1. The vector    is tangent to the surface     at the point (2,1,1).  
     What is   ?

a)  -4           b)  -3             c)   -2          d)  -1           e)  0                    

f)   1            g)   2              h)   3           i)  4             j)   5

Solution: (j)

>	F := (x,y,z) -> x^2-y^8-3*z^4;

>	v := grad(F(x,y,z),[x,y,z]);

>	NORMAL := subs({x=2,y=1,z=1}, evalm(v) );  #Normal to surface

>	solve(dotprod(NORMAL,[a,1,1])=0,a); #Tangent vector is perpendicular to Normal vector

2.  For a particle follows the curve  that is parameterized by .  
For  the particle follows the line that is tangent to  at the point (1,1,1).  
The linear part of the trajectory carries the particle through the point (a, 7, c).  
What is  c - a ?

a)  1          b)  2          c)  3           d)  4             e)  5         
f)   6          g)  7          h) 8             i)  9             j)  10      

Solution:  (f)

>	r1 := t -> [t,t^2,t^3];

>	r2 := s -> [seq(r1(1)[j]+s*map(z->subs(t=1,diff(z,t)), r1(t))[j],j=1..3)];

Warning, `j` in call to `seq` is not local

>	s0 := solve(r2(s)[2] = 7, s);

>	r2(s0)[3] - r2(s0)[1];

3.  A particle's position in space at time  t  is    .  At time    the particle is at point (1,2,3)
   and has velocity .   Calculate   if     
   and .

a)  1            b)  2            c)   3         d)  4           e)  5                     

f)   6           g)  72           h)   8         i)  9            j)  10

Solution: (c)

>	f := (x,y,z) -> 6*x^2-y^2-3*z;

>	u := subs({x=1,y=2,z=3},grad(f(x,y,z),[x,y,z]));

>	v := vector([4,6,7]);

>	dotprod(u,v);

4.  A plane passes through the three points  (1, 0, 1),  (0, 2, 1),  and (2, 3, 2).

Its equation is   Ax + y + Cz = D.   What is   D?

a) -4            b) -3           c)  -2         d) -1          e) 0  
f)   1            g)   2          h)  3           i)  4           j)  5

Solution:  (b)

>	OP := vector([1,0,1]); OQ := vector([0,2,1]);   OR := vector([2,3,2]);

>	v := evalm(OP-OQ);

>	w := evalm(OR-OQ);

>	cp := crossprod(v,w);

>	cp := map(z->-z,cp);

>	eqn1 := cp[1]*(x-OP[1])+cp[2]*(y-OP[2])+cp[3]*(z-OP[3])=0;

>	eqn2 := map(u->u-3, eqn1);

>	D = rhs(eqn2);

5. Let  be the number of saddle points that the function    has.
Let  denote the saddle points.   

What is      ?   

Note: If you determine that , then answer  0  because,  by convention, an empty sum is 0.  

a)  0             b)  1           c)   2          d)  3          e) 4  
f)   5            g)   6           h)  7           i)  8           j)  9

Solution: (e)

>	f := (x,y) -> x^2*y-12*x-x*y^2:

>	solve({diff(f(x,y),x) = 0 , diff(f(x,y),y) = 0}, {x,y} ); #step-by-step below

>	solns := [[4,2],[-4,-2]];

>	discr := (x,y) -> subs( {u=x,v=y} , diff(f(u,v),u$2)*diff(f(u,v),v$2)-(diff(f(u,v),u,v))^2);

>	discr(x,y);

>	for n from 1 to 2 do discr(op(solns[n])); od;

>	abs(solns[1][2])+abs(solns[2][2]);

---------------------------------------------------------------

Step-by-step location of critical points

>	eqn[1] := diff(f(x,y),x) = 0 ; eqn[2] := diff(f(x,y),y) = 0;

From the first equation, observe that there is no real solution for y when x = 0. We may therefore divide by x.

>	eqn[3] := map(z -> simplify(z/x), eqn[2]);

>	eqn[4] := subs( x = 2*y , eqn[1] );

>	solve(eqn[4],y);

6. A particle is accelerating so that its speed at time  is equal to .  At time  the
    acceleration of the particle is  8i + 6j + 2k.  What is the curvature of the particle's path at ?

a)  1/4             b)  1/2             c) 3/4            d) 1                e)  5/4

f)   3/2            g)  7/4              h) 2               i)  9/4             j)  5/2    

Solution: (j)

>	a := [8,6,2];  #acceleration at t = 1

>	speed := t -> t^2+1;

>	a_T := t -> D(speed)(t);

>	a_N := t -> kappa(t) *speed(t)^2;

>	eqn := a_T(1)^2 + a_N(1)^2 = a[1]^2+a[2]^2+a[3]^2;

>	kappa(1) = [solve(eqn,kappa(1))];  #pick positive root

7.  By changing the order of integration, express

          

 in the form         .  What is   ?

a)  10          b)  11          c)  12           d)  13             e)  14         
f)   15          g)  16          h) 17            i)  18              j)  19      

Solution: (c)

