Question

Arc length parameterization Determine whether the following curves use arc length as a parameter. If not, find a description that uses arc length as a parameter. $$\mathbf{r}(t)=\langle 5 \cos t, 3 \sin t, 4 \sin t\rangle, \text { for } 0 \leq t \leq \pi$$

Short answer

Based on the given vector function for the curve, we first calculated the derivative of the curve and found its magnitude. We discovered that the magnitude of the derivative vector is not equal to 1, indicating that the curve is not parameterized by arc length. To reparameterize the curve using arc length, we found a relationship between arc length (s) and the original parameter (t), and rewrote the curve using this new parameter. The new curve, using arc length as a parameter, is given by: $$\mathbf{r}(s)=\langle 5 \cos \frac{s}{5}, 3 \sin \frac{s}{5}, 4 \sin \frac{s}{5}\rangle, \text { for } 0\leq s \leq 5\pi$$

Step-by-step solution

Compute the Derivative of \(\mathbf{r}(t)\)

To find \(\mathbf{r'}(t)\), we need to find the derivatives of each component of the given curve: \(5 \cos t, 3 \sin t, 4 \sin t\). We get: $$\mathbf{r'}(t) = \langle -5 \sin t, 3 \cos t, 4 \cos t\rangle$$

Check If \(|\mathbf{r'}(t)| = 1\)

Next, we'll calculate the magnitude of the derivative vector \(\mathbf{r'}(t)\): $$|\mathbf{r'}(t)| =\sqrt{(-5 \sin t)^2 + (3 \cos t)^2 + (4 \cos t)^2} = \sqrt{25 \sin^2 t + 9 \cos^2 t + 16 \cos^2 t} $$ We then simplify the expression: $$|\mathbf{r'}(t)| = \sqrt{25 \sin^2 t + 25 \cos^2 t} = \sqrt{25(\sin^2 t+\cos^2{t})} = 5$$ Since \(|\mathbf{r'}(t)| \neq 1\), we can conclude that the curve is not parameterized by arc length.

Reparameterize Using Arc Length

As \(|\mathbf{r'}(t)| = 5\), it means that the arc length \(s\) and the parameter \(t\) are related by: $$s = 5t$$ Now we can find a new parameter in terms of s: $$t= \frac{s}{5}$$ Next, we rewrite the original curve using \(t = \frac{s}{5}\). $$\mathbf{r}(s)=\langle 5 \cos \frac{s}{5}, 3 \sin \frac{s}{5}, 4 \sin \frac{s}{5}\rangle, \text { for } 0\leq s \leq 5\pi$$ This new curve uses arc length as a parameter.

Key concepts

Parametric Curves

Parametric curves are a way to describe geometrical shapes using parameters. Instead of using the x and y coordinates specifically, parametric equations use a third variable, often denoted as \( t \), to express the coordinates. For example, for a parametric curve \( \mathbf{r}(t) = \langle x(t), y(t), z(t) \rangle \), each component \( x(t) \), \( y(t) \), and \( z(t) \) is a function of the parameter \( t \). This approach is particularly useful to define curves in spaces beyond two dimensions, allowing us to represent curves in 3D or higher dimensions.

Considerate advantages of parametric equations include:

• Flexibility: They give more freedom to describe motion and paths of objects.
• Versatility: Suitable for representing curves that cannot be defined as functions \( y = f(x) \).

In vector calculus, parametric curves are an essential tool for modeling and solving complex problems in physics, engineering, and computer graphics.

Derivative of Vector Functions

The derivative of a vector function measures how the vector changes as the parameter changes. For a parametric curve \( \mathbf{r}(t) = \langle x(t), y(t), z(t) \rangle \), its derivative \( \mathbf{r'}(t) \) shows how the position of a point on the curve is changing with respect to \( t \).

Here's how you calculate it:

• Take the derivative of each component of the vector function: \( \mathbf{r'}(t) = \langle x'(t), y'(t), z'(t) \rangle \).

In the original problem, given \( \mathbf{r}(t) = \langle 5 \cos t, 3 \sin t, 4 \sin t \rangle \), we derived:

• \( \mathbf{r'}(t) = \langle -5 \sin t, 3 \cos t, 4 \cos t \rangle \).

The magnitude of this derivative tells us the rate at which the curve's length changes with \( t \). It's a valuable concept in determining if a curve is parameterized by arc length or requires reparametrization.

Therefore, understanding the derivative of vector functions helps in problems involving motion along a path or finding velocities and accelerations of objects moving in space.

Reparametrization

Reparametrization is the process of changing the parameter of a curve to a new one that offers a different perspective or information. When a curve is not naturally parameterized by arc length, we might want to reparametrize it so that the traversal of the curve is uniform in terms of distance.

Why is reparametrization important? It helps in simplifying computations related to the arc length and ensures that the speed along the curve is constant, which is particularly useful in animations and physics simulations.

In our exercise, the curve \( \mathbf{r}(t) \) had a derivative's magnitude \( |\mathbf{r'}(t)| = 5 \), indicating it wasn't using arc length as a parameter because the speed isn't constant (it's 5 times greater). So we used the relationship

\[ s = 5t \]
to derive a new parameter \( t = \frac{s}{5} \)
and express the curve in terms of \( s \) as:
\[ \mathbf{r}(s) = \langle 5 \cos \frac{s}{5}, 3 \sin \frac{s}{5}, 4 \sin \frac{s}{5} \rangle \]
for \( 0 \leq s \leq 5\pi \).

This transformation makes the curve arc-length parameterized, meaning that moving along the curve one unit in parameter \( s \) corresponds to moving one unit in space, thus providing a consistent geometric interpretation.