A particle moving along a curve in the $x y$-plane is at position $( x ( t ) , y ( t ) )$ at time $t > 0$. The particle moves in such a way that $\frac { d x } { d t } = \sqrt { 1 + t ^ { 2 } }$ and $\frac { d y } { d t } = \ln \left( 2 + t ^ { 2 } \right)$. At time $t = 4$, the particle is at the point $( 1,5 )$. (a) Find the slope of the line tangent to the path of the particle at time $t = 4$. (b) Find the speed of the particle at time $t = 4$, and find the acceleration vector of the particle at time $t = 4$. (c) Find the $y$-coordinate of the particle's position at time $t = 6$. (d) Find the total distance the particle travels along the curve from time $t = 4$ to time $t = 6$.
A particle moving along a curve in the $x y$-plane is at position $( x ( t ) , y ( t ) )$ at time $t > 0$. The particle moves in such a way that $\frac { d x } { d t } = \sqrt { 1 + t ^ { 2 } }$ and $\frac { d y } { d t } = \ln \left( 2 + t ^ { 2 } \right)$. At time $t = 4$, the particle is at the point $( 1,5 )$.\\
(a) Find the slope of the line tangent to the path of the particle at time $t = 4$.\\
(b) Find the speed of the particle at time $t = 4$, and find the acceleration vector of the particle at time $t = 4$.\\
(c) Find the $y$-coordinate of the particle's position at time $t = 6$.\\
(d) Find the total distance the particle travels along the curve from time $t = 4$ to time $t = 6$.