The temperature difference \( T \) between the inner and outer walls of a pipe carrying cooling water to a large transmitting tube is modeled by: \( T = 38.61 \ln ( D / 5 ) \) degrees where \( D \) is the pipe's outside diameter in inches. Develop the model formula you would use to calculate the rate of change of temperature with respect to the change in diameter dT/dD?
Question
Answer
\(\frac { d } { d D } ( 38.61 \log ( \frac { D } { 5 } ) ) \)
Factor out constants:
\( = 38.61 ( \frac { d } { d D } ( \log ( \frac { D } { 5 } ) ) ) \)
Using the chain rule, \( \frac { d } { d D } ( \log ( \frac { D } { 5 } ) ) = \)
\( \frac { d \log ( u ) } { d u } \frac { d u } { d D } , \) where \( u = \frac { D } { 5 } \) and
\( \frac { d } { d u } ( \log ( u ) ) = \frac { 1 } { u } : \)
\( = 38.61 \cdot \frac { 5 ( \frac { d } { d D } ( \frac { D } { 5 } ) ) } { D } \)
Simplify the expression:
\( 193.05 ( \frac { d } { d D } ( \frac { D } { 5 } ) ) \) /D
Factor out constants:
\( = \frac { 1 } { 5 } ( \frac { d } { d D } ( D ) ) \frac { 193.05 } { D } \)
Simplify the expression:
\( = \frac { 38.61 ( \frac { d } { d D } ( D ) ) } { D } \)
The derivative of \( D \) is \( 1 \) :
\( = 1 \frac { 38.61 } { D } \)
Simplify the expression:
\( = \frac { 38.61 } { D }\)