>	eqn := Int(Int(13*x+17*y,y=0..8*x),x=0..1)+Int(Int(13*x+17*y,y=0..9-x^2),x=1..3) = Int(Int(13*x+17*y,x=y/8..sqrt(9-y)),y=0..8);  Test proposed formula with weird function

>	c := 0;  d := 8; f := y -> y/8; g := y -> sqrt(9-y);

>	d + f(16) + g(5);

8. If      then calculate      at      .

a) -27/13             b) -23/13             c)  -19/13           d) -15/13            e) -11/13               
f)  -7/13               g) -3/13               h)  1/13               i)  5/13               j) 9/13

Solution: (d)

By implicit differentiation:

>	eqn1 := x^3 + y^2 - z - 2*x*z^2 = -16;

>	eqn2 := subs(z = z(x,y), eqn1);

>	eqn3 := map( u -> diff(u,x), eqn2);

>	eqn4 := diff(z(x,y),x) = solve(eqn3 , diff(z(x,y),x));

>	eqn5 := lhs(eqn4) = subs(z(x,y) = z, rhs(eqn4));

>	subs({x=1, y=2, z=3} , rhs(eqn5));

Alternatively:

>	F := (x,y,z) -> x^3 + y^2 - z - 2*x*z^2;

>	expr := -diff(F(x,y,z),x)/diff(F(x,y,z),z);

>	subs({x=1, y=2, z=3} , expr);

9.  Calculate    .   where  C   is the oriented curve parameterized by  
                   

a)                  b)                 c)                  d)                   e)     
f)                g)                  h)               i)                j)      

Solution: (d)

>	r := t -> [cos(t),sin(t),sqrt(t)];

>	Int( 2*r(t)[2]*diff(r(t)[1],t) +  r(t)[1]*diff(r(t)[2],t)+ 2*r(t)[3]*diff(r(t)[3],t),t=0..2*Pi);

>

value(");

10.  The vector field  F( x , y ) = (y + 1) i + (x + 2) j  is the gradient field of some
scalar-valued function U(x,y)   with   U( 1 , -1 )  = 2.    Calculate    U( 2 , 2 ).

a)  2              b)  4             c)   5            d)  8          e)  10                     

f)  12            g)  14            h)  16           i) 18          j)   20  

Solution: (g)

>	M := y+1;   N := x + 2;

>	testeq(diff(M,y) = diff(N,x)); #F is conservative

>	eqn1 := U(x,y) = int(M,x) + g(y);

>	eqn2 := N = diff(rhs(eqn1),y);

>	eqn3 := diff(g(y),y) = solve(eqn2,diff(g(y),y));

>	eqn4 := g(y) = int(rhs(eqn3),y) + C;

>	eqn5 := subs(eqn4,eqn1);

>	eqn6 := C = solve(2 = subs({x=1,y=-1}, rhs(eqn5)), C);

>	eqn7 := subs(eqn6, eqn5);

>	subs( {x=2,y=2}, rhs(eqn7) );

11.  Calculate       where  C   is the oriented curve parameterized by the
vector-valued function     ,    .   

(Although this integration can be done independently of problem 10, you may wish to note
that the integrand is constructed using the vector field of problem 10.)

 a)  0             b)  1             c)   2          d)  3           e)  4  
  f )  5            g)  6             h)  7            i) 8             j)  9  

Solution: (e)

>	r := (t) -> [sqrt(t/2),t*(t^3+6)/(t^2+24)]:

>	r(0); r(2);

>	subs( {x=1,y=1} , (y+1)*x+2*y+4 ) - subs( {x=0,y=0} , (y+1)*x+2*y+4 );

12.

     (Answers a through d signify that only one of the four sets is simply connected.)     

Solution: (g)

13. Consider a vector field

in the region  G  that is shown. (G is a subset of the plane  z = 0 in xyz-space.)  Suppose that  F  is twice-continuously differentiable.  

Consider the following properties that   F   may have.

I )     F   is path-independent.

II )    F   is the gradient of a scalar-valued function.

III )   curl( F ) = 0  

IV )   

Write  "A implies B"   as   "A => B".

At least one of the implications stated in the answers is not correct.
Choose  any   incorrect statement as your answer.

a)  I => II           b)  II => III            c)   III => IV         d)  IV => I             e)  I => III                    
f)  II => I            g)  I => IV              h)  II => IV           i)  IV => III            j)  IV => II

Solution: (c)

Properties I, II, and IV are equivalent. They all imply property III. However, G is not simply connected so

property III does not imply any of the others.

