Components of Vectors — Question 8

Question 8

Direction bounds need not be symmetric about a component

A vector has exact magnitude 10 m. Its standard direction is known only to satisfy $80^\circ\leq\theta\leq100^\circ$. These are hard bounds, with every angle in the interval possible.

1. Find tight bounds on $A_x$ and $A_y$. Include any interior extremum.

2. Can $A_x=0$ and $A_y=9.85$ m occur together exactly? Explain why separate bounds are not independent freedoms.

3. What changes if the angle interval is instead $85^\circ\leq\theta\leq95^\circ$?

Question 8 - Solution

Direction bounds need not be symmetric about a component

(a) Inspect the functions over the entire interval.

$$\boxed{A_x\in[-10\sin10^\circ,10\sin10^\circ]},\quad A_y\in[10\cos10^\circ,10]\quad\mathrm m$$ $A_x=10\cos\theta$ decreases, giving $-1.7365\leq A_x\leq1.7365$ m. The sine peaks at the interior angle $90^\circ$, so $9.8481\leq A_y\leq10$ m. Using only the endpoint sine values would miss the maximum.

(b) Retain the fixed-length constraint.

$$A_x=0\Rightarrow A_y=+\sqrt{100-0}=10\,\mathrm m\ne9.85\,\mathrm m$$ No. Within this angle interval, $A_x=0$ requires $\theta=90^\circ$ and hence $A_y=10$ m. The feasible pairs lie on a circular arc, not throughout the rectangle formed from separate bounds. The proposed exact pair also has length 9.85 m rather than 10 m.

(c) Narrow the arc.

$$|A_x|\le10\sin5^\circ=.8716\,\mathrm m,\quad9.9619\,\mathrm m=10\cos5^\circ\le A_y\le10\,\mathrm m$$ The new bounds are $-0.8716\leq A_x\leq0.8716$ m and $9.9619\leq A_y\leq10$ m. The vertical upper limit remains at 10 m because $90^\circ$ is still allowed. The bounds are deterministic; no statistical confidence level is implied.