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Chapter 12: Problem 27

Find the unit tangent vector at the given value of t for the following parameterized curves. $$\mathbf{r}(t)=\langle\cos 2 t, 4,3 \sin 2 t\rangle, \text { for } 0 \leq t \leq \pi ; t=\pi / 2$$

Short Answer

Expert verified
Answer: The unit tangent vector at \(t=\frac{\pi}{2}\) for the given parameterized curve is \(\mathbf{T}(t) = \langle 0, 0, -1 \rangle\).

Step by step solution

01

Compute the derivative of the parameterized curve

First, we need to compute the derivative of the given parameterized curve with respect to t. The parameterized curve is given by: $$\mathbf{r}(t)=\langle\cos 2 t, 4,3 \sin 2 t\rangle$$ Calculate the derivative: $$\frac{d\mathbf{r}}{dt} = \langle -2\sin{2t}, 0, 6\cos{2t} \rangle$$
02

Find the tangent vector at the given value of t

Now plug in the given value of t, which is \(t=\frac{\pi}{2}\), into the derivative of the parameterized curve. $$\frac{d\mathbf{r}}{dt}\Big|_{t=\frac{\pi}{2}} = \langle -2\sin{2\frac{\pi}{2}}, 0, 6\cos{2\frac{\pi}{2}} \rangle = \langle -2\sin{\pi}, 0, 6\cos{\pi} \rangle = \langle 0, 0, -6 \rangle$$ Now we have the tangent vector at \(t=\frac{\pi}{2}\).
03

Normalize the tangent vector

Normalize the tangent vector by dividing it by its magnitude. $$\text{Magnitude} = \|\langle 0, 0, -6 \rangle\| = \sqrt{0^2 + 0^2 + (-6)^2} = 6$$ Divide the tangent vector by the magnitude to find the unit tangent vector: $$\mathbf{T}(t) = \frac{\langle 0, 0, -6 \rangle}{6} = \langle 0, 0, -1 \rangle$$ So the unit tangent vector at \(t=\frac{\pi}{2}\) for the given parameterized curve is \(\mathbf{T}(t) = \langle 0, 0, -1 \rangle\).

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Most popular questions from this chapter

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