Question

Find the curvature \(K\) of the curve, where \(s\) is the arc length parameter. Helix in Exercise 19: \(\mathbf{r}(t)=\langle 2 \cos t, 2 \sin t, t\rangle\)

Short answer

The curvature \(K\) of the curve \(\mathbf{r}(t)=\langle 2 \cos t, 2 \sin t, t\rangle\) is \(\frac{\sqrt{20}}{27}\).

Step-by-step solution

Compute the Derivatives

The first step will be to compute the first derivative, i.e., the velocity, \(\mathbf{r}'(t)\), and the second derivative, i.e., the acceleration, \(\mathbf{r}''(t)\), of the given parametric equation. \(\mathbf{r}'(t)=\langle-2\sin{t}, 2\cos{t}, 1\rangle\) is the first derivative, while \(\mathbf{r}''(t)=\langle-2\cos{t}, -2\sin{t}, 0\rangle\) is the second derivative.

Compute the Cross Product of the Derivatives

Next, calculate the cross product of the velocity and the acceleration, \(\mathbf{r}'(t) \times \mathbf{r}''(t)\). Using the formula for the cross product, this will be \(\langle 2\sin{t}, 2\cos{t}, -4\rangle\).

Calculate the Magnitude

Now, calculate the magnitude of both vectors \(\mathbf{r}'(t) \times \mathbf{r}''(t)\) and \(\mathbf{r}'(t)\). The magnitude, \(\|\mathbf{r}'(t) \times \mathbf{r}''(t)\| = |\langle 2\sin{t}, 2\cos{t}, -4\rangle| = \sqrt{(2\sin{t})^2+(2\cos{t})^2 + (-4)^2} = \sqrt{20}\). Similarly, \(\|\mathbf{r}'(t)\| = |\langle[-2\sin{t}, 2\cos{t}, 1]\rangle| = sqrt{(-2sin{t})^2 + (2cos{t})^2 + 1^2} = sqrt{9}\).

Find the Curvature

Finally, using the formula for curvature: \(K=\frac{\|\mathbf{r}'(t) \times \mathbf{r}''(t)\|}{\|\mathbf{r}'(t)\|^3}\), substitute the calculated magnitudes to get \(K=\frac{\sqrt{20}}{(sqrt{9})^3}= \frac{\sqrt{20}}{27}\)