final-sp2004.html

1. A line that is perpendicular to the plane 4*x+3*y+2*z=13 has symmetric equations (4*x+1)/a = (3*y-2)/b ` `=4*z What is a? a) 1 b) 2 c) 3 d) 4 e) 5 f) 6 g) 7 h) 8 i) 9 j) 10
2. If `<`*14 , -2*`>`=`<`*a,b*`>`+`<`*c,d*`>` where `<`*a,b*`>` is parallel to `<`*3,1*`>` and `<`*c,d*`>` is perpendicular to `<`*3,1*`>`, then what is b? a) 1 b) 2 c) 3 d) 4 e) 5 f) 6 g) 7 h) 8 i) 9 j) 10
3. A curve is parameterized by r(t)=`<`*t,t^3/3,t^2*`>` . Calculate its principal unit normal vector N=`<`*a,b,c*`>` at the point (1, 1/3, 1). What is a? a) 1/3 b) -1/3 c) 1/sqrt(2) d) -1/sqrt(2) e) 1/sqrt(3) f) -1/sqrt(3) g) 1/sqrt(6) h) -1/sqrt(6) i) 2/sqrt(6) j) -2/sqrt(6)
4. Suppose that x^2 +y*z-z^3=2*x*z-9 . Calculate diff( z ,x) at ``(3,1,2). a) 3/8 b) 4/15 c) 2/9 d) 3/7 e) 5/13 f) 2/17 g) 2/11 h) 3/19 i) 4/17 j) 8/21
5. The function f(x,y)=3*x^2*y- x*y^2- 9*x*y has exactly one critical point P = (a,b) that does not lie on a coordinate axis. Choose the ordered list [b,what] where b is the y-coordinate of the critical point P = (a,b) and "what" describes the behavior of f at P. a) [-1,local minimum] b) [-2,local minimum] c) [-3,local minimum] d) [-1,local maximum] e) [-2,local maximum] f) [-3,local maximum] g) [-1,saddle point] h) [-2,saddle point] i) [-3,saddle point]
6. Let u be the upward unit normal of the plane x-2*y+2*z=7. Calculate the directional derivative of F(x,y,z) = x^3+3*y*z+z^2 in the direction of u at the point ( 2, 2, 3). a) 1 b) 2 c) 3 d) 4 e) 5 f) 6 g) 7 h) 8 i) 9 j) 10
7. Let ( a , b ) be the point in the first quadrant at which f(x,y)=x*y+1 has a maximum value subject to the constraint x^2+2*y^2=1. Calculate a and b. What is the value of a^2+b^2 ? a) 1/8 b) 1/4 c) 3/8 d) 1/2 e) 5/8 f) 3/4 g) 7/8 h) 3/16 i) 5/16 j) 7/16
8. Calculate the integral of x+2*y over the region in the first quadrant that is bounded by the coordinate axes and the circle x^2+y^2=4 . a) 1 b) 2 c) 3 d) 4 e) 5 f) 6 g) 7 h) 8 i) 9 j) 10
9. The vector `<`*a,-3,-1*`>` is tangent to the surface x^2+2*y^3-3*z^2=3 at the point (2,1,1). What is a ? a) 1 b) 2 c) 3 d) 4 e) 5 f) 6 g) 7 h) 8 i) 9 j) 10
10. Calculate where U is the solid half-ball that is described by the inequalities x^2+y^2+z^2<=1 and z>=0. a) 0 b) Pi/12 c) Pi/8 d) Pi/6 e) Pi/4 f) Pi/3 g) Pi/2 h) Pi i) 2*Pi j) 3*Pi
11. Calculate int( y*dx+2*x*dy+(z-x*y+3),z=C..``). where C is the oriented curve parameterized by r(t)=[3*t,t^2,t^3], 0 <=t ``<=1 . a) 1 b) 2 c) 3 d) 4 e) 5 f) 6 g) 7 h) 8 i) 9 j) 10
12. The vector field is the gradient field of some function f(x,y). Calculate f(2,1) - 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.
13. Calculate int( (y+z)*dx+(x+1)*dy+(2*z+x),z=C..``) where C is the oriented curve parameterized by r(t)=[sqrt(t),cos(Pi*sqrt(t)),t*ln(t^2-t+1)], 0 <=t ``<=1 . a) 1 b) -1 c) 2 d) -2 e) 3 f) -3 g) 4 h) -4 i) 5 j) -5
14. What is the surface area of the part of z=1+x^2/2+y^2/2 that lies over the unit disk x^2+y^2<=1 ? a) sqrt(2)/3*Pi b) (sqrt(2)-1)/3*Pi c) (2-sqrt(2))/3*Pi d) 4*sqrt(2)/3*Pi e) 2*(2*sqrt(2)-1)*Pi/3 f) (4*(sqrt(2)-1))/3*Pi g) 2*(4-sqrt(2))/3*Pi h) (4-sqrt(2))*Pi/3 i) 2*(2-sqrt(2) )*Pi/3 j) 4*(2*sqrt(2)-1)*Pi/3
15. Calculate Int(-5*y*dx+11*x,y=C..``) where C is the unit circle traversed counter-clockwise. a) 0 b) Pi c) 2*Pi d) 3*Pi e) 4*Pi f) 6*Pi g) 8*Pi h) 12*Pi i) 16*Pi j) 18*Pi
16. 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 ( 0 , 1 , 2 ) ? a) 0 b) 1 c) 2 d) 3 e) 4 f) 5 g) 6 h) 7 i) 8 j) 9
17. Suppose that f(x,y,z)=3*x^2 - 2*y^2 + 3*z^3 . Calculate div(grad(f(x,y,z)) at the point (3,2,1). a) 1 b) 2 c) 3 d) 4 e) 5 f) 8 g) 12 h) 15 i) 16 j) 20
18. Calculate the surface integral of f(x,y,z)=2*x+z over that part of the plane 2*x+2*y+z=2 that lies in the first octant. a) 1 b) 2 c) 3 d) 4 e) 5 f) 6 g) 7 h) 8 i) 9 j) 10
19. Let U be the cube with equal sides of length one and vertices at ( 0, 0, 0 ) and ( 1, 1, 1 ). Let S be the surface of that part of the cube that lies above the xy plane. (In other words, S has the shape of a box without its bottom.) Let n denote the outward-pointing unit normal on S . Calculate (also denoted by as in our text) where a) 1 b) -1 c) 2 d) -2 e) 3 f) -3 g) 4 h) -4 i) 5 j) -5
20. Let U be the cube with equal sides of length one and vertices at ( 0, 0, 0 ) and ( 1, 1, 1 ) (as in the preceding problem). In this problem, let S be the entire surface of the cube. a) 1 b) -1 c) 2 d) -2 e) 3 f) -3 g) 4 h) -4 i) 5 j) -5

