12.6

Quadric surfaces Review:

rv point on the line, direction  abc , , Vector equation of a line L: r r0 t v 0

Parametric  scalar equation of a line L: x x0  at, y  y 0  bt , z  z 0  ct

r(t ) P0  t ( P 1  P 0 ) with 0  t  1 is the line which goes from P 0 to P 1

r0 on the plane Vector equation of a plane: n r  r0   0 n  abc, , normal to plane Scalar equation of the plane: ax by  cz  d

nn12 Angle between two planes with normal nn12 , : cos  nn12 | (Qrn ) | Distance between a point Q and a plane: 0 n

| (Qrv ) | Distance between a point Q and a line: 0 | v | Surfaces in 3-space: x2 y 2 z 2 Ellipsoid    1 a2 b 2 c 2

Hyperboloid of one sheet x2 y 2 k 2 x2 y 2 z 2 z k:   1  any k a2 b 2 c 2 2 2  2 1 abc horizontal traces are ellipses

Hyperboloid of two sheet x2 y 2 k 2 z k: 2  2  2  1, | k |  1 x2 y 2 z 2 a b c    1 horizontal traces are ellipses a222 b c xy22 (Elliptical paraboloid) z  ab22

Horizontal traces are ellipses (notice z  0)

xy22 (Hyperbolic paraboloid) z  ab22

Horizontal traces are hyperbolas

Vertical traces are parabolas xy22 Cone z2  ab22

Horizontal traces are ellipses

Vertical traces are hyperbolas

y Notice: If x  0 z  straight lines b Cylinders: xy22 yx22 y ax2 1 1 ab22 ab22 VII

IV

II III

VI I

VIII V Draw a picture of the following surfaces:

x y22 z 22or  a) x 2 y 3 z 6 3 2

elliptic paraboloid

opens in the positive x  direction

yz22 b) 4 x y22  4 z  0 or x  41

hyperbolic paraboloid Draw a picture of the surface : 4x2 y 2  4 z 2  4 y  24 z  36  0

4x2 y 2  4 y  ___  4 z 2  6 z  ___9    36  ___4  ___36

4x2  y  222  4 z  3  4

22 x2  yz23     1 1 4 1

ellipsoid center:  0,2,3 notice e.g. y  0 Summary:

x2 y2 z 2   1 Ellipsoid a2 b2 c2 x2 y 2 z 2   1 Hyperboloid of one sheet abc2 2 2 all variables present x2 y 2 z 2    1 Hyperboloid of two sheets a222 b c all variables squared z2 x 2 y 2  Cone  elliptic c2 a 2 b 2

z x22 y  Paraboloid  elliptic c a22 b z x22 y one variable Paraboloid  hyperbolic 22 c a b not squared

one variable not present cylinder opening in the direction of the missing variable 22 22 xy xz 2 1 Elliptic cylinder 1 Hyperbolic cylinder z ax Parabolic cylinder ab22 ab22 13.1

Curves in Space A curve in space (or the plane): x f( t ), y  g ( t ), z  h ( t ) or in the plane: r(),,t f g h  f i  g j  h k r()()()t f t j g t j

This is a vector valued function (as opposed to scalar valued)

Example: r(tt ) cos( ),sin(t), t 

Notice that x ( t )22 y ( t ) 1 i.e. the curve lies on a cylinder

A circular helix: A curve in the plane: r(tt ) 4cos(2 ) i 9sin(2 t) j

xy22 Notice that  1 The particle travels along an ellipse 4922

r(0) 4 i = (4,0) Travels counter clockwise as 0t 2 the particle goes twice around the ellipse

Trefoil knot:

3t 3 t 3 t rt 2  cos 2 cos t , 2  cos 2 sin t ,sin 2 

For an animation go to : http://math.bu.edu/people/paul/225/trefoil_knot.html For a vector valued function: rt  f t,, g t h t

Take limits, derivatives and integrals component wise: limrt  lim f t ,lim g t ,lim h t t a t  a t  a t  a rt  f  t,, g  t h  t b b b b rtdt   ftdt  ,  gtdt  ,  htdt  a a a a sint Example: Let rrt  , e2t ,ln 1 t . Find lim  t t t0 sin t limr te  lim ,0 ,ln 1 1,1,0 ij  tt00t Examples:

(a) Let rt  sin12 t , 1  t ,ln 1  3 t . Find rt

1/2 1 1 2 1 13t rt  ,  12  tt    ,  3   ,, 1t 2 2 13 t 11tt2213 t

2 2 2 2 (b) t2i t t 1 j  t sin t k dt tdttt2 ,  1 dtt ,  sin  tdt 1 1 1 1

2 3 2 t 81 7 t2 dt      33 3 1 3 1 2 1 1 1 3/2 1/2 225/2 3/2 u1 udu u u dt uu t t1 dt    53  0 0 0 1 ut1 du dt 22 6 10 16    tu10   53 15 15 15 tu21   2 tsin  t dt ut dv sin t 1 1 udv uv vdu du dt vt cos    2 t 2 1 cost  cos t dt 1 1

2 t 1 2 1   1 1  costt sin  cos 2 22 sin 2     cos   sin    2     1     21    3 1  0     1  0       

2 7 16 3 t2i t t 1 j  t sin t k dt  , , 1 3 15  Differentiation Rules: Integration rules: d 1. ut  v t  u t  v  t (u(t ) v (t)) dt   u ( t ) dt   v ( t ) dt dt  d 2. cuu t  c t cuu()() t dt c t dt dt   d 3. f tu t  f t u t f t u  t dt  d 4. ut v t  u t  v t  u t  v  t dt  d 5. ut v t  u t  v t  u t  v  t dt  d 6. uu f t  f t  f t dt  Geometry of the derivative vector : z secant vector P rrt h  t

Q r t rth  y

Tangent vector x tangent vector P rrt h  t rt  lim Q rrt h t r t     h0 h h rth  The tangent line to a smooth curve at tt 0 rt  ftgtht ,,     ft  i  gt  j  ht  k

It passes through the point r (t0 )  f t 0 , g t 0 , h t 0  , and has the same direction as the tangent vector at t0

r '(t0 ) f t 0 , g  t 0 , h  t 0  . So the equation of the tangent line is:

s r(t00) s r '( t ) Or in parametric form:

x f( t0 )  sf '( t 0 ), y  g ( t 0 )  sg '( t 0 ), z  h ( t 0 )  sh '( t 0 ) Problem : A drunken bee travels along the path r (tt )  cos(2 ),sin(2 t), t  for 10 seconds. It then travels at constant speed in a straight line for 10 more seconds. At what point does the bee end up?

Question: What is speed and velocity? Velocity is a vector: v (t )  r'(t) Speed is a scalar: | v (t ) |

velocity: v (tt )  2sin(2 ),2cos(2 t),1  Speed = | v (t ) | 9 3

At time t0  10, it travels along the tangent line with speed 3 ft/sec

tangent line: s r(t00 ) s v ( t ) after 10 more seconds, it arrives at

r(tt00) 10 v ( ) cos(20),sin(20),10   10  2sin(20),2cos(20),1   cos(20)  20sin(20),sin(20)  20cos(20),20   17.85, 9.07, 20  Question: What is distance traveled?