Question

Let \(\mathbf{u}(t)=\left\langle 1, t, t^{2}\right\rangle, \mathbf{v}(t)=\left\langle t^{2},-2 t, 1\right\rangle\) and \(g(t)=2 \sqrt{t}\). Compute the derivatives of the following functions. $$\mathbf{u}(t) \times \mathbf{v}(t)$$

Short answer

Answer: The derivative of the cross product of the given vector functions is \(\frac{d\mathbf{w}(t)}{dt} = \langle 1 + 6t^2, 4t^3, 2 - 3t^2 \rangle\).

Step-by-step solution

Compute the cross product of \(\mathbf{u}(t)\) and \(\mathbf{v}(t)\)

The cross product of \(\mathbf{u}(t)\) and \(\mathbf{v}(t)\) can be computed using the determinant of a matrix formed by substituting the given vectors' components as rows. In this case, we need to compute the following determinant: $$ \mathbf{w}(t) = \mathbf{u}(t) \times \mathbf{v}(t) = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & t & t^2 \\ t^2 & -2t & 1 \end{vmatrix} $$

Calculate the components of the cross product vector function

To find the components of \(\mathbf{w}(t)\), we need to compute the determinants of three 2x2 matrices: 1. For the \(\mathbf{i}\) component, we eliminate the first row and first column: $$ w_1 = \begin{vmatrix} t & t^2 \\ -2t & 1 \end{vmatrix} = t(1) - (-2t)(t^2) = t + 2t^3 $$ 2. For the \(\mathbf{j}\) component, we eliminate the first row and second column: $$ w_2 = -\begin{vmatrix} 1 & t^2 \\ t^2 & 1 \end{vmatrix} = -(1 - t^4) = t^4 - 1 $$ 3. For the \(\mathbf{k}\) component, we eliminate the first row and third column: $$ w_3 = \begin{vmatrix} 1 & t \\ t^2 & -2t \end{vmatrix} = -(1)(-2t) - (t^2)(t) = 2t - t^3 $$ Now, the cross product vector function is given by: $$ \mathbf{w}(t) = \langle w_1, w_2, w_3 \rangle = \langle t + 2t^3, t^4 - 1, 2t - t^3 \rangle $$

Compute the derivative of \(\mathbf{w}(t)\)

Now we need to compute the derivative of each component of the vector function \(\mathbf{w}(t)\) with respect to \(t\): 1. For the \(\mathbf{i}\) component: $$ \frac{dw_1}{dt} = \frac{d}{dt} (t + 2t^3) = 1 + 6t^2 $$ 2. For the \(\mathbf{j}\) component: $$ \frac{dw_2}{dt} = \frac{d}{dt} (t^4 - 1) = 4t^3 $$ 3. For the \(\mathbf{k}\) component: $$ \frac{dw_3}{dt} = \frac{d}{dt} (2t - t^3) = 2 - 3t^2 $$ Now, the derivative of the vector function \(\mathbf{w}(t)\) is given by: $$ \frac{d\mathbf{w}(t)}{dt} = \langle \frac{dw_1}{dt}, \frac{dw_2}{dt}, \frac{dw_3}{dt} \rangle = \langle 1 + 6t^2, 4t^3, 2 - 3t^2 \rangle $$