Problem 32 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the curvature \(\kappa\), the unit tangent vector \(\mathbf{T}\), the unit normal vector \(\mathbf{N}\), and the binormal vector \(\mathbf{B}\) at \(t=t_{1}\). \(\mathbf{r}(t)=e^{7 t} \cos 2 t \mathbf{i}+e^{7 t} \sin 2 t \mathbf{j}+e^{7 t} \mathbf{k} ; t_{1}=\pi / 3\)

Step-by-Step Solution

Answer

Curvature \(\kappa\), unit vectors \(\mathbf{T}\), \(\mathbf{N}\), and \(\mathbf{B}\) are computed using derivatives at \( t=\pi/3 \).

Step 1: Find the Derivative of the Vector Function

The first step is to take the derivative of the vector function \( \mathbf{r}(t) \). Given \( \mathbf{r}(t) = e^{7t}\cos(2t) \mathbf{i} + e^{7t}\sin(2t) \mathbf{j} + e^{7t} \mathbf{k} \), calculate \( \mathbf{r}'(t) \), using the product rule: \[ \mathbf{r}'(t) = (7e^{7t}\cos(2t) - 2e^{7t}\sin(2t))\mathbf{i} + (7e^{7t}\sin(2t) + 2e^{7t}\cos(2t))\mathbf{j} + 7e^{7t}\mathbf{k}. \]

Step 2: Evaluate the Derivative at \( t=\pi/3 \)

Substitute \( t=\pi/3 \) into \( \mathbf{r}'(t) \) calculated in the previous step:\[ \mathbf{r}'\left(\frac{\pi}{3}\right) = \left(7e^{7\pi/3}\cos\left(\frac{2\pi}{3}\right) - 2e^{7\pi/3}\sin\left(\frac{2\pi}{3}\right)\right)\mathbf{i} + \left(7e^{7\pi/3}\sin\left(\frac{2\pi}{3}\right) + 2e^{7\pi/3}\cos\left(\frac{2\pi}{3}\right)\right)\mathbf{j} + 7e^{7\pi/3}\mathbf{k}. \]

Key Concepts

Unit Tangent Vector

The unit tangent vector is an essential component of vector calculus. It provides the direction of a curve at a specific point. To obtain this vector, we first calculate the derivative of the position vector function, often denoted as \( \mathbf{r}'(t) \), since this derivative indicates the instantaneous rate of change – or velocity – of the curve. Then, to find the unit tangent vector \( \mathbf{T}(t) \), we normalize this derivative.

Normalization involves dividing the derivative by its own magnitude, which ensures that the resultant vector has a length of one. The formula for the magnitude of a vector \( \mathbf{v}(t) = a\mathbf{i} + b\mathbf{j} + c\mathbf{k} \) is given by:

• \( \|\mathbf{v}(t)\| = \sqrt{a^2 + b^2 + c^2} \)

Applying this to our derivative \( \mathbf{r}'(t) \) and evaluating at \( t = \frac{\pi}{3} \), we find the unit tangent vector. Remember, the unit tangent vector always points in the direction of movement along the curve.

This makes it particularly useful in applications involving motion and paths, as it provides not only direction but also insight into how the curve is traversed at any given point.

Unit Normal Vector

The unit normal vector, often symbolized as \( \mathbf{N}(t) \), is crucial for understanding how a curve changes direction. While the unit tangent vector indicates direction, the unit normal vector tells us how the curve is turning at a specific point. Once we have the unit tangent vector \( \mathbf{T}(t) \), the next step is to find its derivative with respect to \( t \).

The derivative of \( \mathbf{T}(t) \) effectively shows the change in direction along the curve, but it is not yet a unit vector. We again normalize this derivative to extract the unit normal vector:

• Calculate \( \mathbf{T}'(t) \)
• Find the magnitude: \( \|\mathbf{T}'(t)\| \)
• Normalize \( \mathbf{T}'(t) \) by dividing by its magnitude

Executing these steps provides the unit normal vector \( \mathbf{N}(t) \), which is always perpendicular to \( \mathbf{T}(t) \).

This vector is particularly valuable for determining curvature, as it describes how sharply a curve bends. It's a fundamental concept for fields requiring precise understanding of motion paths, such as physics and engineering.

Binormal Vector

The binormal vector \( \mathbf{B}(t) \) completes the orthogonal triad of vectors in the Frenet-Serret formulas, which describe the geometry of a space curve. After obtaining the unit tangent \( \mathbf{T}(t) \) and unit normal vectors \( \mathbf{N}(t) \), the binormal vector is calculated as the cross product between these two vectors:

• \( \mathbf{B}(t) = \mathbf{T}(t) \times \mathbf{N}(t) \)

The result is a vector that is orthogonal to both \( \mathbf{T}(t) \) and \( \mathbf{N}(t) \). This orthogonal nature is what makes \( \mathbf{B}(t) \) such a key player in vector calculus and in understanding the spatial orientation of curves.

Moreover, the binormal vector provides insight into the twisting of the curve around the tangent and normal planes. It is specifically helpful in 3D models where understanding how a curve or path behaves in space is necessary. Utilized in areas ranging from aerodynamics to robotics, the binormal vector helps plot paths and trajectories accurately and effectively.