### Still have math questions?

Ask our expert tutors
Algebra
Question

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?

$$\frac { d } { d D } ( 38.61 \log ( \frac { D } { 5 } ) )$$

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:

$$= \frac { 38.61 } { D }$$

Solution
View full explanation on CameraMath App.