Problem 17 - Calculus Early Transcendentals

Question

Find the unit tangent vector \(\mathbf{T}(t)\) at the point with the given value of the parameter \(t .\) $$ \mathbf{r}(t)=\left\langle t e^{-t}, 2 \text { arctan } t, 2 e^{t}\right\rangle, \quad t=0 $$

Step-by-Step Solution

Answer

The unit tangent vector \( \mathbf{T}(0) \) is \( \left\langle \frac{1}{3}, \frac{2}{3}, \frac{2}{3} \right\rangle \).

Step 1: Differentiate the Position Vector

To find the unit tangent vector, first compute the derivative of the position vector \( \mathbf{r}(t) \). Differentiate each component separately: \( \mathbf{r}'(t) = \left\langle \frac{d}{dt}(t e^{-t}), \frac{d}{dt}(2 \text{ arctan } t), \frac{d}{dt}(2 e^t) \right\rangle \).Calculating these derivatives gives:- The first component: \( \frac{d}{dt}(t e^{-t}) = e^{-t} - te^{-t} \),- The second component: \( \frac{d}{dt}(2 \text{ arctan } t) = \frac{2}{1+t^2} \),- The third component: \( \frac{d}{dt}(2 e^t) = 2e^t \).Thus, \( \mathbf{r}'(t) = \left\langle e^{-t} - te^{-t}, \frac{2}{1+t^2}, 2e^t \right\rangle \).

Step 2: Evaluate the Derivative at \( t=0 \)

Substitute \( t=0 \) into the derivative \( \mathbf{r}'(t) \) to find \( \mathbf{r}'(0) \):\( \mathbf{r}'(0) = \left\langle e^0 - 0, \frac{2}{1+0^2}, 2e^0 \right\rangle = \left\langle 1, 2, 2 \right\rangle \).

Step 3: Compute the Magnitude of \( \mathbf{r}'(0) \)

Find the magnitude of the vector \( \mathbf{r}'(0) = \left\langle 1, 2, 2 \right\rangle \). The formula for the magnitude of a vector \( \left\langle a, b, c \right\rangle \) is \( \sqrt{a^2 + b^2 + c^2} \):\[ \| \mathbf{r}'(0) \| = \sqrt{1^2 + 2^2 + 2^2} = \sqrt{9} = 3. \]

Step 4: Calculate the Unit Tangent Vector \( \mathbf{T}(t) \)

To find the unit tangent vector \( \mathbf{T}(0) \), divide \( \mathbf{r}'(0) \) by its magnitude:\[ \mathbf{T}(0) = \frac{1}{3} \left\langle 1, 2, 2 \right\rangle = \left\langle \frac{1}{3}, \frac{2}{3}, \frac{2}{3} \right\rangle. \]

Key Concepts

Differentiation

Differentiation is a core concept in calculus that involves finding the derivative of a function. Essentially, the derivative gives us the rate at which a function changes at any given point. This is crucial when we are working with vectors because it helps us understand how the position vector of a moving particle changes over time.
In the context of finding a unit tangent vector, differentiation allows us to determine the velocity vector of our original position vector. We do this by taking the derivative of each component of the position vector separately. This means if our position is determined by a vector with components like \((x(t), y(t), z(t))\), we find their respective derivatives \(x'(t), y'(t), \) and \(z'(t)\) to form the velocity vector.
The derivation of these components follows standard calculus rules and can involve simple polynomial differentiation, trigonometric derivatives, or even more complex differentiation techniques.

Position Vector

A position vector is a mathematical entity that represents the position of a point in space relative to an origin. In 3D space, it usually comes in the form of \((x, y, z)\), where each component specifies how far in that direction the point is from the origin.
This vector is crucial in understanding motion in vector calculus, as it provides the starting point for analyzing movement and any changes due to various parameters, like time.
In our specific exercise, the position vector \(\mathbf{r}(t)\) describes a trajectory in 3D space. Its components are functions of the parameter \(t\), which could represent time in a physical situation. By treating each component separately, we can analyze how 'x', 'y', and 'z' coordinates of a point change as \(t\) changes.

• The position vector helps us visualize motion and understand how the spatial location of the object varies over the parameter being analyzed.

Magnitude of a Vector

The magnitude of a vector, often referred to as the length or norm, gives us an idea of how long the vector is. It's calculated using the Pythagorean theorem in \((x, y, z)\) space.
The formula used is \[\|\mathbf{v}\| = \sqrt{x^2 + y^2 + z^2}\] where \(x, y, z\) are the vector components. For instance, if our vector is \((1, 2, 2)\), its magnitude would be calculated as \[\sqrt{1^2 + 2^2 + 2^2} = \sqrt{9} = 3.\]
The magnitude is essential in contextualizing vectors as it tells us about their scale in the space they inhabit.

• We often use magnitudes to normalize vectors, shrinking or expanding them to unit size, which is especially useful when deriving unit vectors.

Understanding the magnitude is key to constructing unit tangent vectors by dividing each component of the vector by this length.

Parametric Equations

Parametric equations involve expressing mathematical functions using a parameter, typically represented by \(t\). These parameters allow for a more flexible description of curves and surfaces in space, especially useful in physics and engineering.
In our exercise, parametric equations define \(x, y, z\) as functions of \(t\).
The beauty of parametric equations is their ability to succinctly represent continuous curves or surfaces where traditional Cartesian equations might be cumbersome or impossible.

• They provide a comprehensive way to describe the path and positioning of an object across time or another variable, offering more detailed information than static coordinates.
• By varying \(t\), we can explore how the vector changes and anticipate the trajectory of the curve.

Understanding parametric equations is fundamental to solving engineering and physics problems that involve object paths.