Problem 45 - Essential Calculus Early Transcendentals

Question

\(45-46\) . 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 Derivative

First, we need to find the derivative of the vector function \( \mathbf{r}(t) = \cos t \mathbf{i} + 3t \mathbf{j} + 2 \sin 2t \mathbf{k} \). Calculate \( \frac{d\mathbf{r}}{dt} \):\[ \frac{d\mathbf{r}}{dt} = \frac{d}{dt}(\cos t) \mathbf{i} + \frac{d}{dt}(3t) \mathbf{j} + \frac{d}{dt}(2 \sin 2t) \mathbf{k} \]Using the differentiation rules, this becomes:\[ \frac{d\mathbf{r}}{dt} = -\sin t \mathbf{i} + 3 \mathbf{j} + 4 \cos 2t \mathbf{k} \]

Step 2: Evaluate the Derivative at t=0

Substitute \( t = 0 \) into the derivative found in the previous step:\[ \mathbf{r}'(0) = -\sin(0) \mathbf{i} + 3 \mathbf{j} + 4 \cos(2 \times 0) \mathbf{k} \]This simplifies to:\[ \mathbf{r}'(0) = 0 \mathbf{i} + 3 \mathbf{j} + 4 \mathbf{k} \]

Step 3: Calculate the Magnitude of the Derivative

Find the magnitude of \( \mathbf{r}'(0) \):\[ \| \mathbf{r}'(0) \| = \sqrt{0^2 + 3^2 + 4^2} \]Solving this gives:\[ \| \mathbf{r}'(0) \| = \sqrt{0 + 9 + 16} = \sqrt{25} = 5 \]

Step 4: Determine the Unit Tangent Vector

The unit tangent vector \( \mathbf{T}(t) \) is found by dividing \( \mathbf{r}'(t) \) by its magnitude:\[ \mathbf{T}(t) = \frac{\mathbf{r}'(t)}{\| \mathbf{r}'(t) \|} \] At \( t = 0 \), this is:\[ \mathbf{T}(0) = \frac{0 \mathbf{i} + 3 \mathbf{j} + 4 \mathbf{k}}{5} = 0 \mathbf{i} + \frac{3}{5} \mathbf{j} + \frac{4}{5} \mathbf{k} \]

Key Concepts

Vector Calculus

Understanding vector calculus is crucial when dealing with vector functions that describe quantities having both magnitude and direction. In our previous example, the vector function \(\mathbf{r}(t)\) provides such a description. This field of mathematics allows us to explore how these vectors change over time or space.

Key concepts in vector calculus include vector fields, line integrals, and the flow of vectors around paths. These concepts allow us to model physical phenomena like electromagnetic fields or fluid dynamics, which require both the direction and magnitude of changes.

• Vector Fields: Descriptions of how vectors are assigned to each point in space.
• Line Integrals: Calculating the integral of a vector field along a path, representing accumulated change.
• Tangent Vectors: Vectors that indicate the direction a curve is heading at a point.

A solid grasp of vector calculus enables you to solve complex problems involving curves and paths, as illustrated in determining the unit tangent vector.

Derivative of Vector Functions

Taking the derivative of a vector function is a fundamental step in determining how the vector changes with respect to time or another parameter, like in our example with \(\mathbf{r}(t)\).

When you differentiate a vector function, you find how each component of the vector changes. In our case, we started with \( \mathbf{r}(t) = \cos t \mathbf{i} + 3t \mathbf{j} + 2 \sin 2t \mathbf{k} \). To find its derivative \( \mathbf{r}'(t) \), we applied basic differentiation rules:

• The derivative of \(\cos t\) with respect to \(t\) is \(-\sin t\).
• The derivative of \(3t\) is a constant, \(3\).
• For \(2 \sin 2t\), apply the chain rule to get \(4 \cos 2t\).

These derivatives provide insights into the velocity of a point moving along the curve described by \( \mathbf{r}(t) \). Understanding derivatives in vector functions is key to analyzing motion and change within a vector field.

Magnitude of Vectors

The magnitude of a vector represents the length or size of the vector, and it is a vital component when normalizing vectors to find unit vectors. In the context of the given example, the magnitude of the derivative \( \mathbf{r}'(0) = 0 \mathbf{i} + 3 \mathbf{j} + 4 \mathbf{k} \) was calculated as:

\[ \| \mathbf{r}'(0) \| = \sqrt{0^2 + 3^2 + 4^2} = \sqrt{0 + 9 + 16} = \sqrt{25} = 5 \]

Finding the magnitude involves summing the squares of each component of a vector and taking the square root. This operation gives us a scalar value indicating the vector's overall size, devoid of direction. Calculating vector magnitudes is essential when converting to unit vectors, which requires dividing a vector by its magnitude.

• A vector with a magnitude of 1 is called a unit vector.
• These vectors preserve direction but standardize size, essential for simplifying calculations.

In our problem, dividing the vector by its magnitude provided the unit tangent vector, giving both direction and scale unity. Understanding magnitude helps students grasp how vectors interact in space and allows for better manipulation and comprehension of vector calculus problems.