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anonymous
 one year ago
1. Modulus of rigidity G=(R^4) T/L where R is the radius, T theta the angle of twist and L the length
a) Determine the approximate percentage error in G when R is increased by
2%, T is reduced by 5% and L is increased by 4%
b) Solve a) by another method.
anonymous
 one year ago
1. Modulus of rigidity G=(R^4) T/L where R is the radius, T theta the angle of twist and L the length a) Determine the approximate percentage error in G when R is increased by 2%, T is reduced by 5% and L is increased by 4% b) Solve a) by another method.

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0For a) i can solve by using small increment formula (partial differential) but for question b i cannot think of other method.

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2the simple and accurate way, & poss what they wanted for part a) \[\frac{1.02^4 \times 1.05}{1.04}  1\] these deltas are far too big for a total differential ... which i assume is the sought b) solution

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0can i ask why you substitute 1.02,1.05 and 1.04 inside the variable and minus with one.

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2you are increasing them (R,T,L) by those percentages, right?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\delta G=\frac{ dG }{ dR } \delta R+\frac{ dG }{ dL } \delta L +\frac{ dG }{ dT } \delta T\] when i substitute all the information given, i get the aswe as G decrease by 1%

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i dont think we can simply substitute the given percentage.. thes the question said R increased by 2% meaning 0.02R not just 0.02.

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2ok, i think we are in agreement but maybe this will show how. what i wrote was shorthand for \[\Delta G = \frac{(1.02R)^4 (0.95T)}{(1.04L)}  \frac{R^4 T}{L} = 0.0112 G\] NB: i had a typo in previous version as T decreases and not increases, but this is the way to do it using differentials gives ∆G = 0.01 G, so we do agree the first method is most accurate and simplest. PS i know i haven't helped by misreading the question; but the principle still holds

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yaaaa....now i understand...thanks for your response..i really appreciated..
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