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Chapter 12: Problem 12

Find the unit tangent vector \(\mathbf{T}\) and the curvature \(\kappa\) for the following parameterized curves. $$\mathbf{r}(t)=\langle 2 \cos t,-2 \sin t\rangle$$

Short Answer

Expert verified
Question: Find the unit tangent vector and the curvature for the parametrized curve given by the vector function \(\mathbf{r}(t) = \langle 2\cos t, -2\sin t\rangle\). Answer: The unit tangent vector \(\mathbf{T}(t)\) is \(\langle -\sin t, -\cos t\rangle\) and the curvature \(\kappa(t)\) is \(\frac{1}{2}\).

Step by step solution

01

Find the derivatives of the vector function

To find the derivatives, we differentiate each component of the vector function with respect to the parameter \(t\). So we have: $$\mathbf{r'}(t) = \left\langle \frac{d}{dt}(2 \cos t), \frac{d}{dt}(-2 \sin t)\right\rangle = \langle -2\sin t, -2\cos t\rangle$$
02

Calculate the magnitude of the first derivative

Before finding the unit tangent vector, we need to determine the magnitude of the first derivative: $$\|\mathbf{r'}(t)\| = \sqrt{(-2\sin t)^2 + (-2\cos t)^2} = \sqrt{4\sin^2t + 4\cos^2t} = 2$$
03

Find the unit tangent vector \(\mathbf{T}(t)\)

Now, we can compute the unit tangent vector using the formula \(\mathbf{T}(t)=\frac{\mathbf{r'}(t)}{\|\mathbf{r'}(t)\|}\): $$\mathbf{T}(t) = \frac{\langle -2\sin t, -2\cos t\rangle}{2} = \langle -\sin t, -\cos t\rangle$$
04

Find the derivative of the unit tangent vector

We differentiate the unit tangent vector with respect to \(t\) to find \(\mathbf{T'}(t)\): $$\mathbf{T'}(t) = \left\langle \frac{d}{dt}(-\sin t), \frac{d}{dt}(-\cos t)\right\rangle = \langle -\cos t, \sin t\rangle$$
05

Calculate the magnitude of the second derivative

We calculate the magnitude of \(\mathbf{T'}(t)\) as follows: $$\|\mathbf{T'}(t)\| = \sqrt{(-\cos t)^2 + (\sin t)^2} = \sqrt{\sin^2t + \cos^2t} = 1$$
06

Find the curvature \(\kappa(t)\)

Finally, we can find the curvature using the formula \(\kappa(t) = \frac{\|\mathbf{T'}(t)\|}{\|\mathbf{r'}(t)\|}\): $$\kappa(t) = \frac{1}{2}$$ Thus, the unit tangent vector \(\mathbf{T}(t)\) is \(\langle -\sin t, -\cos t\rangle\) and the curvature \(\kappa(t)\) is \(\frac{1}{2}\).

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Most popular questions from this chapter

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

\(\mathbb{R}^{2}\) Consider the vectors \(\mathbf{I}=\langle 1 / \sqrt{2}, 1 / \sqrt{2}\rangle\) and \(\mathbf{J}=\langle-1 / \sqrt{2}, 1 / \sqrt{2}\rangle\). Show that \(\mathbf{I}\) and \(\mathbf{J}\) are orthogonal unit vectors.

Determine the equation of the line that is perpendicular to the lines \(\mathbf{r}(t)=\langle 4 t, 1+2 t, 3 t\rangle\) and \(\mathbf{R}(s)=\langle-1+s,-7+2 s,-12+3 s\rangle\) and passes through the point of intersection of the lines \(\mathbf{r}\) and \(\mathbf{R}\).

Zero curvature Prove that the curve $$ \mathbf{r}(t)=\left\langle a+b t^{p}, c+d t^{p}, e+f t^{p}\right\rangle $$ where \(a, b, c, d, e,\) and \(f\) are real numbers and \(p\) is a positive integer, has zero curvature. Give an explanation.

Find the points (if they exist) at which the following planes and curves intersect. $$z=16 ; \mathbf{r}(t)=\langle t, 2 t, 4+3 t\rangle, \text { for }-\infty < t < \infty$$
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