Solving Lesson 46 Page 83 SBT Math 10 – Kite>

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Topic

There are two subways A and REMOVE Running in the inner city also starts from two stations, moving in a straight line. On the radar screen of the control station (referred to as the coordinate plane Oxygen with units on axes in kilometers), after starting t (hour) (t 0), position of the ship A whose coordinates are determined by the formula: \(\left\{ \begin{array}{l}x = 7 + 36t\\y = – 8 + 8t\end{array} \right.\) , position of ships REMOVE whose coordinates are (9 + 8t ; 5 – 36t).

a) Calculate the cosine of the angle between the paths of the two trains A and REMOVE

b) How long after departure are the two closest trains?

Solution method – See details

Step 1: Find the direction vectors of the two lines that are the paths of the two trains A and REMOVE based on the train’s path PT A and the ship’s coordinates REMOVE

Step 2: Calculate the cosine between the two direction vectors found in step 1 and take the positive value to calculate the cosine of the angle between the two paths of the two ships A and REMOVE

Step 3: Find the coordinates of 2 points USA, WOMEN (parameterize the coordinates of 2 points USA, WOMEN) in 2 positions where the ship A and ship REMOVE arrive after departure t hour. Find t to MN achieve GTN

Detailed explanation

a) Train A vector motion \(\overrightarrow {{u_1}} = (36;8)\); ship REMOVE vector motion \(\overrightarrow {{u_2}} = (8; – 36)\)

We have: \(\overrightarrow {{u_1}} .\overrightarrow {{u_2}} = 36.8 + 8.( – 36) = 0\)\( \Rightarrow \overrightarrow {{u_1}} \bot \overrightarrow {{) u_2}} \Rightarrow \cos \left( {\overrightarrow {{u_1}} ,\overrightarrow {{u_2}} } \right) = 0\)

Let \(\alpha \) be the angle between the paths of the two ships. Then \(\cos \alpha = \left| {\cos \left( {\overrightarrow {{u_1}} ,\overrightarrow {{u_2}} } \right)} \right| = 0\)

b) After t time: train A at position \(M(7 + 36t; – 8 + 8t)\); ship REMOVE at the point position \(N(9 + 8t;5 – 36t)\)

We have: \(\overrightarrow {MN} = ( – 28t + 2; – 44t + 13) \Rightarrow MN = \sqrt {{{( – 28t + 2)}^2} + {{( – 44t + 13) }^2}} \)

\( \Leftrightarrow M{N^2} = {( – 28t + 2)^2} + {( – 44t + 13)^2} = 2720{t^2} – 1256t + 173\)

According to the assumption, MN achieve GTNN \( \Leftrightarrow M{N^2}\) gain GTNN

Consider \(M{N^2} = 2720{t^2} – 1256t + 173 = 2720{\left( {t – \frac{{157}}{{680}}} \right)^2} + \ frac{{4761}}{{170}}\)\( \ge \frac{{4761}}{{170}}\) \( \Rightarrow MN \ge \sqrt {\frac{{4761}}{{ 170}}} \)

The sign “=” occurs if and only if t = \(\frac{{157}}{{680}}\)

So after \(\frac{{157}}{{680}}\) hours, the two trains are closest to each other and are separated by a distance of 5.29 km

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