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If x = -1.3 ± 0.002 and y = 2.17 ± 0.0009, find the error in the determination of the term x²-y²

Mathematics

ScanSolve AI Answer

Answer

0.05

Step-by-step solution

5 steps
  1. Identify the given values and their absolute errors.

    We are given:
    x=−1.73x = -1.73 with an absolute error Δx=0.002\Delta x = 0.002
    y=2.17y = 2.17 with an absolute error Δy=0.009\Delta y = 0.009
    We need to find the error in the determination of the term Z=x2−y2Z = x^2 - y^2.

  2. Calculate the nominal value of ZZ.

    Z=x2−y2Z = x^2 - y^2 Z=(−1.73)2−(2.17)2Z = (-1.73)^2 - (2.17)^2 Z=2.9929−4.7089Z = 2.9929 - 4.7089 Z=−1.716Z = -1.716
  3. Determine the absolute error for x2x^2 and y2y^2.

    For a function f(u)=unf(u) = u^n, the absolute error is given by Δf=∣nun−1∣Δu\Delta f = |n u^{n-1}| \Delta u.
    For x2x^2:

    Δ(x2)=∣2x∣Δx\Delta(x^2) = |2x| \Delta x Δ(x2)=2×∣−1.73∣×0.002\Delta(x^2) = 2 \times |-1.73| \times 0.002 Δ(x2)=2×1.73×0.002\Delta(x^2) = 2 \times 1.73 \times 0.002 Δ(x2)=0.00692\Delta(x^2) = 0.00692

    For y2y^2:

    Δ(y2)=∣2y∣Δy\Delta(y^2) = |2y| \Delta y Δ(y2)=2×∣2.17∣×0.009\Delta(y^2) = 2 \times |2.17| \times 0.009 Δ(y2)=2×2.17×0.009\Delta(y^2) = 2 \times 2.17 \times 0.009 Δ(y2)=0.03906\Delta(y^2) = 0.03906
  4. Calculate the total absolute error for Z=x2−y2Z = x^2 - y^2.

    When quantities are added or subtracted, their absolute errors add up to give the maximum possible absolute error.

    ΔZ=Δ(x2)+Δ(y2)\Delta Z = \Delta(x^2) + \Delta(y^2) ΔZ=0.00692+0.03906\Delta Z = 0.00692 + 0.03906 ΔZ=0.04598\Delta Z = 0.04598
  5. Round the error to an appropriate number of significant figures.

    The input errors (0.002 and 0.009) are given to one significant figure. It is common practice to round the final absolute error to one significant figure.

    ΔZ≈0.05\Delta Z \approx 0.05

    The error in the determination of the term x2−y2x^2 - y^2 is 0.05\boxed{0.05}.

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