Brief Sketches of Quiz Solutions
Quizzes
Week 1
No quiz.
Week 2
- Let u = A-B = (0,1,-1), v = C-B = (-1,1,0),
and use the fact that cos(theta) = (u dot v)/|u|*|v| = 1/2.
You get that theta = Pi/3.
- By a computation similar to that in the first problem,
show that the angle between u = Q-P and v = R-P
is zero radians. It follows that the points are on a line.
- (a+b)X(a-b) = (aXa) + (bXa) - (aXb) - (bXb) =
0 + (bXa) + (bXa) + 0 = 2(bXa). Since bXa is perpendicular
to both a and b, and (a+b)X(a-b) = 2(bXa) is a scalar
multiple of that vector, it
follows that (a+b)X(a-b) is perpendicular ot both a and b.
- The points A, B, C and
D lie in a plane if and only if the vectors
u = B-A = (2, 7, 3), v = C-A = (1, 3, 0),
w = D-A = (3, 11, 6) determine a parallelepiped
of volume zero. The determinant whose rows are
u, v and w is indeed zero, so the four
points lie on a plane.
Week 3
- The lines intersect at t = -2. The point of intersection
is (-1,1,5).
- v(t) = 12i + (10*cos(2t))j + (10*sin(2t))k,
and a(t) = (-20*sin(2t))j + (20*cos(2t))k,
- The length is (2/3)*(2^(3/2) - 1).
- The curvature is 1.
Week 4
- As the hint suggests, first check that
v(0) = (1,0,0)
is perpendicular to a(0) = (0,2,0). Therefore
all acceleration is normal to the direction of motion,
and this forces a_T = 0. Also, since
a(0) = a_T*T + a_N*N = a_N*N,
it must be that a_N = |a(0)| = 2.
- x^2+y^2-z^2=1, is an example.
- (square_root(2), Pi/4, Pi/4)
- One could sketch the hyperbolas
x^2-y^2=0, x^2-y^2=1, and x^2-y^2=2. (First hyperbola is degenerate,
i.e. the union of two crossed lines.)
Week 5
- (An odd HW problem) z = 19 + 27(x-3) - 12(y-2).
- You must find (x,y) which minimizes the function
f(x,y) = (x-1)^2 + (y-1)^2 +
(x-2)^2 + (y-3)^2 + (x-4)^2 + (y-1)^2. Setting the x and y partials
equal to zero yields that (x,y) = (7/3,5/3).
- f(x,y) is approximately
f(0,Pi/2) + f_x(0,Pi/2)*(.1) + f_y(0,Pi/2)*(-.07) = 1.1.
Week 6
Week 7
Week 8
- Deduce from grad(F) = lambda*grad(G) that x or y must be zero. Thus
extrema must occur only at (0,1), (0,-1), (square_root(1/2),0),
(square_root(1/2),0). The first two are maxima and the
last two are minima.
- grad(F) = (1 - 1/x, 1 - 1/y), so (1,1) is the only critical
point. The Hessian is the identity matrix, so this critical
point is a minimum.
- Integrate from -2 to -1
the integral from -square_root(4-x^2) to +square_root(4-x^2)
of F(x,y). (This is the first one. There are
three more.)
- By interchanging the order of integration, you
find that the integral can be evaluated, and its value is 2.
Week 9
Week 10
- 1/24
- 3a/4. (Note: This problem has nothing to do with the centroid.
Do not compute x-bar, y-bar or z-bar!)
Week 11
Week 12
Week 13
- div(F) = yz^2+2xyz+x^2y, curl(F) = (x^2z-xy^2, 0, y^2z-xz^2)
- 3*Pi
- f(x,y) = sin(x) + x*ln(y) + e^y (+K). The integral is zero because the field is conservative.
Week 14
Week 15
Week 16
No quiz.
Last modified on Aug 23, 2002