Instructor Hentzel
Office Phone: 515-294-8141
E-mail:   hentzel@iastate.edu
Math Department Fax:      515-294-5454
http://www.math.iastate.edu/hentzel/class.166.ICN

Textbook:  Calculus Early Vectors Preliminary Edition
           By Stewart.

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#                       TEST THIS FRIDAY                         #
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#        Fri, Apr 25 Exam 4:         Sections 11.5-11.8, 12.1    #
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Monday, April 21      12.1 Several Variables     

p730: (1),5,(8),10,15,17,(24),(26),31,(36),37,

      42,45,47,49,53,54,59,(60),61,(62),63,(64)

Main Idea:  You can stick as many unknowns in the function as
you please.  You differentiate each of them separately. 
Key Words: Level Surface, Tangent Plane, Normal Line. 
Goal: Learn basic techniques for working with functions
f(x,y,z,w).
----------------------------------------------------
Previous Assignment.
Page 713 Problem 19
A force with magnitude 20 N acts directly upward from
the xy-plane on an object with mass 4 kg.  The object
starts at the origin with initial velocity V(0) = (1,-1,0).
Find its position function and its speed at time t.
a = (0,0,5)
v = (0,0,5) t + (1,-1,0)
r = (0,0,5) t^2/2 + (1,-1,0) t 
speed(t) = Sqrt[2+25 t^2].
--------------------------------------------------------
r[t_] := {0,0,5} t^2/2 + {1,-1,0} t;
v[t_] = D[r[t],t];
a[t_] = D[v[t],t];
s[t_] = Sqrt[v[t].v[t]];
red[{x_,y_,z_}] := {x,y,z,RGBColor[1,0,0]};
green[{x_,y_,z_}] := {x,y,z,RGBColor[0,1,0]};
blue[{x_,y_,z_}] := {x,y,z,RGBColor[0,0,1]};

aa = ParametricPlot3D[r[t],{t,-1,1}];
bb = Table[ ParametricPlot3D[  red[r[t/5]+x v[t/5]],    {x,0,1}],{t,-5,5}];
cc = Table[ ParametricPlot3D[  green[ r[t/5]+x a[t/5]],  {x,0,1}],{t,-5,5}];
dd = Show[aa,bb,cc,PlotLabel->"Page 713 Problem 19",PlotRange->All];
Display["dd.ps",%];
-----------------------------------------------------------------------------
Page 713 Problem 23
Find the tangential and normal components of the
acceleration vector
r(t) = ( t^3, t^2, t)
v(t) = (3t^2, 2t, 1)
a(t) = (6t, 2, 0)
           18t^3 + 4 t 
At =  ------------------- 
      Sqrt[9t^4 + 4t^2 + 1]

        |   i     j     k  |
        |  6 t    2     0  |
An =    | 3t^2    2t    1  | 
      -------------------------
        Sqrt[9t^4 + 4t^2 + 1]  

       | ( 2, -6 t, 6 t^2 ) |
An =   --------------------------
        Sqrt[9t^4 + 4t^2 + 1]  


        Sqrt[ 36t^4 + 36 t^2 + 4 ] 
An =   --------------------------
            Sqrt[9t^4 + 4t^2 + 1]  

-------------------------------------------------
r[t_] :=  {t^3,  t^2, t}; 
v[t_] = D[r[t],t];
a[t_] = D[v[t],t];
s[t_] = Sqrt[v[t].v[t]];
red[{x_,y_,z_}] := {x,y,z,RGBColor[1,0,0]};
green[{x_,y_,z_}] := {x,y,z,RGBColor[0,1,0]};
blue[{x_,y_,z_}] := {x,y,z,RGBColor[0,0,1]};

aa = ParametricPlot3D[r[t],{t,-1,1}];
bb = Table[ ParametricPlot3D[  red[r[t/5]+x v[t/5]],    {x,0,1}],{t,-5,5}];
cc = Table[ ParametricPlot3D[  green[ r[t/5]+x a[t/5]],  {x,0,1}],{t,-5,5}];
dd = Show[aa,bb,cc,PlotLabel->"Page 713 Problem 23",PlotRange->All];
Display["dd.ps",%];
-----------------------------------------------------------------------------
Page 713 Problem 24
Find the tangential and normal components of the
acceleration vector.
r(t) = (t, Cos[t]^2, Sin[t]^2 )
v(t) = (1, 2 Cos[t](-Sin[t]), 2 Sin[t] Cos[t] )
v(t) = ( 1, -Sin[2t], Sin[2t] )
a(t) = ( 0, -2 Cos[2t], 2 Cos[2t] )
   
