$\text { Suppose the table below shows the temperature } T \left( { } ^ { \circ } \bar { F } \right) \text { at which a given liquid boils }$ at selected pressures $p$ (pounds per square inch). A model that approximates the data is $T = 107.97 + 64.96 \ln p + 9.91 \sqrt { p }$. Find the rate of change of $T$ with respect to $p$ when $P = 20$. Round your answer to two decimal places.
$\begin{array}{|r|r|r|r|r|r|r|r|r|r|}
\hline P & 5 & 10 & \begin{array}{c}
14.696 \
(1 \mathrm{~atm})
\end{array} & 20 & 30 & 40 & 60 & 80 & 100 \
\hline T & 234.68 & 288.88 & 320.55 & 346.89 & 383.19 & 410.28 & 450.7 & 481.26 & 506.22 \
\hline
\end{array}$

A) $113.43 \mathrm { deg } / \mathrm { lb } / \mathrm { in } ^ { 2 }$
B) $4.36 \mathrm { deg } / \mathrm { lb } / \mathrm { in } ^ { 2 }$
C) $5.46 \mathrm { deg } / \mathrm { lb } / \mathrm { in } ^ { 2 }$
D) $112.33 \mathrm { deg } / \mathrm { lb } / \mathrm { in } ^ { 2 }$
E) $1.11 \mathrm { deg } / \mathrm { lb } / \mathrm { in } ^ { 2 }$

Question

$\text { Suppose the table below shows the temperature } T \left( { } ^ { \circ } \bar { F } \right) \text { at which a given liquid boils }$ at selected pressures $p$ (pounds per square inch). A model that approximates the data is $T = 107.97 + 64.96 \ln p + 9.91 \sqrt { p }$. Find the rate of change of $T$ with respect to $p$ when $P = 20$. Round your answer to two decimal places.
$\begin{array}{|r|r|r|r|r|r|r|r|r|r|}
\hline P & 5 & 10 & \begin{array}{c}
14.696 \
(1 \mathrm{~atm})
\end{array} & 20 & 30 & 40 & 60 & 80 & 100 \
\hline T & 234.68 & 288.88 & 320.55 & 346.89 & 383.19 & 410.28 & 450.7 & 481.26 & 506.22 \
\hline
\end{array}$

A) $113.43 \mathrm { deg } / \mathrm { lb } / \mathrm { in } ^ { 2 }$
B) $4.36 \mathrm { deg } / \mathrm { lb } / \mathrm { in } ^ { 2 }$
C) $5.46 \mathrm { deg } / \mathrm { lb } / \mathrm { in } ^ { 2 }$
D) $112.33 \mathrm { deg } / \mathrm { lb } / \mathrm { in } ^ { 2 }$
E) $1.11 \mathrm { deg } / \mathrm { lb } / \mathrm { in } ^ { 2 }$

Quizplus · Accepted Answer

To find the rate of change of $T$ with respect to $p$ when $p = 20$, we differentiate the given model $T = 107.97 + 64.96 \ln p + 9.91 \sqrt { p }$ with respect to $p$, which gives $\frac{dT}{dp} = \frac{64.96}{p} + \frac{9.91}{2\sqrt{p}}$. Substituting $p = 20$ into this derivative, we get $\frac{dT}{dp} = \frac{64.96}{20} + \frac{9.91}{2\sqrt{20}} \approx 4.36 \mathrm { deg } / \mathrm { lb } / \mathrm { in } ^ { 2 }$.