Solutions to Uniform Final Exam for Calculus III –MATH 2421 Fall 2007
(#1) Suppose F and G are vector …elds and f and g are scalar functions. Assume all partial derivatives exist. Which one of these expressions is an illegal operation in vector calculus? For example, we can only dot product two vector -valued functions.
Since r F is a scalar function, we cannot dot it with the “del” operator. We cannot …nd the divergence of a scalar function.
The correct answer is (c) r (r F) :
(#2) Suppose a force vector has components F = h 3; 4i pounds and this force moves an object 20 feet in a straight line. If the angle of separation between the force vector and the displacement vector is 60 ; then how much work does the force accomplish?
W ork = F P Q = kFk! P Q! cos ( ) = q ( 3)2+ ( 4)2 pounds (20 ft) cos (60 ) = (5 lbs) (20 ft) 1 2 = 50 lb-ft = 50 work units. The correct answer is (b).
(#3) Suppose z = f (x; y) and x = x (s; t) and y = y (s; u) : Assume all partial derivatives exist. Which expression below is equivalent to @z
@s? There are TWO paths from f down to s: f
. &
x y
. # # &
t s s u
The correct answer is (b).
@z @s = @f @x @x @s + @f @y @y @s:
(#4) We have diagrams for 4 surfaces and 2 solids. The diagrams follow the usual display conven-tions. 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.
(iii) = 3 =4: This is a right circular cone. Since > =2; the cone is located below the xy-plane. cos ( ) = cos 3 4 ) cos ( ) = p 2 2 cos ( ) = p 2 2 ) z = p 2 2 :
Since must be nonnegative, the surface must reside below the xy-plane (except at the origin). Now square both sides.
z2 = 1 2 2 = 1 2 x 2+ y2+ z2 1 2z 2 = 1 2 x 2+ y2 ) z2 = x2+ y2: This gives us both the upper and lower right circular cones,
z = px2+ y2 = r:
We already know that we must choose the lower cone. (ii) z 4 = r 1 x 2 9 y2
9 : The is the upper half of the ellipsoid x2 9 + y2 9 + z2 16 = 1: (i) z = y2 x2: Hyperbolic paraboloid (saddle shape).
(vi) 2 r 4; 0 z 4: This is the solid region between the right circular cylinders x2+ y2 = 4 and x2+ y2 = 16:
(#5) Let F = y + z; x + z; z3 : The simple closed surface S bounds the solid upper hemisphere Q:
x2+ y2+ z2 1; z 0:
(a) Use the appropriate vector operation formula and show that F is NOT conservative. We must show that the curl r Fis not the zero vector.
r F = hPy Nz; Mz Px; Nx Myi = @ @y z 3 @ @z [x + z] ; @ @z[y + z] @ @x z 3 ; @ @x[x + z] @ @y[y + z] = h0 1; 1 0; 1 1i = h 1; 1; 0i :
(b) Now use the OTHER vector operation and …nd the net OUTWARD ‡ux through S: I
S
(F n) dS;
where n is the unit outward normal on S: Hint: The solid is a part of a SPHERE. This is the upper half of the unit solid hemisphere.
2sin ( ) d d d
: 0 ! 1 : 0 ! =2 : 0 ! 2 :
The Divergence Theorem gives us the net outward ‡ux provided that (r F) is well de…ned on the interior of the simple closed surface S:
I
S
(F n) dS = ZZZ
(r F) dV
Here’s the divergence.
r F = @x@ [y + z] + @
@y[x + z] + @ @z z
3 = 3z2:
In spherical coordinates, this is
3z2= 3 ( cos ( ))2= 3 2cos2( ) : Thus, the triple integral is
N et Outward F lux = Z 2 0 Z =2 0 Z 1 0 3 2cos2( ) 2sin ( ) d d d = 2 5 : This factors into:
Z 1 0 3 4d Z =2 0 cos2( ) sin ( ) d ! Z 2 0 d = 3 5 1 3 (2 ) = 2 5 :X The middle integral is a u-substitution:
u = cos ( ) ; du = sin ( ) d ) sin ( ) d = du: Z =2
0
cos2( ) sin ( ) d = cos
3( ) 3 =2 0 = 0 3 3 13 3 = 1 3: (#6) Let f (x; y) = x y2:
(a) On the axes below, sketch in the two level curves f (x; y) = k for k = 1 AND k = 2: CLEARLY label each graph with its associated value of k:
4 3 2 1 0 - 1 - 2 - 3 - 4 4 3 2 1 0 - 1 - 2 - 3 - 4 x y x y When k = 1; we have x y2 = 1:
x = y2+ 1: This parabola opens out to the right. When k = 2; we have x y2 = 2
x = y2+ 2: We shift the previous graph to the right by one unit.
(b) At the point P (3; 1) ; evaluate the gradient of f; and then sketch that vector at P (3; 1) : Show your work here. Here is the gradient:
rf = hfx; fyi = h1; 2yi
At the point P; we have y = 1; so the gradient vector anchored there is h1; 2i : We see that the gradient vector is normal to the level curve. It points in the direction of the next higher level curve.
