Pat Ross MTH 6 Practce Test # Sprng 999 Name. Fnd the area of the regon bounded by the graph r =acos (θ). Observe: Ths s a crcle of radus a, for r =acos (θ) r =a ³ x r r =ax x + y =ax x ax + y =0 x ax + ³ a + y = ³ a (x a) + y = a Ths s a crcle of radus a centered at (a, 0). To get a better dea of how the graph s lad out, we ll plot a few values of r vs θ. θ r 0 a a 0 a a (% a, B/) (0, B/) (a, 0) (!a, B) (!% a, (B)/) A = R θb θ a [f (θ)] dθ = R 0 [a cos (θ)] dθ = R 0 a cos (θ) dθ =a R 0 cos (θ) dθ = a R 0 +cos(θ) dθ = a R 0 ( + cos (θ)) dθ = a h θ + sn (θ) = 0 a. Note that ths s what we should have come up wth, consderng that the regon was a crcle of radus a. More mportantly, note that we ddn t just blndly ntegrate from 0 to. We had to consder the endponts, or rays of the angle that we parttoned nto sub-angles. Here, we parttoned the angle wth ntal and termnal rays 0 and nto sub-angles. Hence, the lmts of the ntegral are 0 and.
. Fnd the area of the regon that s nsde the cardod r = a ( + cos (θ)), but outsde the crcle r = a. We create a table of values r vs θ. θ r 0 a a + a a a a 0 a a a 7 a + a a (B/, a) (0, B) r R (0, a) (B, a) ( (B)/, a ) To do ths problem, observe that: Area th subregon = Area th large sector - Area th small sector Area of th large sector = θ R = R θ Area of th small sector = θ r = r θ Area of th subregon = R θ r θ = (R r ) θ Area total = P n = (R r ) θ = R 0 (R r ) dθ + R (R r ) dθ = R ³ 0 [a ( + cos (θ))] a dθ + R ³ [a ( + cos (θ))] a dθ = 0 [ cos (θ)+cos (θ)] dθ + R a [ cos (θ)+cos (θ)] dθ = a R R h a 0 cos(θ)+ +cos(θ) dθ + a h sn(θ)+ θ + sn (θ) a h³ + a R (0) + a h ³ + h cos(θ)+ +cos(θ) dθ = + h 0 a sn(θ)+ θ + sn (θ) =a + a =
. Fnd the area bounded by the graph of r =sn(θ). Observe: As θ ncreases from 0 to, the entre graph s generated, as any two values of θ that dffer by a whole number multple of yeld the same value of r. e.g. Suppose θ b = θ a +. Then rb =sn(θ b )=sn((θ a + )) = sn (θ a +) =sn(θ a )=ra. So θ between 0 and generates the entre graph. Note Also: For < θ < we have r < 0, whchsmpossbleforreal values of r. Therefore, θ between and generates nothng - the entre graph s generated by θ between 0 and. We plot some values of r vs θ. θ r r 0 0 0 ± ± 8 8 ± 6 ± ± ± 8 8 ± 0 0 (-, 0) (, 0) Area of one petal = R r dθ = cos () ( cos (0)) = R 0 sn(θ) dθ = R 0 sn (θ) dθ = cos (θ) 0 =
Totalareastheareaoftwopetalswhchs.. Compute the norm of ~v = h,, ~v = + + = Alternately, ~v = ~v ~v = + + =.. Fnd a unt vector n < havng the same drecton as ~v = h,,. From our prevous defnton, ths would be ~u = ~v = h,, ~v = D E,,. 6. Fnd a unt vector n < havng the same drecton as ~v = h,. ~u = ~v = h, ~v = D E, +. The geometrc relatonshp between the two s seen n the pcture below: (, ) 7. Gven vectors ~u = h, and ~v = h,, compute the angle θ between ~u and ~v. cos (θ) = ~u ~v ~u ~v = +6 = 0 = = θ =cos ³. 8. Gven vectors ~u = h, and ~v = h,, compute the projecton of ~u onto ~v, proj ~v ~u, and the projecton of ~u orthogonal to ~v, orth ~v ~u. proj ~v ~u = ~u ~v +6 ~v = h, = D 8, 6 ~v ~v orth ~v ~u = ~u proj ~v ~u = h, D 8, 6 E E D =, E 9. Show that the vector ~v wth ntal pont (, ) and termnal pont (8, 6) s parallel to the vector ~u whose ntal pont s (0, 0) and whose termnal pont (, ). The vectors are pctured below:
(8, 6) (, ) (, ) (0, 0) The component form of the vector v can be computed by subtractng the coordnates of the ntal ponts from the coordnates of the termnal ponts. Hence, ~v = h8, 6 = h,. If we compute the component form of ~u we fnd that ~u = h 0, 0 = h, = ~v. 0. Show that the vectors ~u = h, and ~v = h, are perpendcular. The vectors are shown below: (, ) (0, 0)! (,!) ~u ~v = h, h, =6 6=0. Snce ~u ~v =0, ~u and ~v are perpendcular.. Let ~v =~ 6~j + ~ k and ~u =~ +~j + ~ k. (a) Compute ~v q ~v = h, 6, = +( 6) + = 6 (b) Compute ~v ~u ~v ~u = h, 6, h,, = h9, 8, q h,, = h9, 8, = h8,, 8 = 8 +( ) +8 = 6 = q (6) (7) = 6 7
. Fnd the angle between the lne contanng the ponts (,, ) and (,, 6), and the lne contanng the ponts (,, ) and (, 6, 9). The lne contanng the ponts (,, ) and (,, 6) has the same drecton as the vector wth ntal pont (,, ) and termnal pont (,, 6). In component form, ths s the vector h,, 6 = h,,. Smlarly, the lne contanng the ponts (,, ) and (, 6, 9) has the same drecton as the vector wth ntal pont (,, ) and termnal pont (, 6, 9). In component form, ths vector s h,, 6. Theanglebetweenthetwovectorscanbefoundfromtheequaton Thus we have: θ =cos ³ cos ³ 8 =cos () = 0. ~u ~v = ~u ~v cos (θ) Ã ~u ~v θ =cos ~u ~v h,, h,,6 h,, h,,6. Gven that ~u = h,, and ~v = h,,, compute: (a) comp u (~v)! =cos ³ +8+8 + + + +6 =cos ³ 8 6 = v comp u (v) u Observe: comp u (~v) = ~v cos (θ). Also ~u ~v = ~u ~v cos (θ) cos (θ) = comp u (~v) = ~v cos (θ) = ~v ~u ~v ~u ~v ~u ~v = ~u ~v = h,, h,, = 0 ~u ~v ~u h,, = +( ) + (b) proj u (~v) As for proj u (~v), note that proj u (~v) s the vector whch has the magntude of comp u (~v) and the drecton of ~u. (See below.) 6
v orth u (v) proj u (v) u Therefore, to get proj u (~v), we ll take ts magntude, comp u (~v), and multply t by the unt vector havng the drecton of ~u, to gve proj u (~v) ts drecton. proj u (~v) =comp u (~v) ~u = h,, ~u = h,, +( ) + = D,, E 7. (c) and orth u (~v). From the pcture above, note that ~v = proj u (~v)+orth u (~v) orth u (~v) =~v proj u (~v) =h,, D,, 7 E = D,, 6 7. ~a =7~ +6~j + ~ k and ~ b = ~ +~j ~ k. Compute ~a ~ b. Frst, we set up the matrx dsplayed below, and compute the products gong along the dagonals. E!6k + 0 + (!)j j k j 7 6!! 7 6!!8 + (!)j + k ~a ~ b = ³ 8~ ~j + ~ k ³ 6 ~ k +0~ ~j = 8~ +6~j +0 ~ k. Fnd the volume of the parallelopped wth vertces (,, ), (8,, 7), (0,, ), and (,, ). (Hnt: You may be able to use the results of problem #.) The vector wth ntal pont (,, ) and termnal pont (8,, 7) s equvalent to h7, 6,. The vector wth ntal pont (,, ) and termnal pont (0,, ) s equvalent to h,,. 7
The vector wth ntal pont (,, ) and termnal pont (,, ) s equvalent to h,,. These vectors determne the sdes of a parallelopped. The volume of the parallelopped s the trple product: {z } =h 8,6,0 from prev problem h,, (h7, 6, h,, ) = 8 + 80 0 = = 6. Compute the volume of the parallelogram wth vertces (, ), (, 6), and (6, ). (,6) +,!, (6,) (,) (,) The pcture above depcts the stuaton. The vector wth ntal pont (, 6) and termnal pont (, ) s equvalent to h,. The vector wth ntal pont (, 6) and termnal pont (6, ) s equvalent to h,. The area of the parallelogram s h, h,. To compute ths, we set up the followng matrx: 8
!6k + 0 + 0j j k j!! 0! 0!!! 0 + 0j + k = ~ k ³ 6 ~ k =7 ~ k The area of the parallelogram s gven by: h, h, = 7 ~ k =7 9