        A.V        4 Sin[2t] Cos[2t]
At =  -------- = ---------------------
        |V|         Sqrt[ 1 + 2 Sin[2t]^2 ] 

                 |  i       j            k       | 
                 |  0   -2 Cos[2t]    2 Cos[2t]  |
       |AxV|     |  1   - Sin[2t]     Sin[2t|    |
An =   ------ = -------------------------------------
        |V|              Sqrt[1+2 Sin[2t]^2 ]

      | (  0,  2 Cos[2t],  2 Cos[2t] )|
An = -------------------------------------
            Sqrt[1+2 Sin[2t]^2 ]

             2 Sqrt[2] | Cos[2t] | 
An = -------------------------------------
             Sqrt[1+2 Sin[2t]^2 ]
-------------------------------------------------
r[t_] :=   {t, Cos[t]^2, Sin[t]^2 }
v[t_] = D[r[t],t];
a[t_] = D[v[t],t];
s[t_] = Sqrt[v[t].v[t]];
red[{x_,y_,z_}] := {x,y,z,RGBColor[1,0,0]};
green[{x_,y_,z_}] := {x,y,z,RGBColor[0,1,0]};
blue[{x_,y_,z_}] := {x,y,z,RGBColor[0,0,1]};
aa = ParametricPlot3D[r[t],{t,-Pi,Pi}];
bb = Table[ ParametricPlot3D[  red[ r[t]+x v[t]],    {x,0,1}], {t,-3,3}];
cc = Table[ ParametricPlot3D[  green[ r[t]+x a[t]],  {x,0,1}], {t,-3,3}];
dd = Show[aa,bb,cc,PlotLabel->"Page 713 Problem 23",PlotRange->All,ViewPoint->{0,10,0}];
Display["dd.ps",%];
-----------------------------------------------------------------------------
r[t_] := {t, Cos[t]^2, Sin[t]^2 };
v[t_] = D[r[t],t];
a[t_] = D[v[t],t];
s[t_] := Sqrt[v[t].v[t]];
At = 1/s[t] a[t].v[t];
An = 1/s[t] Sqrt[ Cross[a[t],v[t]].Cross[a[t],v[t]]];

--------------------------------------------------------
Page 713 Problem 25
Find the tangential and normal components of the
acceleration vector.

r(t) = (e^t, Sqrt[2] t, e^(-t) )
v(t) = (e^t, Sqrt[2], -e^(-t) )
a(t) = (e^t, 0, e^(-t) ).

       A.V    e^(2t) -e^(-2t)
At =  ----- = ----------------- = e^t - e^(-t)
       |V|      e^t + e^(-t)

                  |  i        j         k    |                  
                  | e^t       0      e^(-t)  | 
       |AxV|      | e^t    Sqrt[2]  -e^(-t)  |
An = --------- = ------------------------------
        |V|              e^t + e^(-t)

     |( -Sqrt[2] e^(-t),   2,  Sqrt[2] e^t ) |
An = -------------------------------
         e^t + e^(-t)

      Sqrt[ 2 e^(-2t) + 4 + 2 e^(2t)  ] 
An = ----------------------------------
               e^t + e^(-t)

      Sqrt[2]  Sqrt[  e^(-2t) + 2 +  e^(2t)  ] 
An = -------------------------------------------
                 e^t + e^(-t)

      Sqrt[2]  (  e^(-t) + e^t  )
An = ------------------------------- = Sqrt[2]
         e^t + e^(-t)
----------------------------------------------
New Material
============
Visual Example:
y = f(x)   line in a plane.
z = f(x,y)  Surface in space
w = f(x,y,z)  Something like temperature at a point in space.
---------------------------------------------------------
Function Example.
f(x,y,z) = xyz.
f(x,y,z) = x^2 + 6 y^2 + 2 z^3
f(x,y,z) = e^x Cos[y] z^2
------------------------------------------------------------
Domain is the region where the expression for the function works.
f(x,y,z) = Sqrt[ 1 - x^2 - y^2 - z^2].
The domain of f is the region inside the sphere   
1-x^2-y^2-z^2 >= 0  or 1 >= x^2 + y^2 + z^2.
               z
f(x,y,z) = -----------
           x^2 + y^2