(c) Let u be the unit vector which has the same direction as h 1; 1i : The unit vector is
u= qh 1; 1i ( 1)2+ 12
= p1
2h 1; 1i : The directional derivative is
Duf (3; 1) = rf (3; 1) u = h1; 2i 1 p 2h 1; 1i = p1 2(h1; 2i h 1; 1i) = 1 p 2( 1 + ( 2)) = 3 p 2 = 3p2 2 : We expect to lose altitude in this direction.
(#7) Suppose we have projectile motion using English units. The projectile is launched from the origin, the initial speed is v0= 4 ft/sec and angle of elevation is = 30 :
r(t) = v0cos ( ) t; 16t2+ v0sin ( ) t ; t 0:
(a) Find the velocity vector when t = 1
32 seconds. v(t) = hv0cos ( ) ; 32t + v0sin ( )i = h4 cos (30 ) ; 32t + 4 sin (30 )i = D2p3; 32t + 2E We evaluate this at t = 1 32: v 1 32 = 2 p 3; 32 1 32 + 2 = D 2p3; 1 E :
(b) Find the unit tangent vector when t = 1
32 seconds. T 1 32 = v 1 32 v 1 32 = 2 p 3; 1 q 2p3 2+ 12 = p1 13 D 2p3; 1 E :
(c) Since you know the trajectory is parabola, you can now determine the correct direction for the PRINCIPAL unit normal vector when t = 1=32 seconds: N(1=32) = h???; ???i : Write its components here.
We recall that the trajectory must be concave downward, so the principal unit normal vector must point into the curve (downward).
Thus, its vertical component must be negative.
Since N is always orthogonal to T; the only choice for N are: 1 p 13 D 1; 2p3E or p1 13 D 1; 2p3E
[Swap the 2D components and then negate exactly one of them.] The second one is the only reasonable choice.
N 1 32 = 1 p 13 D 1; 2p3E (#8) Here are the parametric equations associated with a 3D line:
x = 2t 5 y = t + 2 z = t + 1
Give the STANDARD form for the plane which is perpendicular to this line at the point P ( 5; 2; 1) :
The direction vector associated with the 3D line can be obtained by reading o¤ the coe¢ cients of t:
This vector will serve as the normal vector to our plane. n= h2; 1; 1i : The standard form for the plane is
2 (x ( 5)) 1 (y 2) + 1 (z 1) = 0 2 (x + 5) 1 (y 2) + 1 (z 1) = 0: (#9) Let f (x; y) = 1
x2+ y2+ 1: We will tell you that the only critical point is the ORIGIN.
Carefully …nd fxx: By symmetry, you can just write down fyy:
fx = @ @x h x2+ y2+ 1 1 i = 1 x2+ y2+ 1 2 @ @x x 2+ y2+ 1 = 2 x (x2+ y2+ 1)2 :
The second partial requires the Quotient Rule.
fxx = 2 0 B @ x2+ y2+ 1 2 @ @x[x] x @ @x h x2+ y2+ 1 2i (x2+ y2+ 1)2 2 1 C A = 2 x 2+ y2+ 1 2 x 2 x2+ y2+ 1 (2x) (x2+ y2+ 1)4 ! = 2 x 2+ y2+ 1 (x2+ y2+ 1)4 ! x2+ y2+ 1 4x2 = 2 3x 2+ y2+ 1 (x2+ y2+ 1)3 By symmetry, we have fyy = 2 3y2+ x2+ 1 (x2+ y2+ 1)3
Carefully …nd fxy: fxy = @ @y[fx] = @ @y 2 x (x2+ y2+ 1)2 = 2x @ @y h x2+ y2+ 1 2i = 2x ( 2) x2+ y2+ 1 3(2y) = 8xy (x2+ y2+ 1)3:
Run the Second Partials Test and classify the critical point. We evaluate all the second partials at (0; 0) : d = fxxfyy (fxy)2 = ( 2) 1 13 ( 2) 1 13 0 2 = 4 > 0:
Both fxx and fyy are negative. Thus, this critical point must be a local maximum.
(#10) Hope you like fractions. Suppose we have a non-conservative ‡uid velocity …eld F(x; y) = y2; 2x :
(a) We have a short path C1 from (0; 0) to (1; 1) along the square root function y =
p
x: Evaluate the circulation along this piece. The parameterization is r (t) = t; pt ; 0 t 1: v(t) = 1; 1 2pt Z C1 F dr = Z 1 0 p t 2; 2t 1; 1 2pt dt = Z 1 0 t + t1=2 dt = t 2 2 + 2 3t 3=2 1 0 = 1 2 + 2 3 = 7 6: (b) Now suppose the C1 from above is a part of this simple closed path:
2 1 0 2 1 0 x y x y
The path starts at (0; 0) ; goes to (1; 1) ; then goes to (0; 2) ; and then returns to the origin, once around counterclockwise! So obviously, for Green’s Theorem, you will need to …nd the equation of the line which runs from (0; 2) to (1; 1) :
We will tell you that the circulation along the other two pieces is ( 4=3) circulation units.
