Components of Vectors — Question 5

Question 5

Can the instrument readings describe one vector?

A two-dimensional instrument prints magnitude $A=10.0$ m and components $A_x=8.0$ m, $A_y=8.0$ m. A second version states hard bounds $9.9\leq A\leq10.1$ m and $7.9\leq A_x,A_y\leq8.1$ m. Hard bounds mean all true values must lie in the stated intervals.

1. Check the exact report and explain why testing each component separately is insufficient.

2. Find the range of magnitudes allowed by the bounded component readings. Can rounding within those bounds reconcile the report?

3. If $A=10.0$ m and $A_x=8.0$ m are trusted exactly, what positive $A_y$ repairs the report?

Question 5 - Solution

Can the instrument readings describe one vector?

(a) Check the joint relation.

$$8^2+8^2=128\ne10^2\Rightarrow\boxed{A_{\mathrm{implied}}=8\sqrt2\,\mathrm m=11.314\,\mathrm m}$$ Each component is less than 10 m, but the pair implies $A=\sqrt{8^2+8^2}=8\sqrt2\approx11.314$ m. Both projections must satisfy $A^2=A_x^2+A_y^2$ together; separate component bounds are only necessary conditions.

(b) Propagate the full allowed rectangle.

$$A\in[7.9\sqrt2,8.1\sqrt2]\,\mathrm m=[11.172,11.455]\,\mathrm m,\quad[11.172,11.455]\cap[9.9,10.1]=\varnothing$$ All components are positive, so the minimum and maximum lengths occur at $(7.9,7.9)$ and $(8.1,8.1)$. Thus $11.172\lesssim A\lesssim11.455$ m. This interval does not intersect $[9.9,10.1]$ m. The specified rounding tolerance cannot explain the inconsistency.

(c) Repair the untrusted entry.

$$A_y=+\sqrt{10^2-8^2}=6\,\mathrm m$$ $A_y=+\sqrt{10^2-8^2}=6.0$ m. The positive sign uses the stated requirement; without it, $-6.0$ m would also satisfy the magnitude. This repair assumes the two trusted entries really are exact.