To prove: d N d s = − k T + τ B using the following steps: (a) Show that d N d s is orthogonal toN. Also, conclude that d N d s lies in the plane spanned by T and B, and hence, d N d s = a T + b B for some scalars a, b. (b) Show that T ⋅ d N d s = − N ⋅ d T d s using N ⋅ T = 0 and compute a andb.

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Answer 1

Question

Chapter 14.4, Problem 90E

To determine

To prove:

$\frac{d N}{d s} = - k T + τ B$ using the following steps:

(a) Show that $\frac{d N}{d s}$ is orthogonal toN. Also, conclude that $\frac{d N}{d s}$ lies in the plane spanned by T and B, and hence, $\frac{d N}{d s} = a T + b B$ for some scalars a, b.

(b) Show that $T \cdot \frac{d N}{d s} = - N \cdot \frac{d T}{d s}$ using $N \cdot T = 0$ and compute a andb.

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Answer 2

Question

Chapter 14.4, Problem 90E

To determine

To prove:

$\frac{d N}{d s} = - k T + τ B$ using the following steps:

(a) Show that $\frac{d N}{d s}$ is orthogonal toN. Also, conclude that $\frac{d N}{d s}$ lies in the plane spanned by T and B, and hence, $\frac{d N}{d s} = a T + b B$ for some scalars a, b.

(b) Show that $T \cdot \frac{d N}{d s} = - N \cdot \frac{d T}{d s}$ using $N \cdot T = 0$ and compute a andb.

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Answer 3

Question

Chapter 14.4, Problem 90E

To determine

To prove:

$\frac{d N}{d s} = - k T + τ B$ using the following steps:

(a) Show that $\frac{d N}{d s}$ is orthogonal toN. Also, conclude that $\frac{d N}{d s}$ lies in the plane spanned by T and B, and hence, $\frac{d N}{d s} = a T + b B$ for some scalars a, b.

(b) Show that $T \cdot \frac{d N}{d s} = - N \cdot \frac{d T}{d s}$ using $N \cdot T = 0$ and compute a andb.

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Answer 4

Question

Chapter 14.4, Problem 90E

To determine

To prove:

$\frac{d N}{d s} = - k T + τ B$ using the following steps:

(a) Show that $\frac{d N}{d s}$ is orthogonal toN. Also, conclude that $\frac{d N}{d s}$ lies in the plane spanned by T and B, and hence, $\frac{d N}{d s} = a T + b B$ for some scalars a, b.

(b) Show that $T \cdot \frac{d N}{d s} = - N \cdot \frac{d T}{d s}$ using $N \cdot T = 0$ and compute a andb.

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Answer 5

Chapter 14.4, Problem 90E

To determine

To prove:

$\frac{d N}{d s} = - k T + τ B$ using the following steps:

(a) Show that $\frac{d N}{d s}$ is orthogonal toN. Also, conclude that $\frac{d N}{d s}$ lies in the plane spanned by T and B, and hence, $\frac{d N}{d s} = a T + b B$ for some scalars a, b.

(b) Show that $T \cdot \frac{d N}{d s} = - N \cdot \frac{d T}{d s}$ using $N \cdot T = 0$ and compute a andb.

Answer 6

Chapter 14.4, Problem 90E

To determine

To prove:

$\frac{d N}{d s} = - k T + τ B$ using the following steps:

(a) Show that $\frac{d N}{d s}$ is orthogonal toN. Also, conclude that $\frac{d N}{d s}$ lies in the plane spanned by T and B, and hence, $\frac{d N}{d s} = a T + b B$ for some scalars a, b.

(b) Show that $T \cdot \frac{d N}{d s} = - N \cdot \frac{d T}{d s}$ using $N \cdot T = 0$ and compute a andb.

Answer 7

Chapter 14.4, Problem 90E

To determine

To prove:

$\frac{d N}{d s} = - k T + τ B$ using the following steps:

(a) Show that $\frac{d N}{d s}$ is orthogonal toN. Also, conclude that $\frac{d N}{d s}$ lies in the plane spanned by T and B, and hence, $\frac{d N}{d s} = a T + b B$ for some scalars a, b.

(b) Show that $T \cdot \frac{d N}{d s} = - N \cdot \frac{d T}{d s}$ using $N \cdot T = 0$ and compute a andb.

Answer 8

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Answer 9

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Answer 10

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Answer 11

Ch. 14.1 - Prob. 1PQ Ch. 14.1 - Prob. 2PQ Ch. 14.1 - Prob. 3PQ Ch. 14.1 - Prob. 4PQ Ch. 14.1 - Prob. 5PQ Ch. 14.1 - Prob. 6PQ Ch. 14.1 - Prob. 1E Ch. 14.1 - Prob. 2E Ch. 14.1 - Prob. 3E Ch. 14.1 - Prob. 4E

Solution Summary: The author demonstrates that d N s is orthogonal to N and concludes that it lies in the plane spanned by T and B.

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Chapter 14 Solutions

Solution Summary: The author demonstrates that d N s is orthogonal to N and concludes that it lies in the plane spanned by T and B.

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To prove: dNds=−kT+τB using the following steps: (a) Show that dNds is orthogonal toN. Also, conclude that dNds lies in the plane spanned by T and B, and hence, dNds=aT+bB for some scalars a, b. (b) Show that T⋅dNds=−N⋅dTds using N⋅T=0 and compute a andb.

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Chapter 14 Solutions