# Math Help - Maximum Error Problem

1. ## Maximum Error Problem

Hi
I am having trouble getting the correct answer to the following question:

1) The sides of a right angle triangle enclosing the right angle are measured as 6.0m and 8.0m respectively. The maximum errors in each measurement are +- 0.1m
i) Find approximately the maximum error in the calculated area of the triangle.

This is what i have done:
$A=\frac{1}{2}bh$

$\partial A = \frac{\partial A}{\partial a} * \partial a + \frac{\partial A}{\partial b} * \partial b
$

$\frac{\partial A}{\partial a}$
$=\frac{1}{2}b$

sub b=8.0m
$=4$

$\frac{\partial A}{\partial b}$

$=\frac{1}{2}h$

sub h=6
$=3$

$\partial a = \frac{0.1}{4}$

$= 0.025$

$\partial b = \frac{0.1}{3}$

$= 0.033$

$\partial A = \frac{4}{0.025} * 0.025 \frac{3}{0.033} * 0.033$

$=7$

P.S

2. Originally Posted by Paymemoney
Hi
I am having trouble getting the correct answer to the following question:

1) The sides of a right angle triangle enclosing the right angle are measured as 6.0m and 8.0m respectively. The maximum errors in each measurement are +- 0.1m
i) Find approximately the maximum error in the calculated area of the triangle.

This is what i have done:
$A=\frac{1}{2}bh$

$\partial A = \frac{\partial A}{\partial a} * \partial a + \frac{\partial A}{\partial b} * \partial b
$
]
Technical point: standard notation for this would be $dA = \frac{\partial A}{\partial a} da + \frac{\partial A}{\partial b} db$. Don't use " $\partial$" with differentials. More importantly, since there is an "h" rather than "a" in the formula, it should be
$dA = \frac{\partial A}{\partial h} dh + \frac{\partial A}{\partial b}db$

$\frac{\partial A}{\partial a}$
$=\frac{1}{2}b$

sub b=8.0m
$=4$

$\frac{\partial A}{\partial b}$

$=\frac{1}{2}h$

sub h=6
$=3$

$\partial a = \frac{0.1}{4}$

$= 0.025$
No, da= 0.1. Where did the "4" come from?

$\partial b = \frac{0.1}{3}$

$= 0.033$
db also equals .1

$\partial A = \frac{4}{0.025} * 0.025 \frac{3}{0.033} * 0.033$

$=7$

P.S
Does this make sense to you? Errors of .1 m in the measurement of the sides will cause an error of 7 square meters in the area? The calculated error would be (1/2)(6)(8)= 24 square meters. 7 square meters would be almost 1/3 of that!

$dA= \frac{1}{2}h db+ \frac{1}{2}b dh= \frac{1}{2}(h db+ b dh)= \frac{1}{2}(6(.1)+ 8(.1))= \frac{1}{2}(1.4)= 0.7$ square meters.

Note that is U= x+ y then dU= dx+ dy.

If A= xy then dA= ydx+ xdy. Dividing both sides by A= xy, $\frac{dA}{A}= \frac{dx}{x}+ \frac{dy}{y}$.

Those give the old engineers "rule of thumb":

"When measurements are added, their errors add. When meaurements are multiplied, their relative errors add."

Here the relative error in height is $\frac{.1}{.6}= \frac{1}{6}$ and the relative error in base is $\frac{.1}{.8}= \frac{1}{8}$. The relative error in the area is $\frac{1}{6}+ \frac{1}{8}= \frac{4}{24}+ \frac{3}{24}= \frac{7}{24}$. Since the calculated area is (1/2)(.6)(.8)= .24, the maximum error is [tex]\frac{7}{24}(.24)= 0.7 square meters.

3. ok thanks i think i understand now.