Geometry Resolving vectors, polar and rectangular forms, bearings problems revisited 2/3/10 Vectors and the trigonometric functions. y Look at the same diagram we used to define the v=<x,y> trigonometric functions. A vector with tail at the =<r, !> origin can be described two ways: r v = <x, y> = <r, " >. y ! These are called the rectangular coordinates of the x x vector and the polar coordinates. ! ! It is easy to switch back and forth between these two representations. You should memorize how to and also learn how your calculator will make the transformation automatically. Given <r, " > you can convert to <x, y> by the formulas: x = rcos( " ) y = rsin( " ). Given <x, y> you can convert to <r, " > by the formulas: y ! " = tan"1 ( ) and r = x 2 + y 2 . ! x ! ! There is a slight problem, however. Just as with sin"1 (x) the calculator gives you only one of two ! ! y values for tan"1 ( ) and you have to consider the other on your own. possible ! x 1. Practice moving back an forth between rectangular and polar coordinates. ! a. Given the vector v = <10 miles, 40º> expressed in terms of a length and a direction. Draw this!vector on an x-y plane. Resolve the vector into its x-y components. This means writing it in terms of <x, y>. All you have to do is calculate x = 10cos(40º) and y = 10sin(40º). Do this manually even if you have ! been shown how the calculator will do this automatically. b. Now suppose you have a vector given by its x-y components, q = <8 miles, 10 miles >. Express this vector in <r, " > form, as a length and a magnitude. 10 Here you simply calculate r = 8 2 + 10 2 " = tan"1 ( ) . Again, don't use the calculator shortcut. 8 ! ! c. Draw the following vectors and resolve them into their x-y components. Explain your result in comparison with! (a). ! ! <10 miles, 140º> <10 miles, -220º> <10 miles, 220º> d. Now deal with the uncertainty of " going from <x, y> to <r, " >. In the case of tan"1 (x) you may not be satisfied with the answer the calculator gives, but will need to add or subtract 180º. Put each of these vectors in <r, " > form being sure to get the angle in the correct quadrant. ! ! ! <–6, 8> <–5, –8> <2, –3> <7, 11> ! 2. Your calculator will convert between modes automatically. The <r, " > form of vectors is called polar form and the <x, y> form is called rectangular or just x-y. Under angle you have choices R " Pr , R " P# , P " Rx , P " Ry whose use is straightforward. If you type R " Pr (3,4) you get 5 you (the length of the vector <3, 4> )and if you type R " P# (3,4) ! get 53.1º (its angle with respect to the positive x-axis). Translate (-5, 5) to <r, " > and back using your calculator. ! ! ! EXAMPLE: Bearing problems can be solved using vectors and resolving them. This avoids completely the laws of sines and cosines. It may no easier but imagine a ship that makes several turns ! on its voyage. The work then becomes much easier this way and can easily be automated. A boat sails 59 miles on a bearing of 41º and then 51 miles on a bearing of 121º. How far from port is it and what bearing from port must be taken to reach the boat in its present position? N 121º 43.72 m -26.27 m 51 59 N 44.53 m 41º 18.26 m 38.71 43.72 m i. Convert bearings to their equivalent representation with respect to the positive x-axis. The vectors are <59 miles, 49º> and <51 miles, –31º>. ii. Then resolve the vectors into their x-y components. <38.71 miles. 44.53 miles> < 43.72 miles , -26.27 miles> iii. Add the vectors in their x-y form. <38.71 miles. 44.53 miles> +< 43.72 miles , -26.27 miles>= <82.43 miles, 18.26 miles> iv. Convert the resultant vector back to <r, " >. <84.43 miles, 12.49º> v. Restate the angle as a bearing. <84.43 miles, 77.51º> ! Here are your old bearing problems. Solve them by thinking of the motions as vectors and performing vector addition. 3. A ship sails from port on a bearing of 80º for 150 miles and then turns and sails another 40 miles due North. How far is the ship from home and what bearing must it sail on to return on a direct line? 4. A ship sails from port on a bearing of 100º for 200 miles. Then it turns and sails on a bearing of 30º for 75 miles. How far is the ship from home and what bearing must it sail on to return on a direct line? 5. More sailing. A ship sails 19 miles on a bearing of 290º and then sails north for 8 miles. It now must sail to a point 20 miles directly north of the starting point. At what bearing must it sail and how far must it travel to reach that point? 6. A ship sails from port on a bearing of 60º for 200 miles and then on a bearing of 98º for another 175. If another ship was to follow from the same port but sail on a direct line to the same position what bearing should it use and how far will it sail?
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