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Eva8 [605]
3 years ago
8

An army base in the desert is located at the origin, O of an x-y coordinate system. A soldier at point P needs to determine his

coordinates with respect to the army base, but he does not have a link to a global positioning system. He is, however, in radio contact with several small towns in the vicinity. Two of the towns are town A and town B. Town A is 32 miles due south of the army base; town B is 18 miles north of the base. At an agreed-upon instant, both towns send out identifying radio signals. The soldier receives the signal from B slightly before the signal from A, so he knows that he is closer to B than to A. Furthermore, by measuring the time difference between receiving the two signals, the soldier is able to compute that he is 40 miles closer to town B than to town A. Explain why the point P must lie on a certain hyperbola.
Mathematics
1 answer:
Nutka1998 [239]3 years ago
8 0

Answer/step-by-step explanation

The soldier at point P lie on a parabola because he determined his position and distances from towns A and B through measurement of the difference in timing (phase) of radio signals received from the two towns.

This analysis of the signal time difference gives the difference in distance of the soldier at P, from the towns.

This process is known as hyperbolic navigation.

These distances of point P from towns A and B is estimated by the soldier at point P, by measuring the delay localizes the receiver to a hyperbolic line on a chart.

Two hyperbolic lines will be drawn by taking timing measurements from the

towns A and B .

Point P will be at the intersection of the lines.

These distances of point P(The soldier's positions) from town A and town B were determined using the timing of the signals received from the two towns, due to the fact that point P was on a certain hyperbola.

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Answer / Explanation

The question is incomplete. It can be found in search engines. However, kindly find the complete question below.

Question

(1) Give an example of functions f : A −→ B and g : B −→ C such that g ◦ f is injective but g is not  injective.

(2) Suppose that f : A −→ B and g : B −→ C are functions and that g ◦ f is surjective. Is it true  that f must be surjective? Is it true that g must be surjective? Justify your answers with either a  counterexample or a proof

Answer

(1) There are lots of correct answers. You can set A = {1}, B = {2, 3} and C = {4}. Then define f : A −→ B by f(1) = 2 and g : B −→ C by g(2) = 4 and g(3) = 4. Then g is not  injective (since both 2, 3 7→ 4) but g ◦ f is injective.  Here’s another correct answer using more familiar functions.

Let f : R≥0 −→ R be given by f(x) = √

x. Let g : R −→ R be given by g(x) = x , 2  . Then g is not  injective (since g(1) = g(−1)) but g ◦ f : R≥0 −→ R is injective since it sends x 7→ x.

NOTE: Lots of groups did some variant of the second example. I took off points if they didn’t  specify the domain and codomain though. Note that the codomain of f must equal the domain of

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