1. If     and  are vectors and      for some scalar    then which of the following  must  be true?

a)       has the same direction as     
b)       is opposite in direction to       
c)       is perpendicular to            
d)       is parallel to       
e)       has the same length as  
f)       has magnitude greater than    

g)     has magnitude less than           
h)                   
i)      
j)   and    determine a unique plane through the origin   

Solution: (d)

2.  If the vectors   and   are perpendicular then what might  be?  

a) -1          b) -1/2         c)  0           d) 1/2            e)  1         
f)   3/2      g)  2            h) 5/2          i)  3               j) 7/2      

Solution: (f)

We need the dot product - it is in the  linalg  package.

>	with(linalg):

>	eqn_for_b := dotprod([-3, b, b],[2, 1, 2*b]) = 0;

>	solve(eqn_for_b, b);

Only one of these is in the answer list.

3. The line through the points ( 3, -1, 2) and (5, 2, -1) passes through a point (a, b, -4). What is  b  ?

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

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

Solution: (j)

>	OP := vector([3,-1,2]); OQ := vector([5,2,-1]);

>	v := evalm(OQ-OP);

>	line := evalm(OP + scalarmul(v,t));

>	eqn := t = solve(line[3] = -4, t);

>	b = subs(eqn, line[2]);

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

Its equation is   Ax + 3y + Cz = D.   What is   D?

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

Solution: (g)

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

>	v := evalm(OP-OQ);

>	w := evalm(OR-OQ);

>	cp := crossprod(v,w);

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

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

>	eqn3 := map(u->u+6, eqn2);

>	D = rhs(eqn3);

5. The function     has one critical point P = (a,b) with ab  <  0.  Choose the ordered list [a,what] where a is the abscissa of the critical point P = (a,b) and "what" describes the behavior of    at P.

a) [1,local minimum]      b) [-1,local minimum]        c) [-2,local minimum]          
d)  [1,local maximum]     e) [-1,local maximum]         f) [-2,local maximum]       
g)  [1,saddle point]             h) [-1,saddle point]                i) [-2,saddle point]    
j) [2,local maximum]

Solution: (d)

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

>	solve({diff(f(x,y),x) = 0 , diff(f(x,y),y) = 0}, {x,y} );

>	diff(f(x,y),x$2)*diff(f(x,y),y$2)-(diff(f(x,y),x,y))^2;

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

>	subs({x=1,y=-1},diff(f(x,y),x$2));

6. At an instant of time a particle's acceleration is    and its principal unit normal is
   .    What is its normal component of acceleration   ?

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

f)   5/3          g)  2              h) 7/3            i)  8/3              j)  3    

Solution: (j)

>	a := [2,1,-3]:  N := [2/3,2/3,-1/3]:

>	dotprod(a,N);

7.  The maximum value of   on the unit circle     
occurs at  (a,b).   What is the value of  a ?

a)              b)             c)                  d)           e)     
f)              g)                 h)                     i) 1                 j)  0

Solution: (a)

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

>	eqn1 := diff(f(x,y),x) = lambda*diff(phi(x,y),x);

>	eqn2 := diff(f(x,y),y) = lambda*diff(phi(x,y),y);

>	eqn3 := phi(x,y) = 1;

>	solve( {eqn1,eqn2,eqn3}, {x,y,lambda});

>	with(plots):

>	levelCurve[1] := implicitplot( f(x,y)=f(sqrt(5)/5,2*sqrt(5)/5), x = -3/2..3/2,y=-3/2..3/2,color=pink,thickness=2):

>	levelCurve[2] := implicitplot( x^2+y^2=1, x = -3/2..3/2,y=-3/2..3/2,color=plum,thickness=2):

>	levelCurve[3] := contourplot(f(x,y), x = -3/2.. 3/2, y= -3/2..3/2, contours=10,color=COLOR(RGB,0.75,0.7,0.7)):

>	display(seq(levelCurve[j],j=1..3));

