MA261-A Calculus III 2006 Fall Practice Test of Midterm 1 Solutions

Transcription

1 MA6-A Calculus III 006 Fall Practice Test of Midterm Solutions (0ts) ( #6) Find an euation of a shere if one of its diameters has end-oints (; ; 4) and (4; ; 0) Since this diameter has end-oints (; ; 4) and (4; ; 0), the midoint of the line segment with endoints (; ; 4) and (4; ; 0) is the center of this shere and the radius is the half of the length of this line segment The midoint is 4 ; ; 40 (; ; 7) The length of the line segment is (4 ) ( ) (0 4) So, the radius is Therefore, the euation is (x ) (x ) (z 7) (0ts) ( #) Suose a vector a makes angles, and with the ositive x-, y-, and z-axes, resectively Find the comonents of a and show that cos cos cos Consider the vector a and the x-axe We can draw a line from the end oint of the vector a to the x-axe which forms a right angle with the x-axe Now, you have a rightangle triangle with the hyotenuse is a and the adjacent side is on the x-axe with the angle in between these two sides So, the length of the adjacent side can be calculated by jaj cos This is the x-coordinate of a Similarly, we can nd the y- and z-coordinates of a We conclude that a hjaj cos ; jaj cos ; jaj cos i The length of a is jaj (jaj cos ) (jaj cos ) (jaj cos ) This imlies that ) ) jaj (jaj cos ) (jaj cos ) (jaj cos ) jaj jaj cos jaj cos jaj cos cos cos cos

2 (0ts) ( #) For the vectors a h; i and b h 4; i, nd orth a b From exercise 7, we have orth a b b roj a b By the formula in the textbook age 6, Thus, roj a b orth a b b roj a b h 4; i a b () ( 4) () () jaj a h; i h; i ; 4 ; 4 ; 4 (0ts) (4 #6) Find the area of the arallelogram with vertices K (; ; ), L (; ; 6), M (; ; 6), and N (; 7; ) After lotting these four oints in a coordinate system, we have a arallelogram KLMN The vector KL ( ; ; 6 ) (0; ; ) and KN ( ; 7 ; ) (; ; 0) Thus, the area of the arallelogram KLMN is KL KN i j k i 0 0 j 0 k j i 6j kj ( ) (6) ( ) 6 (0ts) ( #0) Find the arametric euation and symmetric euation for the line of intersection of the lanes x y z and x z 0 To write down a line euation, we need a directional vector and a oint To get a oint, rst, we assume that z 0 So, we have x y and x 0 This imlies that x 0 and y Thus, the oint (0; ; 0) is in the intersection of our two lanes, which is a oint we need We can read the normal vector of the lane x y z to be (; ; ) and the normal vector of the lane xz 0 to be (; 0; ) Since the intersection line lies on both lanes,

3 the line must be erendicular to both normal vectors So, the directional vector of the line can be (; ; ) (; 0; ) (; 0; ) Therefore, the arametric euation of the line is < : x 0 () t t y (0) t z 0 ( ) t t and the symmetric euation of the line should look like x 0 y z 0 0 But, we cannot divide by 0 Thus, we should write a euation for y individually So, the symmetric euation of the line is x 0 z 0 and y 6 (0ts) ( #0) Find the euation of the lane that asses through the line of intersection of the lanes x z and y z and is erendicular to the lane x y z We can read the normal vector of the lane x z is (; 0; ) and the normal vector of the lane y z is (0; ; ) The directional vector of the line of intersection of both lanes is erendicular to both So, it can be (; 0; ) (0; ; ) (; ; ) Also, by letting z 0 for both lanes, we can have a oint in the line of intersection (; ; 0) The normal vector of the lane x y z is (; ; ) Since our lane is erendicular to the lane x y z, we know that (; ; ) is on our lane Now, we have two vectors (; ; ) and (; ; ) on our lane and a oint (; ; 0) Thus we have the normal vector of our lane is (; ; ) (; ; ) (; ; ) Hence, the euation of the lane is (x ; y ; z 0) (; ; ) 0, which is x y z 4 7 (0ts) (7 #0) Identify the surface r z 4 We see r and z It suggests that we are using cylindrical coordinates In cylindrical coordinates, we have r x y Thus, our euation becomes x y z 4 By dividing by 4, we get x 4 y z 4 By looking at Table in age 6, we identify this surface as a hyerboloid of one sheet (0ts) (0 #0) Find arametric euations for the tangent line to the curve x cos t, y sin t, and z 4 cos t at the oint ; ; Illustrate by grahing both the curve and the tangent line on a common screen

4 4 Let the curve be reresented by a vector function r (t) h cos t; sin t; 4 cos ti When t, we have the oint 6 ; ; So, the directional vector of the tangent line at ; ; is D r 0 h sin t; cos t; sin tij 6 t ; ; 4 E 6 Thus, the arametric euations for the tangent line is < x ( ) t t y t t : z 4 t 4 t (0ts) (0 #) Find the vector T, N, B, and the curvature of r (t) t ; t ; t at the oint ; ; We have r 0 (t) t; t ; This imlies that jr 0 (t)j (t) (t ) 4t 4t 4 Thus, T (t) r0 (t) t jr 0 (t)j t ; t t ; t This imlies that T 0 4t (t ) (t) ; (t ) (t ) ; (t ) When t, we have So, N () T0 () jt 0 ()j T 0 () ; 4 ; Therefore, B () T () N () ; ; ; ; i j k Finally, when t, dt ds jt0 ()j jr 0 ()j 4 (t ) t ; ; () i 6 j k

5 0 (0ts) (0 #) Find a arametric reresentation for the art of the hyerboloid x y z that lies to the right of the xz-lane In hyerboloid, with a xed z, we have a circle This suggests us that we can set x r cos and y r sin, where 0 and r 0 Thus, z x y r cos r sin r The arametric reresentation is r (r; ) r cos ; r sin ; r Since it lies on the right of the xz-lane, it means that y 0 This tells us that r sin 0 Since r 0, we have sin 0 It turns out that 0 Therefore, we have the arametric reresentation r (r; ) r cos ; r sin ; r where r 0 and 0