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Chapter 12: Problem 22

Find the unit tangent vector for the following parameterized curves. $$\mathbf{r}(t)=\langle\cos t, \sin t, 2\rangle, \text { for } 0 \leq t \leq 2 \pi$$

Short Answer

Expert verified
Answer: The unit tangent vector for the given parameterized curve is $$\mathbf{T}(t) = \langle -\sin t, \cos t, 0 \rangle$$.

Step by step solution

01

Find the derivative of the parameterized curve

To find the derivative of the parameterized curve with respect to t, we differentiate each component separately with respect to t. The parameterized curve is given by: $$\mathbf{r}(t) = \langle \cos t, \sin t, 2 \rangle.$$ The derivative of the curve is $$\mathbf{r'}(t) = \langle -\sin t, \cos t, 0 \rangle.$$
02

Compute the magnitude of the derivative

The magnitude of a vector is given by the square root of the sum of the squares of its components. So, the magnitude of the derivative is given by: $$|\mathbf{r'}(t)| = \sqrt{(-\sin t)^2 + (\cos t)^2 + 0^2} = \sqrt{\sin^2 t + \cos^2 t}.$$ Using the trigonometric identity \(\sin^2 t + \cos^2 t = 1\), we have $$|\mathbf{r'}(t)| = \sqrt{1} = 1.$$
03

Normalize the derivative

To normalize the derivative, we divide it by its magnitude. Since the magnitude of the derivative is 1, the normalized derivative is the same as the derivative itself: $$\mathbf{T}(t) = \frac{\mathbf{r'}(t)}{|\mathbf{r'}(t)|} = \langle -\sin t, \cos t, 0 \rangle.$$ The unit tangent vector for the given parameterized curve is$$\mathbf{T}(t) = \langle -\sin t, \cos t, 0 \rangle.$$

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