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Chapter 12: Problem 25

Find the unit tangent vector for the following parameterized curves. $$\mathbf{r}(t)=\langle t, 2,2 / t\rangle, \text { for } t \geq 1$$

Short Answer

Expert verified
Question: Find the unit tangent vector for the parameterized curve r(t) = 鉄╰, 2, 2/t鉄�, where t 鈮� 1. Answer: The unit tangent vector for the given parameterized curve is: $$\mathbf{T}_{unit}(t) = \frac{\langle 1, 0, -\frac{2}{t^2} \rangle}{\sqrt{ 1 + \left(\frac{2}{t^{2}}\right)^2}}$$

Step by step solution

01

Define the given position vector

The position vector r(t) is given by: $$\mathbf{r}(t) = \langle t, 2, \frac{2}{t} \rangle, \quad t \geq 1$$
02

Find the derivative of the position vector

To find the tangent vector, we need to find the derivative of r(t) with respect to t. Let's differentiate each component of r(t): $$\frac{d\mathbf{r}(t)}{dt} = \left\langle \frac{dt}{dt}, \frac{d(2)}{dt}, \frac{d(\frac{2}{t})}{dt} \right\rangle$$ Now, differentiate each component: $$\frac{d\mathbf{r}(t)}{dt} = \left\langle 1, 0, -\frac{2}{t^2} \right\rangle$$ The derivative gives us the tangent vector for the curve.
03

Calculate the magnitude of the tangent vector

To find the unit tangent vector, we first need to find the magnitude of the tangent vector. The magnitude is given by: $$\lVert \mathbf{T}(t) \rVert = \sqrt{(1)^2 + (0)^2 + \left( -\frac{2}{t^2} \right)^2}$$ Calculate the magnitude: $$\lVert \mathbf{T}(t) \rVert = \sqrt{ 1 + \left(\frac{2}{t^{2}}\right)^2}$$
04

Normalize the tangent vector

Now that we have the magnitude of the tangent vector, we can normalize it to obtain the unit tangent vector. Normalize the tangent vector using the formula: $$\mathbf{T}_{unit}(t) = \frac{\mathbf{T}(t)}{\lVert \mathbf{T}(t) \rVert}$$ Substitute the tangent vector and its magnitude into the formula: $$\mathbf{T}_{unit}(t) = \frac{\langle 1, 0, -\frac{2}{t^2} \rangle}{\sqrt{ 1 + \left(\frac{2}{t^{2}}\right)^2}}$$ The unit tangent vector for the given parameterized curve is: $$\mathbf{T}_{unit}(t) = \frac{\langle 1, 0, -\frac{2}{t^2} \rangle}{\sqrt{ 1 + \left(\frac{2}{t^{2}}\right)^2}}$$

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