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Chapter 12: Problem 8

Is the curve \(\mathbf{r}(t)=\langle\cos t, \sin t\rangle\) parameterized by its arc length? Explain.

Short Answer

Expert verified
Answer: Yes, the curve 饾憻(t) = 鉄╟os t, sin t鉄� is parameterized by its arc length, as the length of its tangent vector |r'(t)| is equal to 1 at all points on the curve.

Step by step solution

01

Compute the derivative of the vector function r(t)

First, compute the derivative of the vector function r(t). The derivative of a vector function is found by taking the derivative of each component with respect to the parameter t: $$ \mathbf{r}'(t) = \left\langle \frac{d}{dt}\cos t, \frac{d}{dt}\sin t\right\rangle. $$
02

Calculate the derivatives of cos(t) and sin(t)

Now, we need to find the derivatives of the individual components, starting with the derivative of cos(t): $$ \frac{d}{dt}\cos t = -\sin t. $$ Next, find the derivative of sin(t): $$ \frac{d}{dt}\sin t = \cos t. $$
03

Plug the derivatives into the r'(t) vector

Using the derivatives of the components, plug them back into the r'(t) vector: $$ \mathbf{r}'(t) = \langle -\sin t, \cos t \rangle. $$
04

Find the magnitude of r'(t)

Now, we need to calculate the length of the tangent vector |r'(t)|. The magnitude of a vector is calculated using the formula: $$ |\mathbf{r}'(t)| = \sqrt{(-\sin t)^2 + (\cos t)^2}. $$
05

Simplify the magnitude expression

Now, square each of the components in the magnitude expression and simplify: $$ |\mathbf{r}'(t)| = \sqrt{(\sin^2 t) + (\cos^2 t)}. $$ Using the trigonometric identity sin虏t + cos虏t = 1, the expression simplifies to: $$ |\mathbf{r}'(t)| = \sqrt{1}. $$ So, $$ |\mathbf{r}'(t)| = 1. $$ Since the length of the tangent vector |r'(t)| is equal to 1 at all points on the curve, we can conclude that the curve饾憻(t)=鉄╟os t, sin t鉄� is parameterized by its arc length.

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Most popular questions from this chapter

Consider the curve described by the vector function \(\mathbf{r}(t)=\left(50 e^{-t} \cos t\right) \mathbf{i}+\left(50 e^{-t} \sin t\right) \mathbf{j}+\left(5-5 e^{-t}\right) \mathbf{k},\) for \(t \geq 0\). a. What is the initial point of the path corresponding to \(\mathbf{r}(0) ?\) b. What is \(\lim _{t \rightarrow \infty} \mathbf{r}(t) ?\) c. Sketch the curve. d. Eliminate the parameter \(t\) to show that \(z=5-r / 10\), where \(r^{2}=x^{2}+y^{2}\).

Let \(D\) be a solid heat-conducting cube formed by the planes \(x=0, x=1, y=0, y=1, z=0,\) and \(z=1 .\) The heat flow at every point of \(D\) is given by the constant vector \(\mathbf{Q}=\langle 0,2,1\rangle\) a. Through which faces of \(D\) does \(Q\) point into \(D ?\) b. Through which faces of \(D\) does \(\mathbf{Q}\) point out of \(D ?\) c. On which faces of \(D\) is \(Q\) tangential to \(D\) (pointing neither in nor out of \(D\) )? d. Find the scalar component of \(\mathbf{Q}\) normal to the face \(x=0\). e. Find the scalar component of \(\mathbf{Q}\) normal to the face \(z=1\). f. Find the scalar component of \(\mathbf{Q}\) normal to the face \(y=0\).

a. Show that \((\mathbf{u}+\mathbf{v}) \cdot(\mathbf{u}+\mathbf{v})=|\mathbf{u}|^{2}+2 \mathbf{u} \cdot \mathbf{v}+|\mathbf{v}|^{2}\). b. Show that \((\mathbf{u}+\mathbf{v}) \cdot(\mathbf{u}+\mathbf{v})=|\mathbf{u}|^{2}+|\mathbf{v}|^{2}\) if \(\mathbf{u}\) is perpendicular to \(\mathbf{v}\). c. Show that \((\mathbf{u}+\mathbf{v}) \cdot(\mathbf{u}-\mathbf{v})=|\mathbf{u}|^{2}-|\mathbf{v}|^{2}\).

Given a fixed vector \(\mathbf{v},\) there is an infinite set of vectors \(\mathbf{u}\) with the same value of proj\(_{\mathbf{v}} \mathbf{u}\). Let \(\mathbf{v}=\langle 0,0,1\rangle .\) Give a description of all position vectors \(\mathbf{u}\) such that \(\operatorname{proj}_{\mathbf{v}} \mathbf{u}=\operatorname{proj}_{\mathbf{v}}\langle 1,2,3\rangle\).

Find the domains of the following vector-valued functions. $$\mathbf{r}(t)=\sqrt{4-t^{2}} \mathbf{i}+\sqrt{t} \mathbf{j}-\frac{2}{\sqrt{1+t}} \mathbf{k}$$
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