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 \(t\). $$ \begin{array}{l} \mathbf{r}(t)=2 \cos t \mathbf{i}+2 \sin t \mathbf{j}+\frac{t}{2} \mathbf{k} \\\ t_{0}=\frac{\pi}{2} \end{array} $$

Short answer

The Tangent vector \(\mathbf{T}\) is \(-\frac{2}{\sqrt{5}}\mathbf{i} + \frac{1}{\sqrt{5}}\mathbf{k}\), the Normal vector \(\mathbf{N}\) is \(-\mathbf{i}\), and the unit Binormal vector \(\mathbf{B}\) is \(\frac{1}{\sqrt{5}}\mathbf{j}\).

Step-by-step solution

Compute the first derivative of \(\mathbf{r}(t)\)

The first derivative of \(\mathbf{r}(t) = 2 \cos t \mathbf{i}+2 \sin t \mathbf{j} + \frac{t}{2} \mathbf{k}\) gives us \(\mathbf{r}'(t) = -2\sin t \mathbf{i} + 2\cos t \mathbf{j} + \frac{1}{2} \mathbf{k}\). From here, we evaluate \(\mathbf{r}'(t)\) at \(t_{0}=\frac{\pi}{2}\), yielding \(\mathbf{r}'(t_{0}) = -2\mathbf{i} + \mathbf{k}\).

Find the unit tangent vector \(\mathbf{T}\)

\(\mathbf{T}\) is simply the unit vector of \(\mathbf{r}'(t_{0})\). By taking the magnitude of \(\mathbf{r}'(t_{0})\), we have \(||\mathbf{r}'(t_{0})|| = \sqrt{(-2)^2 + 1^2} = \sqrt{5}\). Hence, \(\mathbf{T} = \frac{\mathbf{r}'(t_{0})}{||\mathbf{r}'(t_{0})||} = -\frac{2}{\sqrt{5}}\mathbf{i} + \frac{1}{\sqrt{5}}\mathbf{k}\).

Compute the second derivative of \(\mathbf{r}(t)\) and find \(\mathbf{N}\)

The second derivative of \(\mathbf{r}(t)\) is \(\mathbf{r}''(t) = -2\cos t \mathbf{i} - 2\sin t \mathbf{j}\). Evaluate \(\mathbf{r}''(t)\) at \(t_{0}\), yielding \(\mathbf{r}''(t_{0}) = -2\mathbf{i}\). Divide by its magnitude to get the unit normal vector \(\mathbf{N} = -\mathbf{i}\).

Compute the unit binormal vector \(\mathbf{B}\)

Now we find \(\mathbf{B}\) by taking the cross product of \(\mathbf{T}\) and \(\mathbf{N}\), i.e. \(\mathbf{B} = \mathbf{T}\times\mathbf{N}\). The cross product of \(-\frac{2}{\sqrt{5}}\mathbf{i} + \frac{1}{\sqrt{5}}\mathbf{k}\) and \(-\mathbf{i}\) results in \(\mathbf{B} = \frac{1}{\sqrt{5}}\mathbf{j}\).

Key concepts

Vector Calculus

Vector calculus is an essential branch of mathematics for analyzing and understanding physical phenomena involving vectors in space. It extends the concepts of calculus to vector fields, which are functions that assign vectors to points in space. These tools allow you to handle curves and surfaces in three-dimensional spaces, giving insights into their properties such as curvature and torsion.

At the heart of vector calculus are the derivative and integral, but these concepts are now applied to vector functions. As we saw in the vector-valued function \(\mathbf{r}(t)\), the derivative can tell us about the function's rate of change at any point, which generates a new vector revealing important aspects of its behavior.

Unit Tangent Vector

The unit tangent vector \(\mathbf{T}\) represents the direction of the curve at a particular point, normalized to a magnitude of 1. Think of it as an arrow skimming along the curve, always pointing in the direction the curve is heading next. It's essentially the vector derivative of a position vector \(\mathbf{r}(t)\), which is \(\mathbf{r}'(t)\), divided by its magnitude to keep the length at one.

To compute \(\mathbf{T}\), it's essential to find the derivative of the vector function, evaluate at the interested point, and then normalize. As in our example, this process yields a vector that shows the direction of the curve at \(t_0\) without the influence of the vector's original magnitude.

Unit Binormal Vector

The unit binormal vector \(\mathbf{B}\) is a bit more abstract. It's perpendicular to both the unit tangent vector \(\mathbf{T}\) and the unit normal vector \(\mathbf{N}\). You can think of it as defining the 'binormal plane' at a point on the curve, which is perpendicular to the curve's instantaneous plane, formed by the unit tangent and unit normal vectors.

Finding \(\mathbf{B}\) involves the cross product of \(\mathbf{T}\) and \(\mathbf{N}\), as seen in the exercise. This particular sequence is important because it maintains an orientation consistent with the right-hand rule, which is conventional in physics and mathematics. Once computed, just like the other vectors, we normalize it to ensure it is of unit length.

Cross Product

The cross product, also known as the vector product, is an operation involving two vectors in three-dimensional space. It produces a third vector that is perpendicular to the plane containing the original vectors. The magnitude of the cross product vector is proportional to the area of the parallelogram with sides defined by the two vectors.

An important property of the cross product is its direction, which abides by the right-hand rule. This rule means if you point the index finger of your right hand along the first vector and the middle finger along the second vector, the thumb will point in the direction of the cross product. As details in the exercise, the cross product of the unit tangent and unit normal vectors yields the unit binormal vector, showcasing its usefulness in vector calculus.