GR9277 #48



Alternate Solutions 
LF 20151016 02:47:49  It\'s possible to guess the answer even without knowing anything about error analysis. \r\n\r\nThere should be nothing special about the equation F = ma, so a priori, there should be no restriction on m or a being negative. However, the standard deviation is always positive. With these two observations we eliminate A, B and E:\r\n\r\nA is eliminated because if m or a are negative, then F (=ma) is negative. However option A is always positive, so sigma_F/F = [something positive] > sigma_F must be negative, something that\'s not possible.\r\n\r\nB is eliminated because if both a and m are negative, then we have a negative number within the square root and the standard deviation can\'t be imaginary.\r\n\r\nE is eliminated because if m is negative and really large, while a is positive and really small, then the second term dominates the first and we have a positive sum. But F (=ma) is again negative, so sigma_F is again negative, and that is impossible.\r\n\r\nThat leaves C and D. Now we can use another observation: if we can measure either m or a with infinite precision, then its standard deviation would be zero. However because of the other variable, we do not know F with infinite precision, and therefore sigma_F is not zero. Plugging in this condition eliminates D, leaving C as the only possible answer.  

Comments 
LF 20151016 02:47:49  It\'s possible to guess the answer even without knowing anything about error analysis. \r\n\r\nThere should be nothing special about the equation F = ma, so a priori, there should be no restriction on m or a being negative. However, the standard deviation is always positive. With these two observations we eliminate A, B and E:\r\n\r\nA is eliminated because if m or a are negative, then F (=ma) is negative. However option A is always positive, so sigma_F/F = [something positive] > sigma_F must be negative, something that\'s not possible.\r\n\r\nB is eliminated because if both a and m are negative, then we have a negative number within the square root and the standard deviation can\'t be imaginary.\r\n\r\nE is eliminated because if m is negative and really large, while a is positive and really small, then the second term dominates the first and we have a positive sum. But F (=ma) is again negative, so sigma_F is again negative, and that is impossible.\r\n\r\nThat leaves C and D. Now we can use another observation: if we can measure either m or a with infinite precision, then its standard deviation would be zero. However because of the other variable, we do not know F with infinite precision, and therefore sigma_F is not zero. Plugging in this condition eliminates D, leaving C as the only possible answer.
LF 20151016 03:46:36 
Attempting to fix the broken code. It looks OK in the preview, hope it displays right as well.\r\n\r\nIt\'s possible to guess the answer even without knowing anything about error analysis. \r\n\r\nThere should be nothing special about the equation F = ma, so a priori, there should be no restriction on m or a being negative. However, the standard deviation is always positive. With these two observations we eliminate A, B and E:\r\n\r\nA is eliminated because if m or a are negative, then F (=ma) is negative. However option A is always positive, so > must be negative, something that\'s not possible.\r\n\r\nB is eliminated because if both a and m are negative, then we have a negative number within the square root and the standard deviation can\'t be imaginary.\r\n\r\nE is eliminated because if m is negative and really large, while a is positive and really small, then the second term dominates the first and we have a positive sum. But F (=ma) is again negative, so is again negative, and that is impossible.\r\n\r\nThat leaves C and D. Now we can use another observation: if we can measure either m or a with infinite precision, then its standard deviation would be zero. However because of the other variable, we do not know F with infinite precision, and therefore is not zero. Plugging in this condition eliminates D, leaving C as the only possible answer.

  MuffinSpawn 20090924 18:57:56  If you don't remember the product rule, you can always derive it quickly using the general function rule:
  petr1243 20080309 18:13:35  Just a simple application of the general rules from error analysis:
For A = B +/ C :
=
For A = B*C or A = B/C:
=
Just treat as our
note 20080821 23:12:47 
There's a typo in your last equation, I think you meant dC/C

 




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