$\text { Find } \frac { d y } { d x } \text { and } \frac { d ^ { 2 } y } { d t ^ { 2 } } \text { if possible, and find the slope and concavity (if possible) at the }$ 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

Question

$\text { Find } \frac { d y } { d x } \text { and } \frac { d ^ { 2 } y } { d t ^ { 2 } } \text { if possible, and find the slope and concavity (if possible) at the }$ 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

Quizplus · Accepted Answer

First, find $\frac{dy}{dx}$ using the chain rule, where $\frac{dy}{dx} = \frac{dy}{dt} \cdot \frac{dt}{dx}$. Given $y = t^2 + 4t$, $\frac{dy}{dt} = 2t + 4$. Given $x = t + 10$, $\frac{dx}{dt} = 1$, so $\frac{dt}{dx} = 1$. Thus, $\frac{dy}{dx} = (2t + 4) \cdot 1 = 2t + 4$.To find $\frac{d^2y}{dx^2}$, differentiate $\frac{dy}{dx}$ with respect to $t$ and then divide by $\frac{dx}{dt}$, which gives $\frac{d^2y}{dx^2} = \frac{d}{dt}(2t + 4) \cdot \frac{1}{\frac{dx}{dt}} = 2$.At $t = 5$, $2t + 4 = 2(5) + 4 = 14$, so the slope is 14. Since $\frac{d^2y}{dx^2} = 2$ is positive, the graph is concave up at $t = 5$.

Point Corresponding to T = 5 x=t+10y=t2+4t\begin{array} { l } x = t + 10 \\y = t ^ { 2 } + 4 t\end{array}x=t+10y=t2+4t​

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