Math 233 Spring 2004 Final Exam

1. A line that is perpendicular to the plane   has symmetric equations



    What is a?

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

Solution: (b)

2. If      where   is parallel to and is perpendicular
   to , then what is b?

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

Solution: (d)

3. A curve is parameterized by . Calculate its principal unit normal vector
     at the point (1, 1/3, 1). What is    ?

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

Solution: (d)

4. Suppose that . Calculate      at    .

a) 3/8               b) 4/15             c) 2/9             d) 3/7             e) 5/13
f)   2/17             g) 2/11             h) 3/19            i) 4/17             j) 8/21

Solution: (f)

> eqn := x^2 +y*z-z^3=2*x*z-9;

> F := (x,y,z) -> lhs(eqn)-rhs(eqn);

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

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

5. The function    has exactly one critical point P = (a,b) that does not lie on a coordinate axis. Choose the ordered list [b,what] where b is the y-coordinate of the critical point P = (a,b) and "what" describes the behavior of at P.

a) [-1,local minimum]         b) [-2,local minimum]          c) [-3,local minimum]
d) [-1,local maximum]       e) [-2,local maximum]         f) [-3,local maximum]
g) [-1,saddle point]            h) [-2,saddle point]              i) [-3,saddle point]

Solution: (f)

> f := (x,y) -> 3*x^2*y - x*y^2- 9*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=-3},");

> subs({x=1,y=-3},diff(f(x,y),x$2));

> f(1,-3); f(.99,-3.1); f(.99,-2.9);

6. Let be the upward unit normal of the plane    . Calculate the directional derivative of       in the direction of    at the point ( 2, 2, 3).

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

Solution: (f)

> with(linalg):

Warning, new definition for norm

Warning, new definition for trace

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

> u := [1/3,-2/3,2/3]; v := grad(F(x,y,z),[x,y,z]);

> dp := dotprod(u,v);

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

7. Let ( a , b )   be the point in the first quadrant at which has a maximum value subject to the constraint    . Calculate a and b. What is the value of ?

