Question

, find the curvature \(\kappa\), the unit tangent vector \(\mathbf{T}\), the unit normal vector \(\mathbf{N}\), and the binormal vector \(\mathbf{B}\) at \(t=t_{1} .\) $$ \mathbf{r}(t)=e^{-2 t} \mathbf{i}+e^{2 t} \mathbf{j}+2 \sqrt{2} t \mathbf{k} ; t_{1}=0 $$

Short answer

Unit Tangent: \( -\frac{1}{2}\mathbf{i} + \frac{1}{2}\mathbf{j} + \frac{\sqrt{2}}{2}\mathbf{k} \), \( \mathbf{N} = \frac{1}{\sqrt{2}}\mathbf{i} + \frac{1}{\sqrt{2}}\mathbf{j} \), \( \mathbf{B} = -\frac{1}{2\sqrt{2}}\mathbf{i} - \frac{1}{2\sqrt{2}}\mathbf{j} - \frac{1}{\sqrt{2}}\mathbf{k} \), Curvature \( \kappa = \frac{1}{\sqrt{2}} \).

Step-by-step solution

Find the Derivatives

To find the unit tangent vector, curvature, and other vectors, we first need to compute the derivatives of the vector function \( \mathbf{r}(t) \). Calculate the first derivative \( \mathbf{r}'(t) \) and the second derivative \( \mathbf{r}''(t) \): \[ \mathbf{r}'(t) = \frac{d}{dt}(e^{-2t})\mathbf{i} + \frac{d}{dt}(e^{2t})\mathbf{j} + \frac{d}{dt}(2\sqrt{2}t)\mathbf{k} = -2e^{-2t}\mathbf{i} + 2e^{2t}\mathbf{j} + 2\sqrt{2}\mathbf{k} \] \[ \mathbf{r}''(t) = \frac{d}{dt}(-2e^{-2t}\mathbf{i} + 2e^{2t}\mathbf{j} + 2\sqrt{2}\mathbf{k}) = 4e^{-2t}\mathbf{i} + 4e^{2t}\mathbf{j} \]

Evaluate Derivatives at \(t_1\)

Plug \( t_1 = 0 \) into the derivatives to get their values at \( t_1 \):\[ \mathbf{r}'(0) = -2e^{0}\mathbf{i} + 2e^{0}\mathbf{j} + 2\sqrt{2}\mathbf{k} = -2\mathbf{i} + 2\mathbf{j} + 2\sqrt{2}\mathbf{k} \] \[ \mathbf{r}''(0) = 4e^{0}\mathbf{i} + 4e^{0}\mathbf{j} = 4\mathbf{i} + 4\mathbf{j} \]

Find the Unit Tangent Vector \(\mathbf{T}\)

The unit tangent vector \( \mathbf{T} \) is the normalized first derivative \( \mathbf{r}'(t) \). Calculate \( \| \mathbf{r}'(0) \| \), the magnitude of \( \mathbf{r}'(t) \) at \( t=0 \):\[ \| \mathbf{r}'(0) \| = \sqrt{(-2)^2 + 2^2 + (2\sqrt{2})^2} = \sqrt{4 + 4 + 8} = \sqrt{16} = 4 \]Now, divide each component of \( \mathbf{r}'(0) \) by its magnitude: \[ \mathbf{T}(0) = \frac{-2}{4}\mathbf{i} + \frac{2}{4}\mathbf{j} + \frac{2\sqrt{2}}{4}\mathbf{k} = -\frac{1}{2}\mathbf{i} + \frac{1}{2}\mathbf{j} + \frac{\sqrt{2}}{2}\mathbf{k} \]

Find the Unit Normal Vector \(\mathbf{N}\)

The unit normal vector \( \mathbf{N} \) is found by normalizing the derivative of \( \mathbf{T}(t) \). First compute \( \mathbf{T}'(t) \), then evaluate at \( t=0 \):The expression for the derivative of \( \mathbf{T}(t) \) at \( t=0 \) is a scalar multiple of \( \mathbf{r}''(0) \), as the function is linear and simplifies directly:Checking alignment, at \( t=0 \), we'd get \( \mathbf{r}''(0) = 4\mathbf{i} + 4\mathbf{j} \) like the derivative only considering alignment onto unit length:Normalize it:\[ \| \mathbf{r}''(0) \| = \sqrt{(4)^2 + (4)^2} = \sqrt{32} = 4\sqrt{2} \]\[ \mathbf{N}(0) = \frac{4}{4\sqrt{2}}\mathbf{i} + \frac{4}{4\sqrt{2}}\mathbf{j} = \frac{1}{\sqrt{2}}\mathbf{i} + \frac{1}{\sqrt{2}}\mathbf{j} \]

Find the Binormal Vector \(\mathbf{B}\)

The binormal vector \( \mathbf{B} \) is the cross product of \( \mathbf{T} \) and \( \mathbf{N} \). Compute \( \mathbf{T}(0) \times \mathbf{N}(0) \):\[ \mathbf{B}(0) = (-\frac{1}{2}\mathbf{i} + \frac{1}{2}\mathbf{j} + \frac{\sqrt{2}}{2}\mathbf{k}) \times (\frac{1}{\sqrt{2}}\mathbf{i} + \frac{1}{\sqrt{2}}\mathbf{j} + 0\mathbf{k}) \] Using the standard determinant method for cross products, this gives:\[ \mathbf{B}(0) = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ -\frac{1}{2} & \frac{1}{2} & \frac{\sqrt{2}}{2} \ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 0 \end{vmatrix} = \left( -\frac{1}{2\sqrt{2}} \right)\mathbf{i} + \left( -\frac{1}{2\sqrt{2}} \right)\mathbf{j} + \left( -\frac{1}{\sqrt{2}} \right)\mathbf{k} \]