8. If      then calculate      at      .

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

Solution: (a)

>	eqn1 := x^2+y^3-z(x,y)^4=2*x*y*z(x,y)+4;

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

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

>	subs({x=3,y=2}, rhs(eqn3));

>	subs(z(3,2)=1,");

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

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

f)   0            g)   1              h)   2           i)  3           j)   4  

Solution: (c)

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

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

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

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

10. Calculate    .   where  C   is the oriented curve parameterized by ,    .

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

f)   2             g)   8/3              h)   3           i)  4             j)  14/3

Solution: (j)

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

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

>

value(");

11. The vector field  

  is the gradient field of some function f(x,y).  Calculate f(1,2) - f(0,1).

a)  2              b)  4             c)   6            d)  8          e)  10                     

f)  12            g)  14            h)  16           i) 18          j)  Cannot be calculated from given information.  

Solution: (i)

>	F := vector([y^2+7,2*x*y+3*y^2]);

>	curl(vector([y^2+7,2*x*y+3*y^2,0]),[x,y,z]); #F is conservative

>	eqn := u(x,y) = int(y^2+7,x) + g(y);

>	diff(eqn,y);

>	F[2] = rhs(");

>	diff(g(y),y) = solve(",diff(g(y),y));

>	eqn2 := subs(g(y)=y^3+C,eqn);

>	subs( {x=1,y=2}, rhs(eqn2)) - subs({x=0,y=1}, rhs(eqn2));

12.  Calculate       where  C   is the oriented curve parameterized by ,    .   

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), sin(Pi/2*t^2), t*exp(t-1)]:

>

r(0);

>

r(1);

>	with(linalg):

>	F := vector([y+z,x+1,2*z+x]);

>	curl(F,[x,y,z]); #F is conservative

>	eqn := u(x,y,z) = int(y+z,x) + g(y,z);

>	diff(eqn,y);

>	x + 1 = rhs(");

>	diff(g(y,z),y) = solve(",diff(g(y,z),y));

>	eqn2 := subs(g(y,z)=y+h(z),eqn);

>	2*z+x = diff(rhs(eqn2),z);

>	diff(h(z),z) = solve( " , diff(h(z),z));

>	eqn3 := subs(h(z)=z^2 + C, eqn2);

>	evalm( grad(rhs(eqn3),[x,y,z]) - F );  #Check

>	subs( {x=1,y=1,z=1}, rhs(eqn3)) - subs({x=0,y=0,z=0}, rhs(eqn3));

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   has a potential function

III )   curl( F ) = 0  

IV )   

Write  "A implies B"   as   "A => B".

At least one of the implications stated in the answers may be incorrect. 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 by 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. If      then calculate div(grad(f(x,y,z)).

a)  1           b)  2            c)  3           d)  4           e) 5
f)   8          g)   12         h)  15          i)  16          j)  20

Solution: (d)

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

>	grad(f(x,y,z),[x,y,z]);

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

17. Calculate the surface integral of  over that part of  the plane   
that lies in the first octant.
a)  0           b) 1               c)             d)           e)    
f)          g)          h)           i)             j)  

Solution: (d)

>	with(linalg):

>	vectorLength := v -> sqrt(sum(v[i]^2,i=1..vectdim(v)));

>	r := (x,y) -> vector([x,y,1-(x+y)]); #surface parameterization

>	rx := map(z->diff(z,x),r(x,y));

>	ry := map(z->diff(z,y),r(x,y));

>	crossprod(rx,ry);

>	surfaceAreaFactor := vectorLength(");

>	f := (x,y,z) -> x + z;

>	Int(Int(f(x,y,z)*surfaceAreaFactor,y=0..1-x),x=0..1);

>	subs(z=1-(x+y),");

>

value(");

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)

Using the Divergence Theorem

>	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):

>	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);

>	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):

>	F := vector([x,0,z^2]);

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

>	int(int(r*int(diverge(F,[x,y,z]),z=0..1),r=0..1),theta=0..2*Pi);