Question

Computing the binormal vector and torsion the unit tangent vector \(\mathbf{T}\) and the principal unit normal vector N were computed for the following parameterized curves. Use the definitions to compute their unit binormal vector and torsion. $$\mathbf{r}(t)=\langle 4 \sin t, 4 \cos t, 10 t\rangle$$

Short answer

Based on the given parameterized curve, we found the unit binormal vector to be: $$\mathbf{B}(t) = \frac{1}{17}\langle 40\cos t, 40 \sin t, -16\sin{t} - 16\cos{t} \rangle$$ And the torsion is: $$\tau(t) = -\frac{40}{17}$$

Step-by-step solution

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

Calculate the derivative of \(\mathbf{r}(t)\) with respect to \(t\). We have: $$\frac{d\mathbf{r}(t)}{dt} = \langle 4 \cos t, -4\sin t, 10 \rangle$$

Normalize the derivative to get \(\mathbf{T}\)

Divide the derivative of \(\mathbf{r}(t)\) by its magnitude to get the unit tangent vector \(\mathbf{T}(t)\): $$\mathbf{T}(t) = \frac{\langle 4 \cos t, -4\sin t, 10 \rangle}{\left\| \langle 4 \cos t, -4\sin t, 10 \rangle \right\|} = \frac{\langle 4 \cos t, -4\sin t, 10 \rangle}{\sqrt{(4\cos t)^2 + (-4\sin t)^2 + 10^2}} = \frac{\langle 4 \cos t, -4\sin t, 10 \rangle}{2\sqrt{17}}$$ Now we want to compute the principal unit normal vector \(\mathbf{N}(t)\). To do so, we need to find the derivative of \(\mathbf{T}(t)\) with respect to \(t\) and normalize it.

Compute the derivative of \(\mathbf{T}\)

Calculate the derivative of \(\mathbf{T}(t)\) with respect to \(t\). We have: $$\frac{d\mathbf{T}(t)}{dt} = \frac{-8\sin t-8 \cos t}{2\sqrt{17}}$$

Normalize the derivative of \(\mathbf{T}\) to get \(\mathbf{N}\)

Divide the derivative of \(\mathbf{T}(t)\) by its magnitude to get the principal unit normal vector \(\mathbf{N}(t)\). We can skip this step since this expression is already normalized. Now that we have \(\mathbf{T}\) and \(\mathbf{N}\), we can compute the binormal vector \(\mathbf{B}\) by taking the cross product of these two vectors:

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

Take the cross product of the unit tangent vector \(\mathbf{T}\) and the principal unit normal vector \(\mathbf{N}\): $$\mathbf{B}(t) = \mathbf{T}(t) \times \mathbf{N}(t) = \frac{1}{17}\langle 40\cos t, 40 \sin t, -16\sin{t} - 16\cos{t} \rangle$$ Now, we can compute the torsion by finding the derivative of \(\mathbf{B}(t)\) with respect to \(t\), and then using the provided formula:

Compute the derivative of \(\mathbf{B}\)

Compute the derivative of \(\mathbf{B}(t)\) with respect to \(t\): $$\frac{d\mathbf{B}(t)}{dt} = \frac{1}{17}\langle -40 \sin t, 40\cos t, -16\cos{t} + 16\sin{t} \rangle$$

Compute the torsion \(\tau\)

Finally, compute the torsion \(\tau(t)\) using the formula above. Note that the numerator is the dot product of \(\frac{d\mathbf{B}(t)}{dt} \) and \( \mathbf{N}(t)\): $$\tau(t) = -\frac{d\mathbf{B}(t)/dt \cdot \mathbf{N}(t)}{\|d\mathbf{B}(t)/dt\|}=-\frac{\langle \frac{1}{17}\langle -40 \sin t, 40\cos t, -16\cos{t} + 16\sin{t} \rangle \cdot \frac{-8\sin t-8 \cos t}{2\sqrt{17}}} {\|\frac{1}{17}\langle -40 \sin t, 40\cos t, -16\cos{t} + 16\sin{t} \rangle \|}=-\frac{40}{17}$$ Thus, the unit binormal vector \(\mathbf{B}(t)\) is given by the expression: $$\mathbf{B}(t) = \frac{1}{17}\langle 40\cos t, 40 \sin t, -16\sin{t} - 16\cos{t} \rangle$$ and the torsion \(\tau(t)\) is: $$\tau(t) = -\frac{40}{17}$$

Key concepts

Torsion

Torsion is a measure of how a space curve turns out of the plane formed by its tangent and normal vectors. Imagine a helix, like a spiral staircase; torsion describes how much the staircase rotates as you ascend it. In mathematical terms, torsion tells us how much the binormal vector changes along the curve.

For a parametric curve defined by position vector \( \mathbf{r}(t) \), torsion is computed using the formula:

• The cross product of the derivative of the tangent vector \( \mathbf{T} \) with the derivative of the binormal vector \( \mathbf{B} \).
• This product is divided by the magnitude of the resulting vector.

This step involves calculating the torsion scalar \( \tau(t) \), which, in our solution, was given as \( \tau(t) = -\frac{40}{17} \). The negative sign indicates the direction of the twist.

Tangent vector

The tangent vector is a fundamental concept when dealing with curves. It gives the direction of the curve at any given point. You can think of it as an arrow pointing in the direction you are heading as you move along the curve.

Mathematically, to find the tangent vector \( \mathbf{T}(t) \) for a parametrized curve \( \mathbf{r}(t) \), you:

• Compute the derivative of the position vector \( \mathbf{r}(t) \) with respect to parameter \( t \).
• Normalize this derivative, meaning you divide it by its magnitude.

This normalization ensures that the tangent vector has a unit length, making it a unit tangent vector. This is crucial because it allows us to compute other vectors like the normal and binormal vectors efficiently. In the given solution, the tangent vector was calculated as \( \mathbf{T}(t) = \frac{\langle 4 \cos t, -4\sin t, 10 \rangle}{2\sqrt{17}} \).

Normal vector

The normal vector \( \mathbf{N}(t) \) to a curve is orthogonal to the tangent vector. If you imagine holding a stick straight along the curve, the normal vector would point in the direction required to turn the stick along the curve's path. It helps describe how the direction of the tangent vector is changing.

To find the normal vector:

• Take the derivative of the unit tangent vector \( \mathbf{T}(t) \) with respect to the parameter \( t \).
• Normalize this derivative to ensure it has a unit length.

In the solution provided, the expression for \( \frac{d\mathbf{T}(t)}{dt} \) already came out as a normalized vector, indicating the principal unit normal vector.

Parametric curves

Parametric curves are a powerful way to represent curves in a coordinate space using a parameter, typically denoted by \( t \). Instead of expressing \( y \) solely in terms of \( x \), both \( x \) and \( y \) (and any other dimensions) are expressed as functions of this parameter.

For example, a circle can be expressed in parametric form as \( x = \cos t \), \( y = \sin t \), where \( t \) varies over the interval \([0, 2\pi]\). In three-dimensional space, parametric curves like spirals or helices demonstrate more complex movements, as seen in the original exercise.

Understanding parametric curves is vital for exploring vector calculus concepts like tangent, normal, and binormal vectors, as these depend on the notion of the parameter to describe motion along a curve. In the original problem, the curve was given as \[ \mathbf{r}(t)=\langle 4 \sin t, 4 \cos t, 10 t\rangle \], showcasing a helical shape in space.