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

question 51

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

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