Practice Uniform Final Exam MATH 2421 (Calculus III)
Kawai
!
2; 3i and the displacement vector be P Q = h7; 3;
!
How many work units will F accomplish moving an object along P Q? Dot product.
!
W ork = F P Q = h1; 2; 3i h7; 3; 1i
(#1) Let the force be (with you) F = h1;
1i :
= (1) (7) + ( 2) ( 3) + (3) ( 1) = 10 work units.
The answer is (d).
(#2) Suppose we have a force F = hcos (20 ) ; sin (20 ) ; 0i and a wrench vector r = hcos (110 ) ; sin (110 ) ; 0i
in the xy-plane. What is the torque, = r F?
Since the two vectors are in the xy-plane, the cross product must be normal to that plane.
Thus, the …rst two component are zero.
By the Right Hand Rule, the shortest angular path from r to F is clockwise (from 110 to
20 ). The cross product vector will point into the xy-plane and the k-component will be
negative.
kr
Fk = krk kFk sin ( )
= (1) (1) sin (90 ) = 1:
Since the component is negative, the correct answer is (a) h0; 0;
1i :
(#3) Suppose we have some surface z = f (x; y) and we know the …rst partials:
fx = 2xy 2
2x2 ; fy = 2x2 y
2y 2 :
Clearly by substitution, we see that (1; 1) is a critical point. Run the Second Partials Test
on it and circle the correct result below.
We have
fxx = 2y 2
fxx =
4x; fxy = 4xy; fyy = 2x2
2; fxy = 4; fyy =
4y:
2:
We note that fyx = 4xy and we see that the equality of mixed partials. The discriminant is
d = fxx fyy
(fxy )2 = ( 2) ( 2)
42 =
12 < 0:
The correct answer is (c) saddle point.
(#4) Evaluate the unit tangent vector if r (t) = cos2 (t) ; sin2 (t) ; 0
t
=2; then:
v (t) = h 2 cos (t) sin (t) ; 2 sin (t) cos (t)i = 2 cos (t) sin (t) h 1; 1i :
The factors in the front are nonnegative since 0 t
=2: Thus, when we take the norm,
we have
kv (t)k = 2 cos (t) sin (t) kh 1; 1ik :
It cancels the factor in numerator.
T (t) =
h 1; 1i
1
= p h 1; 1i :
kh 1; 1ik
2
The correct answer is (c).
1
(#5) Which 3D line below is NORMAL to the plane x
2y
z = 4?
The coe¢ cients lead you to the normal vector, n = h1;
2;
The line in choice (b) has direction vector equal to h2;
4;
1i :
If the 3D line is normal to this plane, then the direction vector must be parallel to n:
2i ; which is two times n:
(#6) Suppose w = f (x; y) is a two-variable function. Here are the dependencies:
x = x (s; t)
y = y (t; )
t = t( )
Thus, w is a function of s and : Which formula below gives us
@w
?
@
If like dependency trees, then here it is:
f
x
. #
s
t
#
There are THREE paths of change when
.
&
y
# &
t
#
changes.
The correct answer is (b):
@w
@f @x @t
@f @y @t
@f @y
=
+
+
@
@x @t @
@y @t @
@y @
(#7) Find the standard equation of the 3D plane which contains the points P (0; 0; 1) ; Q (1; 1; 2) ;
and R (2; 5; 4) :
!
!
We typically choose P Q and P R to create the normal to the plane n:
!
!
P Q = h1; 1; 1i ; P R = h2; 5;
n = h1; 1; 1i
h2; 5;
5i
5i = h 10; 7; 3i :
If we use P as the base point, then the standard form is:
10 (x
0) + 7 (y
0) + 3 (z
and the general form is:
10x + 7y + 3z
2
3 = 0:
1) = 0
If we use the other two points as the base points, then we have:
10 (x
10 (x
1) + 7 (y
2) + 7 (y
1) + 3 (z
5) + 3 (z
2) = 0 and
( 4)) = 0:
These are equivalent to the general form.
