Vectors and Polar Form

Transcription

1 MATH 117 Vectors and Polar Form Dr. Neal, WKU Any point ( x, y ) in the x y plane forms a directed line segment from the origin (0, 0) to the point ( x, y ). Sch a segment is called a ector. When we want to consider the ector and not jst the point, then we generally label it as = ( x, y ). = (7, 4)! Length and Direction A ector = ( x, y ) has a length (or norm) denoted by which is simply the distance to the origin gien by the hypotense. The direction! is the standard angle determined by ( x, y ) as measred from the positie x axis. Ths we hae = x 2 + y 2 and tan! = y x To find the direction!, compte tan!1 (y / x) and then adjst the angle to the proper qadrant. A ector written in terms of its length and direction is then in polar form. Example 1. Let = ( 6, 8) and = ( 2, 10). Find the length and direction of each ector and write the ectors in polar form. Soltion. Vector has length = = 10. Here, tan!1 ( y / x) is tan!1 (8 /!6) 53.13º. So in Qad. II,! =!53.13º +180º !. So = (10, ! ) in polar form. Vector has length = = 104. Its angle in Qadrant III is gien by! = tan!1 (10 / 2) +180º !. Then = ( 104, ! ) in polar form.

2 Conerting Polar Form Back to Rectanglar Form If a ector is gien in polar form (,! ), then we recoer the x and y coordinates by x = cos! and y = sin! Here, is taking the place of the radis r, where x = r cos! and y = r sin!. Example 2. Find the rectanglar form of the ectors = (30, 120º) and = (20, 330º). # Soltion. For = (30, 120º), we hae x = 30 cos120º= 30! " 1 & % ( = "15 and y $ 2' " 3% =30 sin120º= 30! $ ' = 15 3 ; so = ( 15, 15 3 ). # 2 & " 3% For, x = 20 cos330º= 20! $ # 2 & ' = 10 3 and y = 20 sin330º= 20! # " 1 & % $ 2 ( = "10 ; so ' then = (10 3, 10). Adding Vectors Gien two ectors = (x 1, y 1 ) and = (x 2, y 2 ), both in rectanglar form, we obtain the sm of ectors by adding component wise: + = (x 1 + x 2, y 1 + y 2 ). If we make a parallelogram ot of the ectors and, then ector + is the diagonal that starts at the origin. Example 3. Let = (3, 6) and = ( 8, 2). Graph,, and +. What is the length and direction of +? Soltion. First, + = (3 + ( 8), 6 + ( 2)) = ( 5, 4), which we see to be the diagonal of the parallelogram determined by and. (See graphs on next page.) Then + = = 41! 6. 4, and the direction of + is gien by! = tan "1 ("4 / 5) +180º # º.

3 + = (3, 6) + = ( 8, 2) Adding Forces Often, a force is gien in terms of its magnitde and direction. In order to add two forces, we conert each to rectanglar form, add the x and y components to get the sm, then conert the reslt back to polar form. The sm of two forces is called the resltant force. Example 4. Let F 1 be a force of 50 Newtons in the direction 30º East of Soth, and let F 2 be a force of 80 Newtons in the direction 10º Soth of East. Find the magnitde and direction of the resltant F 1 + F 2. Soltion. First, the angle for F 1 is 30º + 270º = 300º, and the angle for F 2 is 360º 10º = 350º. Next, the x and y components for each force and the resltant are: F 1 = (50 cos300º, 50sin 300º ) 30º 10º E F 2 F 2 = (80cos350º, 80sin350º) F 1 So the resltant force is S F 1 + F 2 = (50 cos300º + 80cos350º, 50sin 300º + 80sin350º) = ( , ) So F 1 + F 2 has magnitde F 1 + F 2 = ! Newtons. Its direction is in the 4th Qadrant is tan!1 (! / ) + 360º º, or abot º Soth of East.

4 Sbtraction and Distance Between Vectors Gien two ectors = (x 1, y 1 ) and = (x 2, y 2 ), both in rectanglar form, we obtain the ector from to by the difference! which is obtained by sbtracting component wise:! = (x 2! x 1, y 2! y 1 ). If we make a parallelogram ot of the ectors and, then ector! is eqialent to the diagonal from the end of to the end of after it is picked p and moed to the origin. The length of! (or! ) gies the distance between the endpoints of and, which we call the distance between the ectors. This length of! is eqialent to the common distance formla in the x y plane: = (x 1, y 1 ) = (x 2, y 2 )! = (x 2! x 1, y 2! y 1 ) distance = (x 2! x 1 ) 2 + (y 2! y 1 ) 2 =! Example 5. Let = (3, 6) and = ( 8, 2). Find the ector from to and the distance between and. Graph all the ectors. Soltion. By sbtracting component wise, we get! = ( 11, 8). Now we can pick p the ector ( 11, 8) and place it between the original and to see that! is the diagonal from to. = (3, 6) = (3, 6)! = ( 8, 2) = (!8,!2)!! = ( 3, 6) (!11,!8) The distance between and is! = = 185 " 13.6.

5 Dot Prodct and Angle Between Vectors Gien ectors = (x 1, y 1 ) and = (x 2, y 2 ), the dot prodct is gien by! = x 1 x 2 + y 1 y 2 The dot prodct can be sed to find the angle! between the ectors and. Becase! is the length of the third side of a triangle and is opposite angle!, the Law of Cosines gies! 2 = 2 + 2! 2 cos " We now can sole for cos! and simplify the expression:! " cos! = " " ( ) + x ( y2 ) " ((x 2 " x 1 ) 2 + ( y 2 " y 1 ) 2 ) = x y 1 2 = " ( "2 x 1 x 2 " 2 y 1 y 2 ) 2 = x 1 x 2 + y 1 y 2 = #. Ths, cos! = " and! = cos "1 $ # & % ' ). ( Example 6. Let = (3, 6) and = ( 8, 2). Use the dot prodct to find the angle between and. Use the angles of each ector to erify the reslt.

6 Soltion. For = (3, 6) and = ( 8, 2), the dot prodct is! = x 1 x 2 + y 1 y 2 = ( 24) + ( 12 )= 36. The lengths of the two ectors are = = 45 and = = 68. So the angle between the ectors is! = cos "1 $ # & % ' ) = cos "1 $ "36 ' & ) * º ( % ( We also can find the angles for and separately. The angle for in Qadrant III is tan!1 (2 / 8) +180º º. The angle for in Qadrant I is tan!1 (6 / 3) º. So the angle in between is º º 130.6º º º Exercise Let = (10, 4) and = ( 2, 8). (a) Find the lengths and directions of the ectors, and the length and direction of the resltant +. (b) Find the ector from to and the distance between and. (c) Use the dot prodct to find the angle between and. Verify the reslt sing the angles fond in Part (a).

7 Answers (a) = ( 116, º) and = ( 68, º) + = (8, 4) = ( 80, º) (b)! = ( 12, 12); distance =! = 288 (c)! = 52; Angle in between is! = cos!1 (!52 / ( 116 " 68)) = º Using the angles fond in (a), we hae º! º! = 360º ( º º ) = º