91Ó°ÊÓ

Compute the derivative of the curve function \(\mathbf{r}(t)\) with respect to \(t\): $$\mathbf{r}'(t) = \frac{d \mathbf{r}}{d t} = \left\langle -3 \cos ^{2} t \sin t, 3 \sin ^{2} t \cos t\right\rangle$$ Compute the second derivative of the curve function \(\mathbf{r}(t)\) with respect to \(t\): $$\mathbf{r}''(t) = \frac{d^2 \mathbf{r}}{d t^2} = \left\langle 3 (3 \cos ^{2} t \sin ^{2} t - \cos ^{4} t), 3 (3 \sin ^{2} t \cos ^{2} t - \sin ^{4} t)\right\rangle$$

We want to compute the unit tangent vector, which is defined as the normalized version of the first derivative of the curve: $$\mathbf{T}(t) = \frac{\mathbf{r}'(t)}{\lVert \mathbf{r}'(t) \rVert}$$ First, compute the magnitude of the first derivative: $$\lVert \mathbf{r}'(t) \rVert = \sqrt{(-3\cos^2t\sin t)^2 + (3\sin^2t\cos t)^2} = 3|\sin t\cos t|\sqrt{\sin^2t + \cos^2t}=3|\sin t\cos t|$$ Now, compute the unit tangent vector: $$\mathbf{T}(t) = \frac{\left\langle -3 \cos ^{2} t \sin t, 3 \sin ^{2} t \cos t\right\rangle}{3|\sin t\cos t|}$$ Since, \(\cos^2 t + \sin^2 t = 1\), using that to simplify the above equation: $$\mathbf{T}(t)=\left\langle - \operatorname{sgn}(\sin t\cos t) \cos ^{2} t, \operatorname{sgn}(\sin t\cos t) \sin ^{2} t\right\rangle$$

Now, we will compute the curvature using the formula mentioned above: $$\kappa = \frac{\lVert \mathbf{r}'(t) \times \mathbf{r}''(t) \rVert}{\lVert \mathbf{r}'(t) \rVert^3}$$ First, compute the cross product between the first and second derivatives: $$\mathbf{r}'(t) \times \mathbf{r}''(t) = -9(\cos^5 t\sin^2 t - \cos^2 t\sin^5 t)$$ Compute the magnitude of the cross product: $$\lVert \mathbf{r}'(t) \times \mathbf{r}''(t) \rVert = |-9(\cos^5 t\sin^2 t - \cos^2 t\sin^5 t)|$$ Now, compute the curvature: $$\kappa = \frac{|-9(\cos^5 t\sin^2 t - \cos^2 t\sin^5 t)|}{(3|\sin t\cos t|)^3}=\frac{3|\cos^5 t\sin^2 t - \cos^2 t\sin^5 t|}{|\sin^3 t\cos^3 t|}$$ Finally, the unit tangent vector, \(\mathbf{T}(t)\), and the curvature, \(\kappa\), of the given parameterized curve are: $$\mathbf{T}(t) =\left\langle - \operatorname{sgn}(\sin t\cos t) \cos ^{2} t, \operatorname{sgn}(\sin t\cos t) \sin ^{2} t\right\rangle$$ $$\kappa = \frac{3|\cos^5 t\sin^2 t - \cos^2 t\sin^5 t|}{|\sin^3 t\cos^3 t|}$$