (Hint: Differentiate N = B × T with respect to arc length.)