Point Corresponding to T = 5 $\begin{array} { l } x = t + 10 \\y = t ^ { 2 } + 4 t\end{array}$

Multiple Choice

$\text { Find } \frac { d y } { d x } \text { and } \frac { d ^ { 2 } y } { d t ^ { 2 } } \text { if possible, and find the slope and concavity (if possible) at the }$ point corresponding to t = 5.
$\begin{array} { l } x = t + 10 \\y = t ^ { 2 } + 4 t\end{array}$

A) $\frac { d y } { d x } = \frac { 1 } { 3 } t ^ { 3 } + 2 t ^ { 2 } , \frac { d ^ { 2 } y } { d x ^ { 2 } } = 3 t ^ { 2 } + 4 t$ , at $t = 5$ : slope $- \frac { 325 } { 6 }$ and concave up
B) $\frac { d y } { d x } = - 2 t + 4 , \frac { d ^ { 2 } y } { d x ^ { 2 } } = - 2 ;$ at $t = 5$ : slope $- 6$ and concave down
C) $\frac { d y } { d x } = 2 t + 4 , \frac { d ^ { 2 } y } { d x ^ { 2 } } = 2$ ; at $t = 5$ : slope 14 and concave up
D) $\frac { d y } { d x } = - 2 t + 4 , \frac { d ^ { 2 } y } { d x ^ { 2 } } = - 2$ ; at $t = 5$ : slope 6 and concave down
E) $\frac { d y } { d x } = \frac { 1 } { 3 } t ^ { 3 } + 2 t ^ { 2 } , \frac { d ^ { 2 } y } { d x ^ { 2 } } = 3 t ^ { 2 } + 4 t$ , at $t = 5 :$ slope $\frac { 325 } { 6 }$ and concave up