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Chapter 12: Problem 24

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

Short Answer

Expert verified
Question: Find the unit tangent vector for the given parametric curve in the interval \(0 \leq t \leq 2 \pi\): \(\mathbf{r}(t) = \langle\sin t, \cos t, \cos t\rangle\). Answer: \(\mathbf{T}(t) = \frac{\langle\cos t, -\sin t, -\sin t\rangle}{\sqrt{1 + \sin^2 t}}, \quad 0 \leq t \leq 2 \pi\)

Step by step solution

01

Differentiating the parametric function

Given the parametric function: $$\mathbf{r}(t) = \langle\sin t, \cos t, \cos t\rangle$$ Differentiate each component with respect to \(t\) to get the velocity vector: $$\mathbf{v}(t) = \frac{d\mathbf{r}(t)}{dt} = \langle\cos t, -\sin t, -\sin t\rangle$$
02

Finding the length of the velocity vector

To find the length of the velocity vector \(\mathbf{v}(t)\), use the formula: $$||\mathbf{v}(t)|| = \sqrt{(\cos t)^2 + (-\sin t)^2 + (-\sin t)^2}$$ Substitute the values and simplify: $$||\mathbf{v}(t)|| = \sqrt{\cos^2 t + \sin^2 t + \sin^2 t}$$ Given that \(\sin^2 t + \cos^2 t = 1\), we can simplify the expression further: $$||\mathbf{v}(t)|| = \sqrt{1 + \sin^2 t}$$
03

Normalizing the velocity vector

Now, divide the velocity vector by its magnitude to find the unit tangent vector: $$\mathbf{T}(t) = \frac{\mathbf{v}(t)}{||\mathbf{v}(t)||} = \frac{\langle\cos t, -\sin t, -\sin t\rangle}{\sqrt{1 + \sin^2 t}}$$
04

Expressing the result for the given interval

Finally, the unit tangent vector \(\mathbf{T}(t)\) for the given parametric curve in the interval \(0 \leq t \leq 2 \pi\) is: $$\mathbf{T}(t) = \frac{\langle\cos t, -\sin t, -\sin t\rangle}{\sqrt{1 + \sin^2 t}}, \quad 0 \leq t \leq 2 \pi$$

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