The vector r(t) is the position vector of a particle at time t. Find the angle between the velocity and the acceleration vectors at time t = 0.
-$$\mathbf { r } ( \mathrm { t } ) = \sin ^ { - 1 } ( 7 \mathrm { t } ) \mathbf { i } + \ln \left( 8 \mathrm { t } ^ { 2 } + 1 \right) \mathbf { j } + \sqrt { 7 \mathrm { t } ^ { 2 } + 1 } \mathbf { k }$$
A) $\frac { \pi } { 4 }$
B) $\frac { \pi } { 3 }$
C) $\frac { \pi } { 2 }$
D) 0

Question

Quizplus · Accepted Answer

To find the angle between the velocity and acceleration vectors at $t = 0$, we first need to compute these vectors. The velocity vector $\mathbf{v}(t)$ is the derivative of the position vector $\mathbf{r}(t)$, and the acceleration vector $\mathbf{a}(t)$ is the derivative of the velocity vector.Given $\mathbf{r}(t) = \sin^{-1}(7t)\mathbf{i} + \ln(8t^2 + 1)\mathbf{j} + \sqrt{7t^2 + 1}\mathbf{k}$,1. Differentiate to find the velocity vector $\mathbf{v}(t)$:$$ \mathbf{v}(t) = \frac{d}{dt}[\sin^{-1}(7t)]\mathbf{i} + \frac{d}{dt}[\ln(8t^2 + 1)]\mathbf{j} + \frac{d}{dt}[\sqrt{7t^2 + 1}]\mathbf{k} $$$$ \mathbf{v}(t) = \frac{7}{\sqrt{1-(7t)^2}}\mathbf{i} + \frac{16t}{8t^2 + 1}\mathbf{j} + \frac{7t}{\sqrt{7t^2 + 1}}\mathbf{k} $$2. Differentiate again to find the acceleration vector $\mathbf{a}(t)$:$$ \mathbf{a}(t) = \frac{d}{dt}\left[\frac{7}{\sqrt{1-(7t)^2}}\right]\mathbf{i} + \frac{d}{dt}\left[\frac{16t}{8t^2 + 1}\right]\mathbf{j} + \frac{d}{dt}\left[\frac{7t}{\sqrt{7t^2 + 1}}\right]\mathbf{k} $$At $t = 0$, the velocity vector $\mathbf{v}(0)$ and the acceleration vector $\mathbf{a}(0)$ simplify significantly:$$ \mathbf{v}(0) = 7\mathbf{i} + 0\mathbf{j} + 0\mathbf{k} $$$$ \mathbf{a}(0) = 0\mathbf{i} + 16\mathbf{j} + 0\mathbf{k} $$The angle $\theta$ between $\mathbf{v}(0)$ and $\mathbf{a}(0)$ can be found using the dot product formula:$$ \cos(\theta) = \frac{\mathbf{v}(0) \cdot \mathbf{a}(0)}{|\mathbf{v}(0)||\mathbf{a}(0)|} $$Given that $\mathbf{v}(0) \cdot \mathbf{a}(0) = 0$, the angle $\theta$ is such that $\cos(\theta) = 0$, which implies $\theta = \frac{\pi}{2}$. This means the velocity and acceleration vectors are perpendicular at $t = 0$.

The Vector R(t) Is the Position Vector of a Particle