10b **Position Vector of a Particle in Space**Given the position vector **$\mathbf{r}(t) = \langle t+1, \sqrt{t}, \frac{1}{t} \rangle$** of a particle in space at time $t$.- **Components Explanation**: - The **x-component** $ t + 1 $ represents the position in the x-direction, where the displacement increases linearly as time $ t $ increases. - The **y-component** $ \sqrt{t} $ shows a square root relationship with time, indicating slower growth as time progresses. - The **z-component** $ \frac{1}{t} $ is an inverse relation where the position decreases as time $ t $ increases, moving closer to the x-y plane.Use this vector to analyze the particle's trajectory and behavior over time. 10b. Find the parametric equations of the tangent line to the curve at $ t = 1 $.

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Description: 10b **Position Vector of a Particle in Space**Given the position vector **$\mathbf{r}(t) = \langle t+1, \sqrt{t}, \frac{1}{t} \rangle$** of a particle in space at time $t$.- **Components Explanation**: - The **x-component** $ t + 1 $ represent

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Calculus: Early Transcendentals

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ISBN:9781285741550

Author:James Stewart

Publisher:James Stewart

Chapter1: Functions And Models

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Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...

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10b

**Position Vector of a Particle in Space**

Given the position vector **$\mathbf{r}(t) = \langle t+1, \sqrt{t}, \frac{1}{t} \rangle$** of a particle in space at time $t$.

- **Components Explanation**:
- The **x-component** $ t + 1 $ represents the position in the x-direction, where the displacement increases linearly as time $ t $ increases.
- The **y-component** $ \sqrt{t} $ shows a square root relationship with time, indicating slower growth as time progresses.
- The **z-component** $ \frac{1}{t} $ is an inverse relation where the position decreases as time $ t $ increases, moving closer to the x-y plane.

Use this vector to analyze the particle's trajectory and behavior over time.

10b. Find the parametric equations of the tangent line to the curve at $ t = 1 $.

Expert Solution

Step 1

Given:

The position vector is of a particle in space at a time t :

$\vec{r} (t) = < t + 1, \sqrt{t}, \frac{1}{t} > = < x, y, z > x = t + 1; y = \sqrt{t}; z = \frac{1}{t}$

Differentiate $\vec{r} (t)$ with respect to t,

$\frac{d}{d t} (\vec{r} (t)) = < \frac{d}{d t} (t + 1), \frac{d}{d t} (\sqrt{t}), \frac{d}{d t} (\frac{1}{t}) > \vec{r}' (t) = < 1, \frac{1}{2 \sqrt{t}}, \frac{- 1}{t^{2}} >$

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