(#8) We have diagrams for some solids. The diagrams follow the usual display conventions. The
positive x-axis comes out of the paper, and the positive y-axis points to the right. The origin
is basically located at the center of the display cube.
Write your choice next to the clues given.
(i) D:
x2 y 2 z 2
+
+
4
4
9
(ii) F: 1
3;
1: Solid ellipsoid.
=2
(iii) C: 0
=4;
(iv) E: x2 + y 2
z
: Lower canteloupe.
4: Cone below, sphere above.
4: Volume above elliptic paraboloid. Flat top.
(#9) Find the net UPWARD ‡ux through S: S is the portion of the hyperbolic paraboloid
z = f (x; y) = x2
y 2 above the square 0
x
1; 0
y
1:
The vector …eld is F (x; y; z) = hM; N; P i = hx; y; xy + 2zi :
We need these:
fx = 2x; fy =
2y:
The region R is the square and the net upward ‡ux integral is:
ZZ
( M f N g + P ) dA:
R
The integrand is
(x) (2x)
2x2 + 2y 2 + xy + 2z
(y) ( 2y) + xy + 2z =
and we must have an integrand with only x’s and y’s. Substitute z = x2
2x2 + 2y 2 + xy + 2 x2
y2:
y 2 = xy:
The double integral is:
Z
0
1Z 1
0
xy dy dx =
Z
0
1
y dy
Z
1
x dx
0
3
=
1
2
1
2
=
1
upward ‡ux units.
4
(#10) Find the net OUTWARD ‡ux through S: S is the simple closed surface boundary of the
solid G which is bounded by the elliptic paraboloid z = x2 + y 2 and the plane z = 1:
ez ; x2 y + ez ; xy :
The vector …eld is F (x; y; z) = hM; N; P i = xy 2
Divergence Theorem. We must integrate r F throughout the volume G: We need cylindrical
coordinates.
The volume looks like this:
dz r dr d
lower ! upper surf ace
z: r2 ! 1
r: 0 ! 1
: 0!2 :
We smashed this down onto the xy-plane.
It’s a circle of radius 1.
The divergence is:
r F = Mx + Ny + Pz = y 2 + x2 + 0 = r2 :
The net outward ‡ux integral is:
Z 2 Z 1Z 1
0
r2 dz r dr d =
r2
0
Inner:
Z
1
r2
Next:
Z
1
r
3
1
r
2
6
dz = [z]1r2 = 1
dr =
0
Z
1
r3
outward ‡ux units.
r2
r5 dr =
0
1
4
1
1
=
6
12
The Outer integral multiplies the previous result by 2 :
(#11) We want to …nd the (circulation) value of the closed line integral:
I
(F dr) =???
C
If C is simple closed boundary of R: The region R is bounded by the line y = x; the circle
x2 + y 2 = 2 on the right, and the x-axis on the bottom.
The intersection point is clearly (1; 1) by the graph. The algebra:
y = x ) x2 + x2 = 2 ) x =
4
1:
(a) Using rectangular coordinates, we have no choice but to go horizontal …rst:
dx dy
lef t ! right
p
x: y ! 2 y 2
y: 0 ! 1:
(b) Suppose the vector …eld is F (x; y) =
x2 y; xy 2 : Use Green’s Theorem and a POLAR
double integral to …nd the circulation around C; once around counterclockwise.
This …eld is non-conservative.
My = y 2
Nx
x2 = x2 + y 2 = r2 :
Yes, this looks familiar from (#10), but, at least, the double integral is slightly di¤erent.
