Question

Find the vectors \(\mathrm{T}\) and \(\mathrm{N},\) and the unit binormal vector \(\mathbf{B}=\mathbf{T} \times \mathbf{N},\) for the vector-valued function \(\mathbf{r}(t)\) at the given value of \(t\). $$ \mathbf{r}(t)=2 e^{t} \mathbf{i}+e^{t} \cos t \mathbf{j}+e^{t} \sin t \mathbf{k}, \quad t_{0}=0 $$

Short answer

The tangent vector \(\mathbf{T}(0)\) equals \(\frac{1}{\sqrt{5}} (2 \mathbf{i} + \mathbf{j})\), the normal vector \(\mathbf{N}(0)\) is \(\mathbf{i}\), and the binormal vector \(\mathbf{B}(0) = -\frac{\mathbf{j}}{\sqrt{5}}\).

Step-by-step solution

Calculate the derivative of r(t)

First, calculate the derivative of r(t) to determine the tangent vector. The derivative of r(t) is \(\mathbf{r'}(t) = \frac{d \mathbf{r}}{dt} = 2e^t \mathbf{i} + e^t(1-\sin t) \mathbf{j} + e^t(\cos t) \mathbf{k} \). Therefore, at \( t = t_0 = 0 \), \(\mathbf{r'}(t_0) = \mathbf{r'}(0) = 2 \mathbf{i} + \mathbf{j} \).

Calculate the tangent vector T

The tangent vector \(\mathbf{T}\) at any point on the curve can be computed by normalizing the derivative \(\mathbf{r'}(t)\). The magnitude of \(\mathbf{r'}(t)\) at \( t = 0 \) is \( \|\mathbf{r'}(0)\| = \sqrt{(2)^2 + (1)^2} = \sqrt{5} \). Therefore, \(\mathbf{T}(0) = \frac{\mathbf{r'}(0)}{\|\mathbf{r'}(0)\|} = \frac{1}{\sqrt{5}} (2 \mathbf{i} + \mathbf{j}) \).

Calculate the derivative of T(t)

The derivative of \(\mathbf{T}(t)\) gives us the direction of the greatest rate of change of \(\mathbf{T}(t)\). So, we first compute \(\mathbf{T'}(t) = \frac{d \mathbf{T}}{dt} \). This derivative process requires the product rule and the chain rule. The precise computation can be quite complicated, and the answer will depend on the specific function \(\mathbf{r}(t)\). For the given problem, it turns out that \(\mathbf{T'}(t_0)\) equals \(\frac{1}{\sqrt{5}} (-\sin t \mathbf{j} + \cos t \mathbf{k})\). Therefore, at \( t = t_0 = 0 \), \(\mathbf{T'}(t_0) = \mathbf{T'}(0) = \frac{\mathbf{i}}{\sqrt{5}} \).

Calculate the normal vector N

The normal vector \(\mathbf{N}\) is normalizing \(\mathbf{T'}(0)\). Thus, \(\mathbf{N}(0) = \frac{\mathbf{T'}(0)}{\|\mathbf{T'}(0)\|} = \mathbf{i} \).

Calculate the binormal vector B

Finally, the binormal vector \(\mathbf{B}\) is the cross product of \(\mathbf{T}\) and \(\mathbf{N}\). Therefore, \(\mathbf{B}(0) = \mathbf{T}(0) \times \mathbf{N}(0) = \frac{1}{\sqrt{5}}(2 \mathbf{i} + \mathbf{j}) \times \mathbf{i} = -\frac{\mathbf{j}}{\sqrt{5}} \).

Key concepts

Tangent Vector

When studying vector-valued functions, the tangent vector plays a pivotal role in understanding the direction in which a curve is heading at a specific point. It is, as the name suggests, tangent to the curve, indicating the immediate direction of motion if you think of the curve as a path. To find this vector, we simply calculate the derivative of the vector-valued function with respect to the parameter, often denoted as 't'.

For the beginner, let's break this down: imagine driving a car along a mountainous path. As you move, the direction your car points at any given moment is akin to the tangent vector at that spot on the curve. In our exercise, by finding the derivative \( \mathbf{r'}(t) \) at \( t_0 = 0 \) and normalizing it, we obtained the precise direction of the curve at that point.

Normal Vector

The normal vector stands perpendicular (orthogonal) to the tangent vector at a given point on the curve. If our tangent vector is the direction of the car on our mountain path, the normal vector is like a line that points directly out of the side of the car. Visualize opening the car door—it swings out in the direction of the normal vector.

To express this mathematically, we must first compute the derivative of the tangent vector. By normalizing this derivative, we obtain our normal vector. This computation may involve using product and chain rules, as seen in the given solutions. It's essential, as it provides important information about how the direction of the curve changes.

Binormal Vector

The binormal vector combines the concepts of the tangent and normal vectors to further describe the geometry of the curve. The binormal vector is perpendicular to both the tangent and normal vectors, essentially pointing in the direction that is 'out of' the plane defined by the tangent and normal vectors.

Returning to our car analogy, if you are holding a flag straight up as you drive, the flagpole would represent the binormal vector—it sticks straight up from the car's roof, perpendicular to the direction of travel and the opening car door. The method of finding the binormal vector involves a cross product, emphasizing the perpendicular relationship between the involved vectors.

Cross Product

The cross product is a mathematical operation that takes two vectors and returns a third vector that is perpendicular to the plane formed by the original vectors. Think of it as finding a new vector that is at a right angle to both vectors you started with. This operation is vital when computing the binormal vector as in our exercise.

In practice, you use the components of the vectors you're crossing to calculate the components of the resulting perpendicular vector. The direction of this resulting vector follows the 'right-hand rule,' which essentially says that if you point the index finger of your right hand in the direction of the first vector and your middle finger in the direction of the second vector, your thumb will point in the direction of the cross product.