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

Solution: (f)

> f := (x,y) -> x*y+1: phi := (x,y) -> x^2 + 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});

${x = RootOf(-1+2*_Z^2), y = 1/2, lambda = 1/2*RootOf(-1+2*_Z^2)}$

> c := 1/sqrt(2):

> with(plots):

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

> levelCurve[2] := implicitplot( x^2+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. Calculate the integral of    over the region in the first quadrant that is bounded by the coordinate axes and the circle    .

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

Solution: (h)

> int(int( (r*cos(t)+2*r*sin(t))*r,r=0..2),t=0..Pi/2);

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

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

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,-3,-1])=0,a);
#Tangent vector is perpendicular to Normal vector

10. Calculate


where U is the solid half-ball that is described by the inequalities      and    .
a)           b)               c)               d)               e)
f)                 g)                h)              i)              j)

Solution: (e)

> int(int(int(rho*cos(phi)*rho^2*sin(phi),rho=0..1),phi=0..Pi/2),theta=0..2*Pi);

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

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

Solution: (g)

> r := t -> [3*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(");

12. The vector field


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

> restart: with(linalg):

Warning, new definition for norm

Warning, new definition for trace

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

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

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

Solution: (f)

> r := (t) -> [sqrt(t), cos(Pi*sqrt(t)),t*ln(t^2-t+1)];

> P := r(0);

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

14. What is the surface area of the part of       that lies over the unit disk     ?

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

Solution: (e)

> 2*Pi*int(r*(1+r^2)^(1/2),r=0..1);

> answers := [sqrt(2)/3*Pi,(sqrt(2)-1)/3*Pi,(2-sqrt(2))/3*Pi,4*sqrt(2)/3*Pi,2*(2*sqrt(2)-1)*Pi/3,(4*(sqrt(2)-1))/3*Pi,2*(4-sqrt(2))/3*Pi,(4-sqrt(2))*Pi/3,2*(2-sqrt(2) )*Pi/3,4*(2*sqrt(2)-1)*Pi/3]:

> for i from 1 to 9 do
for j from i+1 to 10 do
if (answers[i]=answers[j]) then print(`repeat`);
fi;
od;
od;

> int(int(sqrt(1+r^2)*r,r=0..1),t=0..2*Pi);

> int(r*sqrt(1+r^2),r=0..1);

> evalf(Int(r*sqrt(1+r^2),r=0..1));

> testeq(diff((1+r^2)^(3/2)/3,r) = sqrt(1+r^2)*r);

> subs(r=1,(1+r^2)^(3/2)/3) - subs(r=0,(1+r^2)^(3/2)/3);

> evalf( " );

15. 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( -5*r(t)[2]*diff(r(t)[1],t) + 11*r(t)[1]*diff(r(t)[2],t),t=0..2*Pi);

> value(");

More easily by Green's Theorem:

> Int(Int( r*(diff(11*x,x)-diff(-5*y,y)),r=0..1),theta=0..2*Pi);

> value(");

16. 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 ( 0 , 1 , 2 ) ?

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

Solution: (e)

> 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},");

17. Suppose that     . Calculate div(grad(f(x,y,z)) at the point (3,2,1).

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

Solution: (j)

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

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

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

18. Calculate the surface integral of over that part of the plane
that lies in the first octant.

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

Solution: (b)

> with(linalg):

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

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

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

> subs(z=2-2*(x+y),");

> value(");

19. Let U be the cube with equal sides of length one and vertices at ( 0, 0, 0 ) and ( 1, 1, 1 ). Let S be the surface of that part of the cube that lies above the plane. (In other words, S has the shape of a box without its bottom.)   Let n denote the outward-pointing unit normal on S . Calculate


(also denoted by


as in our text)   where



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

Solution: (a)

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(2*y,y=0..1);

20. Let   U be the cube with equal sides of length one and vertices at ( 0, 0, 0 ) and ( 1, 1, 1 ) (as in the preceding problem). In this problem, let S be the entire surface of the cube.



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

Solution: (f)

Using the Divergence Theorem

> with(linalg):

> F := vector([y*z^5,12*z^3*y,y^3]);

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

> int(int(int(p,z=0..1),y=0..1),x=0..1);