Compute the Curvature \(\kappa\)

Curvature \( \kappa \) is given by the formula:\[ \kappa = \frac{\| \mathbf{r}'(t) \times \mathbf{r}''(t) \|}{\| \mathbf{r}'(t) \|^3} \]Calculate the cross product \( \mathbf{r}'(0) \times \mathbf{r}''(0) \):\[ \mathbf{r}'(0) = -2\mathbf{i} + 2\mathbf{j} + 2\sqrt{2}\mathbf{k}, \quad \mathbf{r}''(0) = 4\mathbf{i} + 4\mathbf{j} \]\[ \mathbf{r}'(0) \times \mathbf{r}''(0) = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ -2 & 2 & 2\sqrt{2} \ 4 & 4 & 0 \end{vmatrix} = (-8\sqrt{2})\mathbf{i} + (-8\sqrt{2})\mathbf{j} + (-16)\mathbf{k} \]The magnitude:\[ \| \mathbf{r}'(0) \times \mathbf{r}''(0) \| = \sqrt{(-8\sqrt{2})^2 + (-8\sqrt{2})^2 + (-16)^2} = \sqrt{128 + 128 + 256} = \sqrt{512} = 16\sqrt{2} \]Plug into formula:\[ \kappa = \frac{16\sqrt{2}}{4^3} = \frac{16\sqrt{2}}{64} = \frac{1}{\sqrt{2}} \]

Conclusion

We have calculated the required vectors and curvature at \( t_1 = 0 \):- Unit Tangent Vector \( \mathbf{T}(0) = -\frac{1}{2}\mathbf{i} + \frac{1}{2}\mathbf{j} + \frac{\sqrt{2}}{2}\mathbf{k} \)- Unit Normal Vector \( \mathbf{N}(0) = \frac{1}{\sqrt{2}}\mathbf{i} + \frac{1}{\sqrt{2}}\mathbf{j} \)- Binormal Vector \( \mathbf{B}(0) = -\frac{1}{2\sqrt{2}}\mathbf{i} - \frac{1}{2\sqrt{2}}\mathbf{j} - \frac{1}{\sqrt{2}}\mathbf{k} \)- Curvature \( \kappa = \frac{1}{\sqrt{2}} \)

Key concepts

Unit Tangent Vector

The unit tangent vector is crucial in understanding the direction of a curve at any given point. It gives us a way to describe the path of the curve as a motion along the curve. To find the unit tangent vector, start by calculating the first derivative of the position vector, \( \mathbf{r}'(t) \). This derivative represents the velocity of a point moving along the curve, giving us the direction of the curve at each point.
Next, the vector is normalized to ensure it has a magnitude of 1. This means dividing each component of the derivative by its magnitude. For example, if \( \mathbf{r}'(0) \) is \(-2\mathbf{i} + 2\mathbf{j} + 2\sqrt{2}\mathbf{k} \), first compute the magnitude: \( \| \mathbf{r}'(0) \| = 4 \). Then, divide each component by 4 to obtain the unit tangent vector, \( \mathbf{T}(0) = -\frac{1}{2}\mathbf{i} + \frac{1}{2}\mathbf{j} + \frac{\sqrt{2}}{2}\mathbf{k} \).
The unit tangent vector is essential in curvature calculus as it helps in defining the direction of motion along the curve precisely.

Unit Normal Vector

The unit normal vector, \( \mathbf{N} \), is orthogonal to the unit tangent vector and gives insight into the direction in which the curve is bending at a specific point. This vector is particularly useful in determining curvature.
To obtain the unit normal vector, you need to differentiate the unit tangent vector, \( \mathbf{T}(t) \), with respect to \( t \). However, since \( \mathbf{T}(t) \) is already normalized, you simply compute this derivative and then normalize it again to get \( \mathbf{N} \).
For instance, given the problem's result where \( \mathbf{r}''(0) = 4\mathbf{i} + 4\mathbf{j} \), normalizing this derivative leads to the unit normal vector \( \mathbf{N}(0) = \frac{1}{\sqrt{2}}\mathbf{i} + \frac{1}{\sqrt{2}}\mathbf{j} \).
Learning about the unit normal vector is critical for students to understand how curves change direction.

Binormal Vector

The binormal vector, \( \mathbf{B} \), forms a complete orthonormal set with the tangent and normal vectors, creating what is known as the TNB frame. It is perpendicular to both the unit tangent vector and the unit normal vector.
To find the binormal vector, calculate the cross product of the tangent and normal vectors. This cross product is vital because it guarantees the binormal vector's orthogonality to the other two. In our example, within the TNB frame, the binormal vector is given by \( \mathbf{B}(0) = -\frac{1}{2\sqrt{2}}\mathbf{i} - \frac{1}{2\sqrt{2}}\mathbf{j} - \frac{1}{\sqrt{2}}\mathbf{k} \).
Understanding the binormal vector is an essential part of three-dimensional curve analysis, as it helps describe how a curve lies in space.

Derivatives in Calculus

Derivatives are a core tool in calculus, allowing us to investigate changes and patterns in various mathematical contexts. In the topic of curve analysis, derivatives are used to compute important vectors like the unit tangent, normal, and binormal vectors.
The first derivative \( \mathbf{r}'(t) \) tells us about the rate and direction of change along a curve; it's essentially the velocity vector when the curve is interpreted as motion. The second derivative, \( \mathbf{r}''(t) \), provides information about the acceleration and how the curve is bending.
Calculating derivatives effectively is fundamental in curvature calculus, as these calculations form the basis for finding other vectors and parameters that describe the curve. Mastery of derivatives empowers students to explore the dynamic behavior of curves and other functions fully.