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Chapter 11: Problem 5

Give a practical formula for computing the principal unit normal vector.

Short Answer

Expert verified
Answer: The practical formula for computing the principal unit normal vector, denoted as N(s), is N(s) = dT/ds / ||dT/ds||, where dT/ds is the derivative of the unit tangent vector T'(s) with respect to arc length s, and ||dT/ds|| is the magnitude of the vector dT/ds.

Step by step solution

01

Find the tangent vector T(s)

To compute the principal unit normal vector, first find the tangent vector T(s) of the curve. The tangent vector is the derivative of the curve parameterized by arc length s, which means T(s) = r'(s), where r(s) is the position vector function that describes the curve.
02

Normalize the tangent vector

Ensure the tangent vector is a unit vector, meaning it has a magnitude of 1. Denote the unit tangent vector by T'(s) = T(s) / ||T(s)||, where ||T(s)|| is the magnitude of T(s). In general, T(s) is already a unit vector if parameterized by arc length, so this step might not be necessary.
03

Compute the derivative of the unit tangent vector T'(s) with respect to arc length s

Calculate the derivative of the unit tangent vector T'(s) with respect to the arc length parameter s, i.e., find dT/ds.
04

Find the principal unit normal vector N(s)

The principal unit normal vector N(s) is obtained by normalizing the derivative of the unit tangent vector T'(s). Calculate N(s) = dT/ds / ||dT/ds||, where ||dT/ds|| is the magnitude of the vector dT/ds. Now we have the practical formula for computing the principal unit normal vector, N(s) = dT/ds / ||dT/ds||, given a curve parameterized by arc length s.

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Most popular questions from this chapter

Suppose the vector-valued function \(\mathbf{r}(t)=\langle f(t), g(t), h(t)\rangle\) is smooth on an interval containing the point \(t_{0} .\) The line tangent to \(\mathbf{r}(t)\) at \(t=t_{0}\) is the line parallel to the tangent vector \(\mathbf{r}^{\prime}\left(t_{0}\right)\) that passes through \(\left(f\left(t_{0}\right), g\left(t_{0}\right), h\left(t_{0}\right)\right) .\) For each of the following functions, find an equation of the line tangent to the curve at \(t=t_{0} .\) Choose an orientation for the line that is the same as the direction of \(\mathbf{r}^{\prime}\). $$\mathbf{r}(t)=\left\langle e^{t}, e^{2 t}, e^{3 t}\right\rangle ; t_{0}=0$$

Cauchy-Schwarz Inequality The definition \(\mathbf{u} \cdot \mathbf{v}=|\mathbf{u}||\mathbf{v}| \cos \theta\) implies that \(|\mathbf{u} \cdot \mathbf{v}| \leq|\mathbf{u}||\mathbf{v}|\) (because \(|\cos \theta| \leq 1\) ). This inequality, known as the Cauchy-Schwarz Inequality, holds in any number of dimensions and has many consequences. Triangle Inequality Consider the vectors \(\mathbf{u}, \mathbf{v},\) and \(\mathbf{u}+\mathbf{v}\) (in any number of dimensions). Use the following steps to prove that \(|\mathbf{u}+\mathbf{v}| \leq|\mathbf{u}|+|\mathbf{v}|\) a. Show that \(|\mathbf{u}+\mathbf{v}|^{2}=(\mathbf{u}+\mathbf{v}) \cdot(\mathbf{u}+\mathbf{v})=|\mathbf{u}|^{2}+\) \(2 \mathbf{u} \cdot \mathbf{v}+|\mathbf{v}|^{2}\) b. Use the Cauchy-Schwarz Inequality to show that \(|\mathbf{u}+\mathbf{v}|^{2} \leq(|\mathbf{u}|+|\mathbf{v}|)^{2}\) c. Conclude that \(|\mathbf{u}+\mathbf{v}| \leq|\mathbf{u}|+|\mathbf{v}|\) d. Interpret the Triangle Inequality geometrically in \(\mathbb{R}^{2}\) or \(\mathbb{R}^{3}\).

Consider the curve \(\mathbf{r}(t)=(a \cos t+b \sin t) \mathbf{i}+(c \cos t+d \sin t) \mathbf{j}+(e \cos t+f \sin t) \mathbf{k}\) where \(a, b, c, d, e,\) and \(f\) are real numbers. It can be shown that this curve lies in a plane. Assuming the curve lies in a plane, show that it is a circle centered at the origin with radius \(R\) provided \(a^{2}+c^{2}+e^{2}=b^{2}+d^{2}+f^{2}=R^{2}\) and \(a b+c d+e f=0\).

Let \(\mathbf{u}(t)=\left\langle 1, t, t^{2}\right\rangle, \mathbf{v}(t)=\left\langle t^{2},-2 t, 1\right\rangle\) and \(g(t)=2 \sqrt{t}\). Compute the derivatives of the following functions. $$g(t) \mathbf{v}(t)$$

Prove the following vector properties using components. Then make a sketch to illustrate the property geometrically. Suppose \(\mathbf{u}, \mathbf{v},\) and \(\mathbf{w}\) are vectors in the \(x y\) -plane and a and \(c\) are scalars. $$\mathbf{u}+\mathbf{v}=\mathbf{v}+\mathbf{u}$$
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