We integrate (Nx My ) over the interior of the C:
r dr d
p h
p 2i
r: 0 ! 2 x2 + y 2 =
2
: 0 ! =4:
Net Circulation =
Z
0
Inner:
Z
p
2
0
=4 Z
p
2
r2 r dr d =
0
r4
r3 dr =
4
p
4
:
2
= 1:
0
The Outer integral multiplies the previous result by =4:
(#12) Consider the scalar function f (x; y) = x2
y2:
(a) On the axes below, sketch in the level curves for k = 1: The curve is a hyperbola.
y
3
2
1
(b) rf (x; y) = hfx ; fy i = h2x;
rf (2; 1) = h4; 2i :
0
-3
-2
-1
0
1
2
3
x
2yi
-1
-2
-3
(c) Evaluate the directional derivative: Du f (2; 1)
when u is the unit direction vector which has the same direction as h3; 4i :
The unit vector is
u =
Du f (2;
p
1
1
h3; 4i = h3; 4i :
2
5
+4
32
1) = rf (2;
1) u = h4; 2i
5
1
1
h3; 4i = (12 + 8) = 4
5
5
(#13) We already know that F is conservative in Quadrant I where x and y are both positive.
F (x; y) =
x2
x
y
; 2
2
+ y x + y2
SIMPLIFY the value of (independent of path in the region stated above)
Z (3; 4)
F dr =???;
(1; 1)
for any simple path starting at (1; 1) and ending at (3; 4) :
The potential function is
' (x; y) =
=
Z
Z
M dx =
N dy =
Z
Z
x2
x
1
dx = ln x2 + y 2 +
2
+y
2
(y)
x2
1
y
dx = ln x2 + y 2 +
2
+y
2
(x) :
1
ln x2 + y 2 + C:
2
The entire family of functions is ' (x; y) =
They match, so we can go straight to the Fundamental Thm. of Line Integration.
Z (3; 4)
1
(3; 4)
F dr = ' (3; 4) ' (1; 1) =
ln x2 + y 2 (1; 1)
2
(1; 1)
=
1
ln 32 + 42
2
ln 12 + 12
1
ln
2
If we combine the logarithms, then we have
25
2
1
(ln (25)
2
=
ln (2))
:
(#14) Find the mass of the UPPER solid sphere
x2 + y 2 + z 2
if the density function is
x2
1; z
0
1
1
= 2:
2
2
+y +z
Spherical coordinates:
2 sin (
) d d d
: 0!1
: 0 ! =2 [upper half]
: 0!2 :
Z 2 Z
0
0
=2 Z 1
0
1
2
2
sin ( ) d d d
=
Z
2
0
=
Z
0
Z
0
1
d
=2 Z 1
sin ( ) d d d
0
Z
=2
0
= (1) (1) (2 ) = 2 :
6
sin ( ) d
! Z
0
2
d
(#15) EVALUATE the arc length. Suppose we have the parametric curve associated with
r (t) = ht sin (t) + cos (t) ; t cos (t)
sin (t)i ; 0
t
:
The arc length di¤erential is:
q
q
ds =
(x0 (t))2 + (y 0 (t))2 dt = (t cos (t))2 + (t sin (t))2 dt
q
p
=
t2 cos2 (t) + sin2 (t) dt = t2 dt = jtj dt:
Since t is nonnegative, we can drop the absolute value bars.
Z
s=
2
t dt =
0
2
units.
(#16) Projectile vomiting.
16t2 + v0 sin ( ) t + y0
r (t) = v0 cos ( ) t;
feet.
A poisonous Tuna Sandwich is thrown from the point (0; y0 ) = (0; 6) feet.
Its initial velocity is v0 = 32 ft/sec and the angle of elevation is
= 30 :
At what value of t (seconds) does the sandwich attain its highest altitude?
We want the time when y 0 = 0:
32t + v0 sin ( ) = 0 ) t =
v0 sin ( )
32 sin (30 )
1
=
= second.
32
32
2
Now plug this back into y:
y
1
2
1
2
2
=
16
=
4 + 8 + 6 = 10 ft.
(#17) Implicit. We have a surface x2 z
cos (xy) +
Find an expression for the implicit derivative
+ 32 sin (30 )
1
2
+6
y
= 1:
z
@z
:
@x
We create the “parent” function by moving everything to the left side.
F
@z
@x
= x2 z
=
=
cos (xy) +
Fx
Fz
=
y
z
1
0
1
@ 2xz + y sin (xy) A =
y
x2
z2
(2xz + y sin (xy)) z 2
x2 z 2 y
7
(2xz + y sin (xy)) z 2
y
x2
z2
z2