Problem 18 - Calculus Early Transcendentals

Question

Find the unit tangent vector \(\mathbf{T}(t)\) at the point with the given value of the parameter \(t .\) $$ \mathbf{r}(t)=4 \sqrt{t} \mathbf{i}+t^{2} \mathbf{j}+t \mathbf{k}, \quad t=1 $$

Step-by-Step Solution

Answer

The unit tangent vector at \(t=1\) is \(\mathbf{T}(1) = \frac{2}{3} \mathbf{i} + \frac{2}{3} \mathbf{j} + \frac{1}{3} \mathbf{k}.\)

Step 1: Differentiate the Vector Function

The first step is to differentiate the vector function \(\mathbf{r}(t)\) with respect to \(t\). The derivative gives us \(\mathbf{r}'(t)\): \[ \mathbf{r}'(t) = \frac{d}{dt} [4 \sqrt{t} \mathbf{i} + t^2 \mathbf{j} + t \mathbf{k}] = \frac{d}{dt} (4t^{1/2}) \mathbf{i} + \frac{d}{dt} (t^2) \mathbf{j} + \frac{d}{dt} (t) \mathbf{k}. \] \[ \mathbf{r}'(t) = 2t^{-1/2} \mathbf{i} + 2t \mathbf{j} + \mathbf{k}. \]

Step 2: Evaluate the Derivative at t=1

Next, we substitute \(t=1\) into the derivative \(\mathbf{r}'(t)\). \[ \mathbf{r}'(1) = 2(1)^{-1/2} \mathbf{i} + 2(1) \mathbf{j} + \mathbf{k} = 2 \mathbf{i} + 2 \mathbf{j} + \mathbf{k}. \]

Step 3: Find the Magnitude of \(\mathbf{r}'(1)\)

To find the unit tangent vector, we first need the magnitude of \(\mathbf{r}'(1)\). Calculate \|\mathbf{r}'(1)\|. \[ \|\mathbf{r}'(1)\| = \sqrt{(2)^2 + (2)^2 + (1)^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3. \]

Step 4: Calculate the Unit Tangent Vector

The unit tangent vector \(\mathbf{T}(1)\) is found by dividing \(\mathbf{r}'(1)\) by its magnitude. \[ \mathbf{T}(1) = \frac{\mathbf{r}'(1)}{\|\mathbf{r}'(1)\|} = \frac{2 \mathbf{i} + 2 \mathbf{j} + \mathbf{k}}{3} = \frac{2}{3} \mathbf{i} + \frac{2}{3} \mathbf{j} + \frac{1}{3} \mathbf{k}. \]

Key Concepts

Vector Differentiation

Vector differentiation is a process used to find the rate of change of a vector function with respect to a parameter, usually denoted as "t". In calculus, vectors can describe position, velocity, or acceleration. Differentiation helps us understand how these quantities change over time. Given a vector function \(\mathbf{r}(t) = 4 \sqrt{t} \mathbf{i} + t^2 \mathbf{j} + t \mathbf{k}\), differentiating each component gives us the derivative \(\mathbf{r}'(t)\). Here is how:

• The derivative of \(4\sqrt{t}\) with respect to \(t\) results in \(2t^{-1/2}\).
• For \(t^2\), the derivative is \(2t\).
• The derivative of \(t\) is simply \(1\).

This process is straightforward, but remember to handle each vector component separately. Differentiation gives the vector \(\mathbf{r}'(t) = 2t^{-1/2} \mathbf{i} + 2t \mathbf{j} + \mathbf{k}\), illustrating changes in the direction of the vector as \(t\) changes.

This step underpins many calculus problems involving motion, as these derivatives often represent velocities or similar rates of change.

Tangent Vector

A tangent vector is crucial in understanding the behavior of a curve at a certain point. For a given parameter value \(t\), the tangent vector \(\mathbf{r}'(t)\) provides the direction in which the curve is heading.

• The tangent vector doesn't just point in a direction; it also has a magnitude, which represents the speed of movement if \(\mathbf{r}(t)\) denotes a path.
• Finding the tangent vector is a step towards obtaining the unit tangent vector, which is crucial in making calculations regarding angles or directions dimensionless.

In the given problem, the tangent vector at \(t=1\) is \(2 \mathbf{i} + 2 \mathbf{j} + \mathbf{k}\). This vector tells us the path's instantaneous direction at this parameter value. Understanding tangent vectors helps in multiple calculus problems and applications like calculating derivatives in 3D motion.

Calculus Problem Solving

Solving calculus problems often involves systematic steps to reach the desired solution. Each step builds upon the last and needs precision. This problem is a perfect example of the workflow needed in calculus problem solving.

Firstly, differentiate the vector function to get the tangent vector. Secondly, evaluate this vector at the specified parameter value. Next, compute the magnitude to normalize the vector, helping translate it into a unit tangent vector.

The unit tangent vector moves us from precise mathematical values to a vector of length one, which keeps the direction but loses the magnitude. Here are key steps used:

• Differentiate to find \(\mathbf{r}'(t)\).
• Evaluate \(\mathbf{r}'(t)\) at \(t=1\).
• Calculate the magnitude with \(\|\mathbf{r}'(1)\| = \sqrt{9}\).
• Divide \(\mathbf{r}'(1)\) by its magnitude to get \(\mathbf{T}(1)\).

These steps form a strategy not just for this problem but as a general approach to tackling similar calculus scenarios.

Parameter Evaluation

Parameter evaluation involves substituting specific values into differentiated expressions to analyze particular scenarios or capture instantaneous rates of change.

In the given exercise, the aimed value is \(t=1\). Inserting this into the tangent vector \(\mathbf{r}'(t)\) leads to \(\mathbf{r}'(1) = 2 \mathbf{i} + 2 \mathbf{j} + \mathbf{k}\). This step establishes the tangent vector's exact direction at \(t=1\).

• Parameter evaluation in functions converts general expressions into specific values, offering clearer insights.
• This technique proves essential to many scientific and engineering applications, where exact values representation is necessary.

Evaluating parameters assist not only in making sense of a single point on a curve or trajectory but also lays the groundwork for more complex concepts like curvature, which further explore how these curves behave.