Calculating Decimals for Expressions

[dagg3r]

can somebody tell me how many decimals places for this question and tell me if i am doing this right

1. Calculate a value for the expression

p=a^-2

a=27.3 +- 0.05 metres ( the plus and minus symbol is on top of each other)

b) calculate the maximum error in your answer



this is what i did

for a) p=27.3^-2 = 0.001342 is that how many decimals i put down how many do i put down

b) to calculate the maximum error i differintiated the questio so i got -2*a^(-3) therefore
-2(27.3)^(-3) * 0.05 = 4.91 * 10^(-6)
is that how i put the form in how many decimal places?

do i state the maximum error is 4.91*10^(-6)?

ok if somebody can answe those questions thanks!

 

[stunner5000pt]

your datat is given to three significant digits
so your answer should be given to three significant digits (i.e. you should get 1.34 x 10^-3)
same applies for you error

 

[HallsofIvy]

Did you consider just computing it directly? That is, since a=27.3 +- 0.05 metres, the largest a can be is 27.35 metres so a^-2= 0.001337, the smallest possible value for a^-2. The smallest a can be is 27.25 so that a²= 0.001347, the largest possible value for a^-2. Since 27.3^-2= .001342, that would be a simple "guess" for (a) which said only "calculate a value". 0.001347- 0.001342= 0.000005 while 0.001342-0.001337= 0.000005. Since those are the same the maximum error for that "guess" is 0.000005.

The problem did not say anything about "significant digits".

By the way, there is a "rule of thumb" that says "If measurements are added, add the errors. If measurements are multiplied, add the relative errors."

Since the "measurement" is 27.3 and the error is 0.05, the relative error is 0.05/27.3= 0.001832. squaring (even -2 power) is multiplying the number by itself so "adding relative errors" gives a relative error for the result of 0.003663... That, times (27.3)^-2= 0.001342 gives 0.000005 as before.