Problem 19 - Calculus Early Transcendentals

Question

Find the unit tangent vector \(\mathbf{T}(t)\) at the point with the given value of the parameter \(t .\) $$ \mathbf{r}(t)=\cos t \mathbf{i}+3 t \mathbf{j}+2 \sin 2 t \mathbf{k}, \quad t=0 $$

Step-by-Step Solution

Answer

The unit tangent vector at \( t = 0 \) is \( \mathbf{T}(0) = 0 \mathbf{i} + \frac{3}{5} \mathbf{j} + \frac{4}{5} \mathbf{k}. \)

Step 1: Find the velocity vector

To find the unit tangent vector, we first need the velocity vector, which is the derivative of the position vector \( \mathbf{r}(t) \). Differentiate each component of \( \mathbf{r}(t) \) with respect to \( t \). The given vector is \( \mathbf{r}(t) = \cos t \mathbf{i} + 3t \mathbf{j} + 2 \sin 2t \mathbf{k} \).The derivative is:\[\mathbf{r}'(t) = -\sin t \mathbf{i} + 3 \mathbf{j} + 4 \cos 2 t \mathbf{k}.\]

Step 2: Evaluate at the given t-value

Substitute \( t = 0 \) into the velocity vector \( \mathbf{r}'(t) = -\sin t \mathbf{i} + 3 \mathbf{j} + 4 \cos 2 t \mathbf{k} \) to find the velocity vector at \( t = 0 \).\[\mathbf{r}'(0) = -\sin(0) \mathbf{i} + 3 \mathbf{j} + 4 \cos(0) \mathbf{k} = 0 \mathbf{i} + 3 \mathbf{j} + 4 \mathbf{k}.\]

Step 3: Find the magnitude of the velocity vector

The magnitude \( \|\mathbf{r}'(0)\| \) of the velocity vector \( \mathbf{r}'(0) = 0 \mathbf{i} + 3 \mathbf{j} + 4 \mathbf{k} \) is found using the formula:\[\|\mathbf{r}'(0)\| = \sqrt{0^2 + 3^2 + 4^2} = \sqrt{0 + 9 + 16} = \sqrt{25} = 5.\]

Step 4: Find the unit tangent vector

The unit tangent vector \( \mathbf{T}(t) \) is the normalized velocity vector at \( t = 0 \). Normalize \( \mathbf{r}'(0) = 0 \mathbf{i} + 3 \mathbf{j} + 4 \mathbf{k} \) by dividing it by its magnitude:\[ \mathbf{T}(0) = \frac{1}{5} (0 \mathbf{i} + 3 \mathbf{j} + 4 \mathbf{k}) = 0 \mathbf{i} + \frac{3}{5} \mathbf{j} + \frac{4}{5} \mathbf{k}. \]

Key Concepts

Understanding the Velocity Vector

In calculus, a velocity vector gives you insightful information about an object's speed and direction at any point in time. When you have a position vector like \( \mathbf{r}(t) = \cos t \mathbf{i} + 3t \mathbf{j} + 2 \sin 2t \mathbf{k} \), the velocity is found by taking the derivative with respect to time, denoted as \( \mathbf{r}'(t) \).

Taking the derivative essentially means applying differentiation to each component of the vector separately. For our position vector, the process yields \( \mathbf{r}'(t) = -\sin t \mathbf{i} + 3 \mathbf{j} + 4 \cos 2t \mathbf{k} \). This velocity vector shows us exactly how our position vector \( \mathbf{r}(t) \) is changing over time.

• The component \( -\sin t \mathbf{i} \) tells us how the \( \mathbf{i} \) (or \( x \)) direction changes.
• The \( 3 \mathbf{j} \) component is a constant, indicating a steady increase in the \( \mathbf{j} \) (or \( y \)) direction.
• Lastly, \( 4 \cos 2t \mathbf{k} \) informs the change in the \( \mathbf{k} \) or \( z \) direction over time.

At \( t = 0 \), calculating the velocity vector gives you a snapshot of these directional components: \( 0 \mathbf{i} + 3 \mathbf{j} + 4 \mathbf{k} \). This means at that precise moment, there's no change in the \( x \)-direction, while the \( y \) and \( z \) directions have direct and significant rates of change.

Magnitude Calculation Explained

Once you have the velocity vector, the next step is often to find its magnitude, especially if you're aiming to obtain the unit tangent vector. The magnitude tells us about the vector's length or size.

To determine magnitude, use the formula \( \|\mathbf{v}\| = \sqrt{x^2 + y^2 + z^2} \), where \( x, y, \text{ and } z \) are the components of the vector.

For the velocity vector at \( t=0 \), \( \mathbf{r}'(0) = 0 \mathbf{i} + 3 \mathbf{j} + 4 \mathbf{k} \), calculate as follows:

• First, square each component: \( 0^2=0 \), \( 3^2 = 9 \), and \( 4^2 = 16 \).
• Add these squared values up to get \( 0 + 9 + 16 = 25 \).
• Finally, take the square root of 25 to find the magnitude: \( 5 \).

The result, 5, represents the overall speed or rate of movement of the vector at that moment. Calculating the magnitude allows you to later normalize the vector to find the unit tangent vector, crucial in studying vector directions.

The Role of Derivative of Vector Functions

Differentiating vector functions, like \( \mathbf{r}(t) \), helps us grasp how variables shift with time or other factors. This process underlies the calculation of the velocity vector, which is pivotal for comprehending movement within 3D spaces.

Here's how differentiation of each component works:

• The vector \( \mathbf{r}(t)=\cos t \mathbf{i}+3t \mathbf{j}+2 \sin 2t \mathbf{k} \) becomes \( \mathbf{r}'(t)=-\sin t \mathbf{i}+3 \mathbf{j}+4 \cos 2t \mathbf{k} \) upon differentiation.
• For \( \cos t \) in the \( \mathbf{i} \)-component, the derivative is \( -\sin t \).
• For \( 3t \), implying constant velocity, it's simply \( 3 \) in the \( \mathbf{j} \)-component.
• For \( 2 \sin 2t \) in the \( \mathbf{k} \)-component, apply the chain rule to achieve \( 4 \cos 2t \).

Through the derivative, we attain precise insights on how each segment of our vector-led journey evolves. Whether it's the connection between position and velocity, or the transition to finding acceleration, understanding derivatives in vector functions is an integral skill.