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Chapter 12: Problem 21

Find the unit tangent vector for the following parameterized curves. $$\mathbf{r}(t)=\langle 2 t, 2 t, t\rangle, \text { for } 0 \leq t \leq 1$$

Short Answer

Expert verified
Answer: The unit tangent vector for the given parameterized curve is $$\mathbf{T}(t) = \langle \frac{2}{3}, \frac{2}{3}, \frac{1}{3} \rangle$$.

Step by step solution

01

Find the derivative of the curve with respect to t

To find the tangent vector, we need to differentiate the given curve with respect to t: $$\mathbf{r}(t)=\langle 2 t, 2 t, t\rangle$$ Differentiating each component with respect to t, $$\frac{\text{d}\mathbf{r}}{\text{d}t} = \langle 2, 2, 1 \rangle$$
02

Find the magnitude of the tangent vector

To normalize the tangent vector, we need to find its magnitude. The magnitude of a vector 鉄╝, b, c鉄� is given by: $$||\mathbf{v}|| = \sqrt{a^2 + b^2 + c^2}$$ For our tangent vector 鉄�2, 2, 1鉄�, the magnitude is: $$||\frac{\text{d}\mathbf{r}}{\text{d}t}|| = \sqrt{2^2 + 2^2 + 1^2} = \sqrt{9} = 3$$
03

Normalize the tangent vector to get the unit tangent vector

To normalize the tangent vector, we divide it by its magnitude: $$\mathbf{T}(t) = \frac{\dfrac{\text{d}\mathbf{r}}{\text{d}t}}{||\dfrac{\text{d}\mathbf{r}}{\text{d}t}||} = \frac{\langle 2, 2, 1 \rangle}{3}$$ So the unit tangent vector for the given parameterized curve is: $$\mathbf{T}(t) = \langle \frac{2}{3}, \frac{2}{3}, \frac{1}{3} \rangle$$

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