Collection of Maths Problems

Measurement precision

Task number: 2746

• Variant 1

  determine the perimeter of the rectangle with a precision of 1 \(mm\)?

• Resolution

  Assume that our measurements are in millimeters and that we have the same absolute error in both measurements: \(a=80\pm\varepsilon\), \(b=60\pm\varepsilon\).

  \(p=2(a+b)=28\pm4\varepsilon\). If the resulting error is to be at most \(\frac12\, mm\), we have \(4\varepsilon< \frac12\) or \(\varepsilon<\frac18\).

• Result

  We must measure the sides with a precision of at least \(\frac18\, mm\).

• Variant 2

  determine the area of the rectangle with a precision of 1 \(mm^2\)?

• Resolution

  \(P=ab=(80\pm\varepsilon)(60\pm\varepsilon)=4800\pm140\varepsilon\pm\varepsilon^2\).
  If the resulting error is to be at most \(\frac12\,mm^2\), we have \(140\varepsilon+\varepsilon^2< \frac12\) or \(\varepsilon<\frac{\sqrt{19602}-140}2\doteq 0{,}0035713\).

• Result

  We must measure the sides with a precision of at least \(3{,}5\,\mu{}m\).

• Variant 3

  determine the length of the rectangle's diagonals with a precision of 1 mm?
  (Ignore the fact that we could measure the diaonals directly.)

• Resolution

  \(d=\sqrt{a^2+b^2}=\sqrt{(80\pm\varepsilon)^2+(60\pm\varepsilon)^2}= \sqrt{6400\pm160\varepsilon+\varepsilon^2+3600\pm120\varepsilon+\varepsilon^2}= \sqrt{10000\pm280\varepsilon+2\varepsilon^2} \)

  Now we compare with \(100\pm\frac12\), or \(\sqrt{10000\pm280\varepsilon+2\varepsilon^2}< \sqrt{(100\pm\frac12)^2}\).

  This gives us \(280\varepsilon+2\varepsilon^2-100\frac14 \le 0\) and \(\varepsilon<\frac{\sqrt{280^2+802}-280}4\doteq 0{,}35712\)

  (The inequality \(-280\varepsilon+2\varepsilon^2-99\frac34 \le 0\) gives the poorer estimate \(\varepsilon<\frac{280-\sqrt{280^2-798}}4\doteq 0{,}35716\) )

• Result

  We must measure the sides with a precision of at least \(0{,}35\, mm\).

• Variant 4

  determine the area of the rectangle with a precision of 1 \(cm^2\)?

• Resolution

  Taking centimeters as our units,

  \(P=ab=(8\pm\varepsilon)(6\pm\varepsilon)=48\pm14\varepsilon\pm\varepsilon^2\).
  If the resulting error is to be at most \(\frac12\,cm^2\), we have \(14\varepsilon+\varepsilon^2< \frac12\) or \(\varepsilon<\frac{\sqrt{198}-14}2\doteq 0{,}035623\).

• Result

  We must measure the sides with a precision of at least \(0{,}35\, mm\).