/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 19 Find the unit tangent vector \(\... [FREE SOLUTION] | 91Ó°ÊÓ

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Find the unit tangent vector \(\mathbf{T}\) and the curvature \(\kappa\) for the following parameterized curves. $$\mathbf{r}(t)=\left\langle\int_{0}^{t} \cos \left(\pi u^{2} / 2\right) d u, \int_{0}^{t} \sin \left(\pi u^{2} / 2\right) d u\right\rangle, t>0$$

Short Answer

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Question: Given the parameterized curve of a plane, \(\mathbf{r}(t) = \left\langle \int_0^t \cos\left(\frac{\pi u^2}{2}\right) du, \int_0^t \sin\left(\frac{\pi u^2}{2}\right) du \right\rangle\), find the expressions for the unit tangent vector \(\mathbf{T}\) and the curvature \(\kappa\). Answer: For the given parameterized curve, the unit tangent vector \(\mathbf{T}\) and curvature \(\kappa\) are: $$\mathbf{T} = \left\langle \cos\left(\frac{\pi t^2}{2}\right), \sin\left(\frac{\pi t^2}{2}\right) \right\rangle$$ $$\kappa = \pi t$$

Step by step solution

01

Find \(\mathbf{r}'(t)\) and \(\mathbf{r}''(t)\)

To calculate the first and second derivatives of the parameterized curve, we differentiate the components of the curve with respect to \(t\). $$\mathbf{r}'(t)=\left\langle \frac{d}{dt}\int_{0}^{t} \cos \left(\pi u^{2} / 2\right) d u, \frac{d}{dt}\int_{0}^{t} \sin \left(\pi u^{2} / 2\right) d u\right\rangle$$ By the Fundamental Theorem of Calculus, the derivatives with respect to \(t\) are: $$\mathbf{r}'(t)=\left\langle\cos(\pi t^{2} / 2), \sin(\pi t^{2} / 2)\right\rangle$$ Similarly, we differentiate once more to find \(\mathbf{r}''(t)\): $$\mathbf{r}''(t)=\left\langle -\pi t \sin(\pi t^{2} / 2), \pi t\cos(\pi t^{2} / 2)\right\rangle$$
02

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

To find the unit tangent vector \(\mathbf{T}\), we have to normalize the first derivative: $$\mathbf{T}=\frac{\mathbf{r}'(t)}{||\mathbf{r}'(t)||}$$ Calculate the magnitude of \(\mathbf{r}'(t)\): $$||\mathbf{r}'(t)||=\sqrt{\cos^2(\pi t^{2} / 2)+\sin^2(\pi t^{2} / 2)}=1$$ Then, the unit tangent vector \(\mathbf{T}\) is: $$\mathbf{T}=\mathbf{r}'(t)=\left\langle\cos(\pi t^{2} / 2), \sin(\pi t^{2} / 2)\right\rangle$$
03

Find the curvature \(\kappa\)

To find the curvature \(\kappa\), we use the following formula: $$\kappa = \frac{ || \mathbf{r}'(t) \times \mathbf{r}''(t) || }{ ||\mathbf{r}'(t)||^{3} }$$ First, compute the cross product of \(\mathbf{r}'(t)\) and \(\mathbf{r}''(t)\), noting that these are 2-dimensional vectors, and so the cross product reduces to the scalar quantity, given by the determinant of the matrix formed from inserting the vectors as columns. $$\mathbf{r}'(t) \times \mathbf{r}''(t) = \begin{vmatrix} \cos(\pi t^{2} / 2) & -\pi t\sin(\pi t^{2} / 2) \\ \sin(\pi t^{2} / 2) & \pi t\cos(\pi t^{2} / 2) \end{vmatrix}=\pi t$$ Now, compute the magnitude of \(||\mathbf{r}'(t)||^{3}\): $$||\mathbf{r}'(t)||^{3}=(1)^{3}=1$$ Finally, compute the curvature \(\kappa\): $$\kappa=\frac{|\pi t|}{1}=\pi t$$ So, for the given parameterized curve, the unit tangent vector \(\mathbf{T}\) and curvature \(\kappa\) are: $$\mathbf{T}=\left\langle\cos(\pi t^{2} / 2), \sin(\pi t^{2} / 2)\right\rangle$$ $$\kappa=\pi t$$

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