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Find the unit tangent vector at the given value of t for the following parameterized curves. $$\mathbf{r}(t)=\langle 6 t, 6,3 / t\rangle, \text { for } 0

Short Answer

Expert verified
Answer: The unit tangent vector at t=1 is: ⟨2/√5, 0, -1/√5⟩.

Step by step solution

01

Find the velocity vector

Differentiate r(t) with respect to t to get the velocity vector: \(\mathbf{r}(t)=\langle 6 t, 6, \frac{3}{t} \rangle\) \(\frac{d\mathbf{r}(t)}{dt} = \langle\frac{d(6t)}{dt},\frac{d(6)}{dt},\frac{d(3/t)}{dt}\rangle\)
02

Evaluate the velocity vector at the given value of t

The given value of t is 1: \(\frac{d\mathbf{r}(t)}{dt} = \langle 6, 0, -3 \rangle\) at \(t=1\)
03

Normalize the resulting vector

To normalize the vector, find its magnitude and divide each component by the magnitude: Magnitude of the velocity vector: \(||\mathbf{v}|| = \sqrt{(6^2 + 0^2 + (-3)^2)} = 3\sqrt{5}\) Normalized velocity vector (unit tangent vector): \(\mathbf{T}(t) = \frac{\mathbf{v}}{||\mathbf{v}||}=\left\langle\frac{6}{3\sqrt{5}},\frac{0}{3\sqrt{5}},\frac{-3}{3\sqrt{5}}\right\rangle\) So, the unit tangent vector at \(t=1\) is: \(\mathbf{T}(1) = \langle \frac{2}{\sqrt{5}}, 0, -\frac{1}{\sqrt{5}}\rangle\).

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