Question

Find the curvature \(K\) of the curve, where \(s\) is the arc length parameter. Curve in Exercise 20: \(\mathbf{r}(t)=\left\langle 4(\sin t-t \cos t), 4(\cos t+t \sin t), \frac{3}{2} t^{2}\right\rangle\)

Short answer

The curvature \(K\) of the curve is \( K = \frac{\sqrt{144t^2+256t^4}}{125t^3} \)

Step-by-step solution

Find the first derivative \( \mathbf{r}'(t) \)

Differentiate each component of \( \mathbf{r}(t) \) with respect to \( t \): \( \mathbf{r}'(t)=\langle 4(\cos t - \cos t+ t\sin t), 4(-\sin t+\sin t+ t\cos t), 3t \rangle = \langle 4t\sin t, 4t\cos t, 3t \rangle

Find the second derivative \( \mathbf{r}''(t) \)

Again, differentiate each component of \( \mathbf{r}'(t) \) with respect to \( t \): \( \mathbf{r}''(t) = \langle 4(\cos t +t\cos t), -4(\sin t +t\sin t), 3 \rangle = \langle 4(1+t)\cos t, -4(1+t)\sin t, 3 \rangle

Cross Product

Calculate the cross product of \( \mathbf{r}'(t) \) and \( \mathbf{r}''(t) \): \( \mathbf{r}'(t) \times \mathbf{r}''(t) = \langle 12t\cos t, 12t\sin t, -16t^2 \rangle

Magnitude of cross product

Calculate the magnitude of the cross product vector: \( \|\mathbf{r}'(t) \times \mathbf{r}''(t)\| = \sqrt{(12t\cos t)^2+(12t\sin t)^2 +(-16t^2)^2} = \sqrt{144t^2+256t^4} \)

Magnitude of first derivative

Calculate the magnitude of the first derivative: \( \|\mathbf{r}'(t)\| = \sqrt{(4t\sin t)^2+(4t\cos t)^2+(3t)^2} = \sqrt{16t^2+9t^2} = \sqrt{25t^2} = 5t \)

Cube of magnitude of first derivative

Calculate the cube of the magnitude of the first derivative: \( \|\mathbf{r}'(t)\|^3 = (5t)^3 = 125t^3 \)

Calculate Curvature

Calculate the curvature:\( K = \frac{\|\mathbf{r}'(t) \times \mathbf{r}''(t)\|}{\|\mathbf{r}'(t)\|^3} = \frac{\sqrt{144t^2+256t^4}}{125t^3} \)