14.  Calculate  where C is the unit circle traversed counter-clockwise.
a)                  b)                 c)                  d)                   e)     
f)                g)                  h)               i)                j)      

Solution: (i)

>	r := t -> [cos(t),sin(t)]:

>	Int( -r(t)[2]^3*diff(r(t)[1],t) + r(t)[1]^3*diff(r(t)[2],t),t=0..2*Pi);

>

value(");

More easily, using Green's Theorem:

>	Int(Int(r*(diff(x^3,x)-diff(-y^3,y)),r=0..1),theta=0..2*Pi);

>	subs(x^2=r^2-y^2,");

>

value(");

15. Let  G  be a solid region in space. Suppose that the boundary  S  of  G  is an orientable closed surface with outward unit normal n. Vector calculus tells us that

        

for some function   f.  If  

then what is  f(1,2,3) ?

a)  0           b)  1             c)   2          d)  3           e)  4                     

f)  5            g)  6             h)  7           i) 8             j)  9  

Solution: (g)

>	with(linalg):

>	diverge([x*y,2*y*z,x^2+x^2*y^3-y*z],[x,y,z]);

>	subs({x=1,y=2,z=3},");

16.  Suppose that   F (x, y, z)  = 2z i - xz j +   k . Let  G (x, y, z)  = x curl(F )(x,y,z) .
Calculate div(G)(x, y, z).

a)  x + y            b)  x + 2y          c)  2x + y          d)  x + z           e)  x + 2z
f)   2x + z          g)   y + z            h)  y + 2z          i)  2y + z           j)  x + y + z

Solution: (b)

>	F := (x,y,z) -> [2*z, -x*z, y^2];

>	expr := curl(F(x,y,z), [x,y,z] );

>	G := (x,y,z) -> map(w ->x*w, curl(F(x,y,z), [x,y,z] ));

>

G(x,y,z);

>	diverge(G(x,y,z),[x,y,z]);

17.

(Orient  S  by the unit normal  n  that is upward at the point for
which  u =1/2 and v = 1/2.)

a)  2/3           b)  5/6            c)  1            d)  4/3         e)  5/3
f)   2             g)   7/6            h)  3/2         i)  8/3           j)  3

Solution: (a)

>	with(linalg):

>	r := (u,v) -> [2*u+v, u+v^2,u^2+v];

>	F := (x,y,z) -> [0,0,x-z];

>	F(op(r(u,v)));

>	r_u := (u,v) -> subs({s=u,t=v},map(z->diff(z,s),r(s,t)));

>	r_v :=  (u,v) -> subs({s=u,t=v},map(z->diff(z,t),r(s,t)));

>	N := (u,v) -> convert(crossprod(r_u(u,v),r_v(u,v)),list):   N(u,v)[3]; N(1/2,1/2); dotprod(F(op(r(u,v))), N(u,v));

>	int(int(dotprod(F(op(r(u,v))), N(u,v)),u=0..1),v=0..1);

18. Let S  be the surface formed by the four triangular sides of the pyramid whose vertex is at (0,0,1) and whose base is the square with vertices (1,1,0), (-1,1,0), (-1,-1,0), and (1,-1,0).

If  n  denotes the outward pointing unit normal on  S  then calculate

a)  0             b)  2             c)  4            d)  6           e) 8
f)   10          g)   12          h)  14           i)  16          j)  20
 

Solution: (c)

Using Stokes's Theorem the required surface integral can be calculated by the line integral

integrated counterclockwise over the boundary of the square in the xy plane. Since   z = dz = 0  on all four sides and dy = 0 on the two sides parallel to the x-axis we have

>	int(1,y=-1..1)+int(-1,y=1..-1);

19.

a)                    b)                 c)                d)                  e)     
f)              g)              h)              i)            j)      

Solution: (h)

Directly:

>	with(linalg):

>	F := vector([x,0,-z]);

>	surface_eqn := z + x^2 + y^2 = 2;

>	N := grad(lhs(surface_eqn),vector([x,y,z]));

>	dotprod(F,N);

>	Integrand := subs(z=2-x^2-y^2, " );

>	Integrand := subs({x=r*cos(theta),y=r*sin(theta)},Integrand);

>	int(int(r*Integrand,r=0..1), theta = 0 .. 2*Pi);

Alternative calculation using the Divergence Theorem

>	with(plots):

Warning, the name changecoords has been redefined

>	parab := plot3d(2-x^2-y^2, y = -sqrt(1-x^2)..sqrt(1-x^2),x=-1..1,color=aquamarine):

>	disk := plot3d([r*cos(theta),r*sin(theta),1],r=0..1,theta=0..2*Pi,color=plum):

>	display(parab,disk,orientation=[35,120]);

>	diverge(F,[x,y,z]);

Therefore the sum of the required integral (integrated over the aquamarine paraboloid) and the flux integral through the bottom plum disk is equal to 0.

On the plum disk the unit outward normal is  - k   and

>	dotprod(F,vector([0,0,-1]));

But on the plum disk   z = 1.  Therefore:

>	required_integral = -int(int(1*r,r=0..1),theta=0..2*Pi);

20.

a)             b)                c)                  d)               e)       
f)                g)                h)            i)               j)      

Solution: (i)

Using the Divergence Theorem

>	with(linalg):

Warning, new definition for norm

Warning, new definition for trace

>	F := vector([0,2*y,z^3]);

>	divergence := diverge(F,[x,y,z]);

>	int(int(int(r*divergence,z=-1..2),r=0..1),theta=0..2*Pi);