Last question in my assignment

[Rusty789]

Firstly I have no idea how to tackle this one, I am not sure if we've been taught it yet or if its sort of a lead on to the next set of topics..

Homework Statement

The shear stress t in a shaft of diameter D under torque T is given by t = KT / \PiD³.

Determine the approximate error in calculating t if T is measure 3% too small and D is measured 1.5% to large.

Homework Equations

None

The Attempt at a Solution

As previously stated i don't know how to handle this one, so any advice will be great.

Also i don't want to know the answer, i would like to know how to work it out..

EDIT:

My first theory was to put numbers in do the actual calculation twice with physical number and compare the difference but that didnt work. e.g. (although it did this time around, so I am now completely confused)

Completely random values.

t = 3*10/3.14*10³ = 9.554*10^-3

then do:

t = 3*9.7/3.14*10.15³ = 8.862*10^-3

This gives a percentage difference of 7.243%

t = (3*154)/(3.14*56³) = 8.378*10^-3

t = (3*149.38)/(3.14*56.84³) ) =7.772*10^-3

this gives a percentage difference of 7.233%

Is this right?

 

[Mark44]

Firstly I have no idea how to tackle this one, I am not sure if we've been taught it yet or if its sort of a lead on to the next set of topics..

Homework Statement

The shear stress t in a shaft of diameter D under torque T is given by t = KT / \PiD³.

Determine the approximate error in calculating t if T is measure 3% too small and D is measured 1.5% to large.

Homework Equations

None

The Attempt at a Solution

As previously stated i don't know how to handle this one, so any advice will be great.

Also i don't want to know the answer, i would like to know how to work it out..

EDIT:

My first theory was to put numbers in do the actual calculation twice with physical number and compare the difference but that didnt work. e.g. (although it did this time around, so I am now completely confused)

Completely random values.

t = 3*10/3.14*10³ = 9.554*10^-3

then do:

t = 3*9.7/3.14*10.15³ = 8.862*10^-3

This gives a percentage difference of 7.243%

t = (3*154)/(3.14*56³) = 8.378*10^-3

t = (3*149.38)/(3.14*56.84³) ) =7.772*10^-3

this gives a percentage difference of 7.233%

Is this right?

Am I correct in assuming that you are in a calculus course? The right way to do this type of problem is to calculate the total differential of t, which will involve taking the partials with respect to T and with respect to D.

IOW,
dt = \frac{\partial}{\partial T}\left(\frac{KT}{\pi D^3}\right)dT + \frac{\partial}{\partial D}\left(\frac{KT}{\pi D^3}\right)dD

I have started this for you, but you will need to use other differentiation rules to finish it. Your textbook should have some similar examples.

 

[Rusty789]

I'm doing HNC in electronics, and this unit is called analytical methods.

 

[Mark44]

I don't know what HNC is. What I showed in my previous post is the usual way of going about problems like this. You can substitute -.03 for dT and +.015 for dD, but you will need to know how to differentiate to get the partial derivatives I showed.

 

[Rusty789]

Its a higher national certificate, a british thing, I'm right in thinking this is mainly american?

Differentiation is definitely part of the next topic so maybe it was just to see how we'd handle it..

 

[Mark44]

Its a higher national certificate, a british thing, I'm right in thinking this is mainly american?

Do you mean "mainly British"? AFAIK, we don't have such a thing here in the States.

Differentiation is definitely part of the next topic so maybe it was just to see how we'd handle it..

You will first be presented ordinary differentiation (differentiation of functions with one variables). After that would come partial differentiation (differentiation of functions with two or more variables).