domain is everything except the z axis.
---------------------------------------------------------------
Level Curves and Level Surfaces.
One way to study functions is to plot the level surfaces.  That is
plot all points satisfying f(x,y) = k.  then vary k and do it again.
This gives a contour map.
f(x,y) = (x^2+y^2) Sin[x].
-------------------------------------------------------------------
f[x_,y_] := (x^2+y^2) Sin[x];
ContourPlot[f[x,y],{x,-4 Pi,4 Pi},{y,-4 Pi,4 Pi}, Contours->30,
ContourShading->False];
Plot3D[f[x,y],{x,-4 Pi,4 Pi},{y,-4 Pi,4 Pi},PlotPoints->100]
-----------------------------------------------------------------


When the function is a surface, the contours are lines.  So
z = f(x,y) has contour lines.
when the function is of three variables, 
w = f(x,y,z), then the f(x,y,z) = c will be a surface and is
called a level surface.  

For higher number of variables, v = f(x,y,z,w,u) one can
talk about the "higher dimensional" level curves or the
"higher dimensional" level surfaces.  They are both the same.

f(x,y,z) =   2x^2 + 3 y^2 + z^2 
The level surfaces are ellipsoids.   Take the level surface through
(1,1,2).  This is the surface on which f maintains the same value
as it does at (1,1,2).  The equation of this level surface is
gotten by evaluating f(1,1,2) = 9.  The level surface has equation
2x^2 + 3y^2 + z^2 = 9. 

To find the tangent plane to the surface we need the normal to the
surface.  We find the normal to the surface by differentiating the
function first with respect to x, then with respect to y, then
with respect to z.
del f = (4x,6y,2z)
del f(1,1,2) = (4,6,4) and this is the normal at the point (1,1,2).

                       x = 1 + 4t 
The normal line is     y = 1 + 6t
                       z = 2 + 4t

The tangent plane at (1,1,2) is 4x+6y+4z = 18.
-----------------------------------------------------------
2x^2 + 3y^2 + z^2 = 9. 
f[x_,y_] := Sqrt[ 9 - 2 x^2 - 3 y^2]
p[x_,y_] := (18-4x-6y)/4;
n[t_] := {1+4t,1+6t, 2+4t};
a = Plot3D[f[x,y],{x,-Sqrt[9/2],Sqrt[9/2]},{y,-Sqrt[3],Sqrt[3]},PlotPoints->50];
b = Plot3D[p[x,y],{x,0,Sqrt[9/2]},{y,0,Sqrt[3]}];
c = ParametricPlot3D[n[t],{t,-3,3}];
d = Show[a,b,c,PlotLabel->"2x^2 + 3 y^2 + z^2 = 9 at (1,1,2)"];
Display["d.ps",d];
-------------------------------------------------------------------
Find the tangent plane and normal line to the level surface of
f(x,y,z) = xyz at (1,2,3).
del f = (yz,xz,xy)
delf(1,2,3) = (6,3,2)
                       x = 1 + 6t
The normal line is     y = 2 + 3t
                       z = 3 + 2t

The tangent plane is   6x+3y+2z = 18
-----------------------------------------------------------
xyz = 6;
f[x_,y_] :=  6/(x y); 
p[x_,y_] := (18-6x-3y)/2; 
n[t_] := {1+6t,2+3t,3+2t};
a = Plot3D[f[x,y],{x,0.5, 10},{y,2, 10},PlotPoints->50];
b = Plot3D[p[x,y],{x,1, 10},{y,2, 3}];
red[{a_,b_,c_}] := {a,b,c,RGBColor[1,0,0]};
c = ParametricPlot3D[red[n[t]],{t,0,2}];
d = Show[a,b,c,PlotLabel->"2x^2 + 3 y^2 + z^2 = 9 at (1,1,2)",
PlotRange->All];
Display["d.ps",d];
-------------------------------------------------------------
Find del f for these functions.
1.  f(x,y,z) = Sin[xy] z
2.  f(x,y,z) = x^2 y z
3.  f(x,y,z) = 2xy
-----------------------------------------------------------