*Remarks in italics are by me for extra explanation. These comments would not be necessary for full marks in an exam situation. Exercises taken from p.63 of these notes.*

## Question 1

*Use differentials to estimate the amount of tin in a closed tin closed tin can with diameter 8 cm and height 12 cm if the can is 0.04 cm thick.*

### Solution

Assuming that the measurements of 8 cm and 12 cm are taken from the outside of the can, then we could estimate the change in volume of a cylinder if the radius were increased by 0.04 cm to 4 cm and the height increased by 0.08 cm to 12 cm (convince yourself of this with a picture.). Now the tin in the can comprises the difference between a cylinder and a cylinder.

Now the volume of a cylinder is given by

.

We can use the differential of , (evaluated at — *although the other way around would also be a good estimate*) to estimate the change in volume:

,

where cm and cm. Now

, and

.

cm.

## Question 2

*Use differentials to estimate the amount of metal in a closed cylindrical can that is 10 cm high and 4 cm is diameter if the metal in the wall is 0.05 cm thick and the metal in the top and bottom is 0.1 cm thick.*

### Solution

Using the same method, the differential is a good estimate:

,

where in this case we take , and and . Now

,

.

Hence

cm .

## Question 3

*If is the total resistance of three resistors, connected in parallel, with the resistances and , then *

*. *

*If the resistances are measured as , and , with possible errors of 5 % in each case, estimate the maximum error in the calculated value of .*

### Solution

Now first we want to get an expression for :

,

We can approximate the error in , by:

,

where corresponds to the error in , .

Now the errors are 5 % hence:

Also, using the quotient rule:

Similarly,

.

.

Note that these will be evaluated at , using a calculator;

Similarly,

Hence,

.

## Question 4

*The moment of inertia of a body about an axis is given by where is a constant and and are the dimensions of the body. If and are measured as 2 m and 0.8 m respectively, and the measurement errors are 10 cm in and $8$ mm in , determine the error in the calculated value of the moment of inertia using the measured values, in terms of .*

### Solution

In class we did an example where we estimated the change in due to changes in and (from the sample test). The only difference between that example and this one is that errors are always positive and we take absolute values; i.e:

.

## Question 5

*The volume, , of a liquid of viscosity coefficient delivered after a time when passed through a tube of length and diameter by a pressure is given by *

* *

*If the errors in , and are 1 %, 2 % and 3 % respectively, determine the error in . HINT: If the error in is % then the error is when $A=A_0$.*

### Solution

First we solve for a function :

Again we use the differential to estimate the error (as errors in and were not mentioned we will assume they don’t have errors):

Now we must look at the partial derivatives (verify the last steps yourself):

.

.

Now by the hint, the errors in and :

,

, and

.

Hence,

.

That is the error in is 6 %.

## 2 comments

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March 15, 2011 at 7:11 pm

Graham KielyHi JP, Partial derivative with respect to L is negative for n/l. This then cancels out error in V and P. Do we ignore this negative and take absolute error?

March 15, 2011 at 7:52 pm

J.P. McCarthyYes.

Ordinarily when we do differentials everything is signed so that if then

(*)

and the and are either positive or negative (respectively.)

However, when we do error analysis things are different because we always take our error to be positive, say — therefore the corresponding change in , lies between plus and minus this:

.

we attribute a potential error in our measurements of and — say (and ). These errors can be positive

ornegative. Hence, suppose that is negative and is positive; and suppose further that theactualerrors (we only know that we are correct to within a specified range, but in theory there is a measured value and an actual value) and are negative and positive, and moreover, at their maxima. Using the definition of a differential (how we estimate the change in due to changes in and ), our estimation of the error:,

that is we don’t necessarily have cancellations so we have to take absolute values.

For estimating changes in the dependent variable due to changes in the independent variables we use (*). For error analysis we use the following:

where , and .