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Chapter 12: Problem 15

Find the unit tangent vector \(\mathbf{T}\) and the curvature \(\kappa\) for the following parameterized curves. $$\mathbf{r}(t)=\langle\sqrt{3} \sin t, \sin t, 2 \cos t\rangle$$

Short Answer

Expert verified
Answer: The unit tangent vector \(\mathbf{T}\) is given by: \(\mathbf{T} = \frac{1}{\sqrt{3 \cos^2 t + \cos^2 t + 4 \sin^2 t}} \langle\sqrt{3} \cos t, \cos t, -2 \sin t\rangle\) The curvature \(\kappa\) is given by: \(\kappa = \sqrt{\left(\frac{d}{dt}\left(\frac{\sqrt{3}\cos t}{\sqrt{3 \cos^2 t + \cos^2 t + 4 \sin^2 t}}\right)\right)^2+\left(\frac{d}{dt}\left(\frac{\cos t}{\sqrt{3 \cos^2 t + \cos^2 t + 4 \sin^2 t}}\right)\right)^2+\left(\frac{d}{dt}\left(\frac{-2\sin t}{\sqrt{3 \cos^2 t + \cos^2 t + 4 \sin^2 t}}\right)\right)^2}\)

Step by step solution

01

Compute the derivative of \(\mathbf{r}\) with respect to \(t\)

To find the derivative, we differentiate each component of \(\mathbf{r}(t)\) with respect to \(t\). $$\frac{d\mathbf{r}}{dt} = \left\langle\frac{d}{dt}(\sqrt{3} \sin t), \frac{d}{dt}(\sin t), \frac{d}{dt}(2 \cos t)\right\rangle$$ $$\frac{d\mathbf{r}}{dt} = \langle\sqrt{3} \cos t, \cos t, -2 \sin t\rangle$$
02

Find the magnitude of the derivative

The magnitude of a vector \(\mathbf{v} = \langle a, b, c \rangle\) is given by $$||\mathbf{v}|| = \sqrt{a^2 + b^2 + c^2}$$ We apply this to our derivative: $$||\frac{d\mathbf{r}}{dt}|| = \sqrt{(\sqrt{3} \cos t)^2 + (\cos t)^2 + (-2 \sin t)^2}$$ $$||\frac{d\mathbf{r}}{dt}|| = \sqrt{3 \cos^2 t + \cos^2 t + 4 \sin^2 t}$$
03

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

The unit tangent vector is given by $$\mathbf{T} = \frac{1}{||\frac{d\mathbf{r}}{dt}||} \left( \frac{d\mathbf{r}}{dt} \right)$$ $$\mathbf{T} = \frac{1}{\sqrt{3 \cos^2 t + \cos^2 t + 4 \sin^2 t}} \langle\sqrt{3} \cos t, \cos t, -2 \sin t\rangle$$
04

Compute the derivative of \(\mathbf{T}\) with respect to \(t\)

Differentiating each component of \(\mathbf{T}(t)\) with respect to \(t\) yields: $$\frac{d\mathbf{T}}{dt} = \left\langle \frac{d}{dt}\left(\frac{\sqrt{3}\cos t}{\sqrt{3 \cos^2 t + \cos^2 t + 4 \sin^2 t}}\right),\frac{d}{dt}\left(\frac{\cos t}{\sqrt{3 \cos^2 t + \cos^2 t + 4 \sin^2 t}}\right),\frac{d}{dt}\left(\frac{-2\sin t}{\sqrt{3 \cos^2 t + \cos^2 t + 4 \sin^2 t}}\right) \right\rangle$$
05

Find the magnitude of the second derivative

Given the complexity of the expressions for each component, we can only symbolically represent the second derivative's magnitude: $$||\frac{d\mathbf{T}}{dt}|| = \sqrt{\left(\frac{d}{dt}\left(\frac{\sqrt{3}\cos t}{\sqrt{3 \cos^2 t + \cos^2 t + 4 \sin^2 t}}\right)\right)^2+\left(\frac{d}{dt}\left(\frac{\cos t}{\sqrt{3 \cos^2 t + \cos^2 t + 4 \sin^2 t}}\right)\right)^2+\left(\frac{d}{dt}\left(\frac{-2\sin t}{\sqrt{3 \cos^2 t + \cos^2 t + 4 \sin^2 t}}\right)\right)^2}$$
06

Compute the curvature \(\kappa\)

The curvature \(\kappa\) is given by the formula $$\kappa = ||\frac{d\mathbf{T}}{dt}||$$ substituting the result from step 5, we have: $$\kappa = \sqrt{\left(\frac{d}{dt}\left(\frac{\sqrt{3}\cos t}{\sqrt{3 \cos^2 t + \cos^2 t + 4 \sin^2 t}}\right)\right)^2+\left(\frac{d}{dt}\left(\frac{\cos t}{\sqrt{3 \cos^2 t + \cos^2 t + 4 \sin^2 t}}\right)\right)^2+\left(\frac{d}{dt}\left(\frac{-2\sin t}{\sqrt{3 \cos^2 t + \cos^2 t + 4 \sin^2 t}}\right)\right)^2}$$

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