Total Differential of Gauge Function g

[PhysDrew]

Homework Statement

Manufacturing plants create rolls of metal of a desired gauge (thickness)
by feeding the metal through very large rollers. The thickness, g (mm),
of the resulting metal depends on the gap, r (mm), between the working
rollers, the speed, s (m/s), at which the rollers turn and the temperature,
t (oC), of the metal.
(i) Write down an expression for the total differential of the gauge
function g. In a few words, explain what this total differential
represents.

(ii) For a certain metal, a gauge of 4mm is produced by a roller gap
of 4mm, a speed of 10m/s and a temperature of 900oC. Experi-
ments show that for the same metal, an increase in speed of 0.2m/s
increases the gauge by 0.06mm and an increase in temperature of
10oC decreases the gauge by 0.04mm. Use a linearization of the
gauge function to estimate the gauge of this metal at a roller gap of
4mm, a roller speed of 10.1m/s and a metal temperature of 880oC.

Homework Equations

For this the total differential would be: dg=(dg/dr)*dr + (dg/ds)*ds + (dg/dt)*dt (Eq. 1)

The Attempt at a Solution

I have been trying to get the general equation relating all the variables to the gauge thickness 'g.' Haven't been very successful though. Am I on the right track if I equate dg in the instance where an increase in speed of 0.2m/s as; 0.06= (dg/ds)*10, as all the other variables are not changing, hence they are replaced with 0. This is in relation to Eq. 1 (the total differential). Please help, I'm running around in circles here it seems!

 

[Stephen Tashi]

Am I on the right track if I equate dg in the instance where an increase in speed of 0.2m/s as; 0.06= (dg/ds)*10, as all the other variables are not changing, hence they are replaced with 0. This is in relation to Eq. 1 (the total differential). Please help, I'm running around in circles here it seems!

You're on the right track insofar as you must solve for dg/ds and dg/dt. (No need to know dg/dr since dr =0 in the last part of the question.) But why did you use 10 for ds instead of 0.2 ?

 

[PhysDrew]

You're on the right track insofar as you must solve for dg/ds and dg/dt. (No need to know dg/dr since dr =0 in the last part of the question.) But why did you use 10 for ds instead of 0.2 ?

Ah darn it yes that was a stupid mistake! Should have been 0.2. And I've just realized I don't need the entire equation as dg/dr will be multiplied by 0, just like you've said. Thanks for that!