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Chapter 12: Problem 28

Find the unit tangent vector at the given value of t for the following parameterized curves. $$\mathbf{r}(t)=\left\langle\sin t, \cos t, e^{-t}\right\rangle, \text { for } 0 \leq t \leq \pi ; t=0$$

Short Answer

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

Step by step solution

01

Find the derivative of the parameterized curve with respect to t

To find the tangent vector, we'll take the first derivative of the position vector function, \(\mathbf{r}(t)\), with respect to t as follows: $$\mathbf{r}'(t)=\left\langle\frac{d}{dt}\sin t, \frac{d}{dt}\cos t, \frac{d}{dt}e^{-t}\right\rangle$$ We'll now find the derivative of each component: $$\frac{d}{dt}\sin t = \cos t$$ $$\frac{d}{dt}\cos t = -\sin t$$ $$\frac{d}{dt}e^{-t} = -e^{-t}$$ Thus, $$\mathbf{r}'(t)=\left\langle\cos t, -\sin t, -e^{-t}\right\rangle$$
02

Evaluate the tangent vector at t=0

We will now evaluate the tangent vector \(\mathbf{r}'(t)\) at t=0: $$\mathbf{r}'(0)=\left\langle\cos 0, -\sin 0, -e^{0}\right\rangle$$ Simplifying the values, we get: $$\mathbf{r}'(0)=\left\langle 1, 0, -1 \right\rangle$$
03

Normalize the tangent vector

Lastly, we will normalize the tangent vector by dividing it by its magnitude. First, let's find the magnitude of the tangent vector: $$\|\mathbf{r}'(0)\|= \sqrt{(1)^2 + (0)^2 + (-1)^2} = \sqrt{2}$$ Now, we divide the tangent vector by its magnitude to obtain the unit tangent vector: $$\mathbf{T}(t=\!0)=\frac{\mathbf{r}'(0)}{\|\mathbf{r}'(0)\|}$$ Substituting the values, we obtain the unit tangent vector: $$\mathbf{T}(t=\!0)=\frac{\left\langle 1, 0, -1 \right\rangle}{\sqrt{2}}=\left\langle \frac{1}{\sqrt{2}}, 0, -\frac{1}{\sqrt{2}} \right\rangle$$ So, the unit tangent vector at t=0 for the given parameterized curve is: $$\mathbf{T}(t=\!0)=\left\langle \frac{1}{\sqrt{2}}, 0, -\frac{1}{\sqrt{2}} \right\rangle$$

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Most popular questions from this chapter

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