Maximum Error Problem

Dec 2008
509
2
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:
\(\displaystyle A=\frac{1}{2}bh\)

\(\displaystyle \partial A = \frac{\partial A}{\partial a} * \partial a + \frac{\partial A}{\partial b} * \partial b
\)

\(\displaystyle \frac{\partial A}{\partial a}\)
\(\displaystyle =\frac{1}{2}b\)

sub b=8.0m
\(\displaystyle =4\)

\(\displaystyle \frac{\partial A}{\partial b}\)

\(\displaystyle =\frac{1}{2}h\)

sub h=6
\(\displaystyle =3\)

\(\displaystyle \partial a = \frac{0.1}{4} \)

\(\displaystyle = 0.025\)

\(\displaystyle \partial b = \frac{0.1}{3}\)

\(\displaystyle = 0.033\)

\(\displaystyle \partial A = \frac{4}{0.025} * 0.025 \frac{3}{0.033} * 0.033\)

\(\displaystyle =7\)

P.S
 

HallsofIvy

MHF Helper
Apr 2005
20,249
7,909
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:
\(\displaystyle A=\frac{1}{2}bh\)

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

\(\displaystyle \frac{\partial A}{\partial a}\)
\(\displaystyle =\frac{1}{2}b\)

sub b=8.0m
\(\displaystyle =4\)

\(\displaystyle \frac{\partial A}{\partial b}\)

\(\displaystyle =\frac{1}{2}h\)

sub h=6
\(\displaystyle =3\)

\(\displaystyle \partial a = \frac{0.1}{4} \)

\(\displaystyle = 0.025\)
No, da= 0.1. Where did the "4" come from?

\(\displaystyle \partial b = \frac{0.1}{3}\)

\(\displaystyle = 0.033\)
db also equals .1

\(\displaystyle \partial A = \frac{4}{0.025} * 0.025 \frac{3}{0.033} * 0.033\)

\(\displaystyle =7\)

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!

\(\displaystyle dA= \frac{1}{2}h db+ \frac{1}{2}b dh= \frac{1}{2}(h db+ b dh)= \frac{1}{2}(6(.1)+ 8(.1))= \frac{1}{2}(1.4)= 0.7\) 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, \(\displaystyle \frac{dA}{A}= \frac{dx}{x}+ \frac{dy}{y}\).

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