Answer:
[tex]\mathbf{v}(\pi/4)=-3\mathbf{i}+3\mathbf{j}+3 \mathbf{k}[/tex]
Step-by-step explanation:
Instantaneous Velocity
Given r(t) as the vector function of the position for time t, the instantaneous velocity is computed as:
[tex]\mathbf{v}=\frac{d\mathbf{r}}{dt}=\mathbf{r}'(t)[/tex]
We are given:
[tex]\mathbf{r}=\sin^2(3t)\mathbf{i}+3t\mathbf{j}-\cos^2(3t)\mathbf{k}[/tex]
Thus:
[tex]\mathbf{v}(t)=\mathbf{r}'(t)=[\sin^2(3t)\mathbf{i}+3t\mathbf{j}-\cos^2(3t)\mathbf{k}]'[/tex]
Computing the derivative:
[tex]\mathbf{v}(t)=2\sin(3t)\cos(3t)(3)\mathbf{i}+3\mathbf{j}+2\cos(3t)\sin(3t)(3)\mathbf{k}[/tex]
[tex]\mathbf{v}(t)=6\sin(3t)\cos(3t)\mathbf{i}+3\mathbf{j}+6\cos(3t)\sin(3t)\mathbf{k}[/tex]
Evaluating for t=π/4:
[tex]\mathbf{v}(\pi/4)=6\sin(3\pi/4)\cos(3\pi/4)\mathbf{i}+3\mathbf{j}+6\cos(3\pi/4)\sin(3\pi/4) \mathbf{k}[/tex]
[tex]\mathbf{v}(\pi/4)=6(\sqrt{2}/2)(-\sqrt{2}/2)\mathbf{i}+3\mathbf{j}+6(-\sqrt{2}/2))(\sqrt{2}/2)) \mathbf{k}[/tex]
[tex]\mathbf{v}(\pi/4)=-3\mathbf{i}+3\mathbf{j}-3 \mathbf{k}[/tex]