Perform Green’s Theorem on the interior of this simple closed path. With the information above, you can now check to see if both answers are consistent with each other. Good luck!
For the interior region, we have dy dx lower ! upper y: px ! 2 x x: 0 ! 1: Evaluate I C
F dr; once around counterclockwise using Green’s Theorem. The circulation density is
Nx My = @ @x[2x] @ @y y 2 = 2 2y:
This is the integrand for the Green’s Theorem double integral. N et Circulation = Z 1 0 Z 2 x p x (2 2y) dy dx = 1 6: Inner: Z 2 x px (2 2y) dy = 2y y 2 2 xp x = 2 (2 x) (2 x) 2 2px px 2 = 4 2x (2 x)2 2x1=2+ x = 4 x (2 x)2 2x1=2: Outer: Z 1 0 4 x (2 x)2 2x1=2 dx = " 4x x 2 2 + (2 x)3 3 2 2 3x 3=2 #1 0 = 4 1 2+ 1 3 4 3 0 0 + 8 3 0 = 1 6: This agrees with the information presented in the problem. The total circulation (from the …rst part) is 7 6+ 4 3 = 1 6:X (#11) Evaluate the surface integral (a mass integral)
m = ZZ
S
(x; y; z) dS;
if the surface S is the elliptic paraboloid z = x2+ y2; over the circular region R: x2+ y2 1: The density function is (x; y; z) =p1 + 4x2+ 4y2=p1 + 4r2:
The surface area di¤erential happens to match the density function expression. dS = q 1 + (fx)2+ (fy)2dA = q 1 + (2x)2+ (2y)2dA = p1 + 4x2+ 4y2dA =p1 + 4r2 r dr d :
The circular disk R is: r dr d
r: 0 ! 1 : 0 ! 2
The mass integral is m = Z 2 0 Z 1 0 p 1 + 4r2p1 + 4r2 r dr d = Z 1 0 1 + 4r2 r dr Z 2 0 d = r 2 2 + r 4 1 0 (2 ) = 3 2 (2 ) = 3 mass units.
(#12) Use Calculus to estimate the change in z if x moves from 2.0 to 2.2, and y moves from
4 to 4 0:1 radians.
z = f (x; y) = cos (xy) We have dx = +0:2 and dy = 0:1:
dz = fxdx + fydy = y sin (xy) dx x sin (xy) dy
= 4sin 2 4 (+0:2) 2 sin 2 4 ( 0:1) = 4(0:2) + 0:2 = 20 + 1 5 = 4 20 : It should be a small positive number.
z = cos 2:2
4 0:1 cos 2 4
:
= +0:063:
(#13) The formula for the implicit derivative is @y @x =
Fx
Fy
where F (x; y) = 0: Suppose we have the surface xyz eyz+ x3 = 7: F = xyz eyz+ x3 7 Find the implicit derivative @z
@y: We evaluate @z @y = Fy Fz = xz ze yz xy yeyz :
(#14) The region of integration R (planar lamina) is the portion of the circle in Quadrant I whose center is located at (0; 1) with radius 1.
If the density function is (x; y) = 3 8
p
x2+ y2; …nd the mass of the planar lamina.
In polar, the region is: r dr d r: 0 ! 2 sin ( ) : 0 ! =2: m = 3 8 Z =2 0 Z 2 sin( ) 0 r r dr d = 2 3 mass units. Inner: Z 2 sin( ) 0 r2dr = 8 sin 3( ) 3 Outer: 3 8 8 3 Z =2 0 sin3( ) d = Z =2 0 1 cos2( ) sin ( ) d = Z =2 0
sin ( ) cos2( ) sin ( ) d
= cos ( ) + cos 3( ) 3 =2 0 = (0 + 0) 1 +1 3 = 2 3:X
(#15) Reverse the order of integration, but DO NOT EVALUATE the double integral. Z 1
0
Z e ey
f (x; y) dx dy =??? We have the horizontal …rst order.
dx dy lef t ! right x: ey ! e y: 0 ! 1 x = ey ) y = ln (x) : dy dx lower ! upper y: 0 ! ln (x) x: 1 ! e
(#16) The curvature formula is = kr0(t) r00(t)k kr0(t)k3 : If r (t) = t 4 4; t; t
2 ; …nd the curvature when t = 1:
r0(t) = t3; 1; 2t ) r0(1) = h1; 1; 2i r00(t) = 3t2; 0; 2 ) r00(1) = h3; 0; 2i Here is the cross product:
r0(1) r00(1) = i j k 1 1 2 3 0 2 = i 1 2 0 2 j 1 2 3 2 + k 1 1 3 0 = i ( 2 0) j(2 6) + k (0 ( 3)) = h 2; 4; 3i : Thus, the curvature there is
= kh 2; 4; 3ik kh1; 1; 2ik3 = q ( 2)2+ 42+ 32 q 12+ ( 1)2+ 22 3 = p 29 6p6 = p 174 36 :
