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. $$\mathbf{r}(t)=\left\langle\cos t^{2}, \sin t^{2}\right\rangle, \text { for } 0 \leq t \leq \sqrt{\pi}$$

Short answer

Based on the given problem, the curve is given as \(\mathbf{r}(t)=\left\langle\cos t^{2}, \sin t^{2}\right\rangle\) for \(0 \leq t \leq \sqrt{\pi}\). Upon differentiation, we obtained \(\frac{d\mathbf{r}}{dt} = \left\langle-\sin{(t^2)}(2t), \cos{(t^2)}(2t)\right\rangle\). The magnitude of the derivative was calculated as \(2|t|\), which is not equal to 1. Therefore, the curve does not use arc length as a parameter. We then found the arc length function as \(s(t) = t^2\). By inverting this function, we got \(t(s) = \sqrt{s}\). Substituting it back into the original function, we obtained the arc length parameterization of the curve: \(\mathbf{r}(s)=\left\langle\cos s, \sin s\right\rangle\) for \(0 \leq s \leq \pi\). Hence, we have successfully parameterized the curve using arc length.

Step-by-step solution

Calculate the derivative

To calculate the derivative of the given curve with respect to \(t\), we differentiate each component with respect to \(t\): $$\frac{d\mathbf{r}}{dt} = \left\langle-\sin{(t^2)}(2t), \cos{(t^2)}(2t)\right\rangle$$

Check the magnitude of the derivative

Now, let's calculate the magnitude of this derivative: $$\left\lVert\frac{d\mathbf{r}}{dt}\right\rVert = \sqrt{(-\sin{(t^2)}(2t))^2 + (\cos{(t^2)}(2t))^2} = \sqrt{(-4t^2)\sin^2{(t^2)} + (4t^2)\cos^2{(t^2)}}$$ After simplifying the result further, we get: $$\left\lVert\frac{d\mathbf{r}}{dt}\right\rVert = \sqrt{4t^2(\sin^2{(t^2)} + \cos^2{(t^2)})} = \sqrt{4t^2} = 2|t|$$ Since the magnitude of the derivative is \(2|t|\) and not equal to 1, it does not use arc length as a parameter.

Find the arc length parameterization

Since \(\mathbf{r}(t)\) does not use arc length as a parameter, we should find an arc length parameterization. We will start by finding the total arc length function \(s(t)\): $$s(t) = \int_{0}^{t} \left\lVert\frac{d\mathbf{r}}{dt}\right\rVert dt = \int_{0}^{t} 2|t|\,\, dt = \left\{ \begin{array}{ll} \int_{0}^{t} 2t\, dt & \quad 0 \leq t \\ \int_{0}^{t} -2t\, dt & \quad t < 0 \end{array} \right.$$ Now, for the given range of the parameter, \(0 \leq t \leq \sqrt{\pi}\) the integration becomes: $$s(t) = \int_{0}^{t} 2t\, dt = t^2\Bigg|_0^t = t^2$$ Therefore, the arc length function \(s(t) = t^2\). Now, we need to invert this function to find \(t(s)\): $$t(s) = \sqrt{s}$$ Substitute this back into the original \(\mathbf{r}(t)\) function: $$\mathbf{r}(s) = \left\langle\cos (\sqrt{s})^2, \sin (\sqrt{s})^2\right\rangle = \left\langle\cos s, \sin s\right\rangle$$ Now we have the arc length parameterization of the curve: \(\mathbf{r}(s)=\left\langle\cos s, \sin s\right\rangle\) for \(0 \leq s \leq \pi\).

Key concepts

Derivative of Parametric Curves

Understanding parametric curves starts with differentiating each component of the vector function individually. For a curve described by \(\mathbf{r}(t) = \langle f(t), g(t) \rangle\), the derivative with respect to \(t\) is formed by differentiating the component functions: \(\frac{d\mathbf{r}}{dt} = \langle f'(t), g'(t) \rangle\). It's like just applying your regular differentiation rules to each part of the vector separately.
You can think of this derivative as tracking the change in direction and speed of the point moving along the curve. In our exercise:

• We differentiated \(\cos(t^2)\) and \(\sin(t^2)\) using the chain rule.
• Both derivatives included an extra \(2t\) from differentiating \(t^2\).

By calculating these derivatives, we gather essential information about how the curve behaves at each point.

Magnitude of Derivative

Once you have differentiated the parametric curve, the next step is to determine the magnitude of this derivative. This helps us understand the speed at which the curve is being traced. Think of the magnitude of the derivative as the "speedometer" for our curve.
We calculate the magnitude by taking the square root of the sum of the squares of the components of the derivative:\(\left\lVert \frac{d\mathbf{r}}{dt} \right\rVert = \sqrt{(f'(t))^2 + (g'(t))^2}\). In the exercise:

• We simplified the magnitude to \(\sqrt{4t^2}\).
• This ultimately resulted in \(2|t|\), indicating that the parameterization was not using arc length, as this value is not constant.

Having constant magnitude equal to 1 specifies the original curve uses arc length, but in our example, \(2|t|\) shows it doesn’t.

Arc Length Function

The arc length function \(s(t)\) is key in reparameterizing curves using their arc length. We derive this function by integrating the magnitude of the derivative over the curve. This tells us how much 'distance' is covered on the curve as the parameter \(t\) changes.
For our given curve:

• The arc length \(s(t)\) was calculated by integrating \(2|t|\) from \(0\) to \(t\), since \(0 \leq t \leq \sqrt{\pi}\).
• This yielded \(s(t) = t^2\).

This function now represents the length of the curve from the starting point up to \(t\). In essence, it transforms our parameter \(t\) into one that describes the actual distance along the curve, providing a more intuitive and consistent way to parameterize the curve.

Parameter Inversion

The final concept involves inverting the arc length function to obtain a new parameter in terms of \(s\), the arc length itself. This step turns the arc length function on its head—literally! By solving for \(t\) in terms of \(s\), the distance traveled becomes the main parameter.

• We found \(t(s) = \sqrt{s}\) from the arc length \(s(t) = t^2\).

Putting this back into the original parametric form, the curve becomes:\(\mathbf{r}(s) = \langle \cos s, \sin s \rangle\).
This means that \(s\) now directly measures the length along the curve, giving us a reparameterization of the original curve. Now, the parameterization is smooth and solely dependent on the 'distance,' which is why it's called arc length parameterization.