P2_soln - 


Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 
 Prelim
II
–
Solutions
–
Spring
2009
–
Math
1920
 
 Comments:
The
solution
to
problem
3
is
directly
from
the
homework
solutions
for
16.2.11
 
 
 
 
 
 Problem
1
 
 (a) –
The
normal
to
the
hyperboloid
is
found
by
taking
the
gradient
of
G(x,y,z)
 
 ∇G = ∇( 4 x 2 + 4 y 2 − 4 − z 2 ) =< 8 x, 8 y, −2 z > 
 
 
 At
the
point
P(1,2,4)
,
 ∇G =< 8,16, −8 > 
 
 € The
equation
of
the
plane
is
 ∇G ⋅ ( P − P0 ) = 0 
 
 8( x − 1) + 16( y − 2) − 8( z − 4 ) = 0 
 So,

 € 
 € (b) –
The
direction
of
max
increase
in
z
at
the
point
P(1,2,4)
is
found
by
taking
grad(z).
 
 From
our
equation
and
the
point,
we
find
that

 €
 z = 4x + 4y − 4 
 
 
 Therefore,
 
 € ∇z = 
 2 2 2 4x 4x + 4y − 4 = 1, 2 P (1, 2, 4 ) 2 , 4y 4x + 4y2 − 4 2 
 ∇z 
 (c) –
The
change
in
z
at
point
P(1,2,4)
is
found
by
taking
the
directional
derivative.
 
 € 
 ˆ Du z = ∇z ⋅ u = 1, 2 ⋅ 0 ,1 = 2 
 Problem 16.1.26 √ v =< 1, 2, t > M= 2 0 ⇒ |v| = √ 5+t 2 0 √ √ 3 3 5 + t 5 + t dt = 15t + 2 t2 2 0 2 0 2 0 = 36 Myz = Mxz = Mxy = x= ¯ y= ¯ z= ¯ Myz M Mxz M Mxy M 3t(5 + t) dt = 6t(5 + t) dt = 2 15 2 2t + t3 2 0 = 38 2 0 15 2 2t + t3 = 76 2 0 2 2t3/2 (5 + t) dt = 2 2t5/2 + 7 t7/2 38 36 76 36 144 7 = 144 7 √ 2 = = = = = √ 2 19 18 19 9 36 = √ 42 7 Problem 16.2.11 a) C : r =< t, t, t > W= 1 0 ⇒ v =< 1, 1, 1 > 1 0 F =< 3t2 − 3t, 3t, 1 > 1 0 F · vdt = 3t2 + 1 dt = t3 + t =2 F =< 3t2 − 3t, 3t4 , 1 > 1 0 b) C : r =< t, t2 , t4 > W= 1 0 ⇒ v =< 1, 2t, 4t3 > F · vdt = 1 0 3 3t2 − 3t + 6t5 + 4t3 dt = t3 − 2 t2 + t6 + t4 = 3 2 c) C1 : r1 =< t, t, 0 > C2 : r2 =< 0, 0, t > W= 1 0 ⇒ v1 =< 1, 1, 0 > ⇒ v2 =< 0, 0, 1 > 1 0 F1 =< 3t2 − 3t, 0, 1 > F2 =< 0, 3t, 1 > 1 0 [F1 · v1 + F2 · v2 ] dt = 3t2 − 3t + 1 dt = t3 − 3 t2 + t 2 = 1 2 Problem 16.2.15 C : r =< sint, cost, t > W= 2π 0 ⇒ v =< cost, −sint, 1 > t cost − sin2 t + cost dt F =< t, sint, cost > F · v dt = 2π 0 The integral of first and third terms will equal 0 since we’re integrating over a full period. W= 2π 0 −sin2 t dt = − 2π 1−cos(2t) 2 0 t dt = − t 2 + sin2t 2π 4 0 = −π Problem 16.2.19 C : r =< t2 , t > W= −1 2 ⇒ v =< 2t, 1 > −1 2 F =< t4 , −t > 16 3t F · vdt = 2t5 − t dt = − 1 t2 2 −1 2 1 = −6 − 128 6 + 12 6 = −117 6 = − 39 2 3 Problem
5
 
 
 
 Part
(a):
 
 (Shown
on
Graph)
 The
gradient
points
in
the
direction
of
fastest
increase.
 It
is
orthogonal
to
the
level
curve
at
(2,3)
 
 Part
(b):
 
 Moving
one
unit
in
the
direction
of
the
gradient
takes
us
from
the
5
contour
to
the
7
contour,
so
the
change
 in
z
is
about
2.
 
 Part
(c):
 
 (See
graph)
 The
height
of
the
sketch
is
determined
by
the
contours
along
the
line
y=1
in
the
plane.
 
 Part
(d):
 
 The
two
constraints
are
the
parabola
and
disk
shown
on
the
graph.
 The
maximum
subject
to
both
constraints
occurs
at
approximately
(0,4)
and
has
a
z
value
of
about
6.8.
 
 Part
(e):
 
 The
three
minimums
and
two
saddle
points
are
marked
on
the
graph.
 
 Problem 6 a) Set up a triple integral in spherical coordinates for the mass above z = 3x2 + 3y 2 and below z = 4. The density is δ = x2 + y 2 . Solution The density is x2 + y 2 = r2 = (ρ sin(φ))2 = ρ2 sin2 (φ). The object is symmetric around the z -axis and therefore we have θ going from 0 to 2π . z = 3x2 + 3y 2 √ represents a cone with slope 3 and simple trigonometry gives that our φ should 1 go from 0 to arctan √3 = π/6. Our ρ should go from zero to z = 4. We have z = ρ cos φ so z = 4 ⇒ ρ cos φ = 4 ⇒ ρ = 4/ cos φ. The integral is therefore 2π 0 0 2π 0 0 π /6 0 π /6 0 4/ cos φ 4/ cos φ ρ2 sin2 (φ)ρ2 sin(φ)dρdφdθ = ρ4 sin3 (φ)dρdφdθ b) Describe the object whose mass is 2π 0 π /2 π/4 1 2/sinφ θρ sin φρdρdφdθ Solution Mass integrals in spherical coordinates are of the form δ ρ2 sin(φ)dρdφdθ θ This gives us that the density is ρ We have the θ bound going from 0 to 2π so a whole circle around the z -axis and since the other bounds don’t have θ in them we have that the object is symmetric around the z -axis. The lower bound for φ is π/4 so we get that our object is below the cone z = r. The upper bound for φ is π/2 so our object lies above the xy -plane. The lower bound for ρ is 1 so our object lies outside the sphere of radius 1 and the upper bound for ρ is 2/sinφ so our object lies inside ρ = 2/sinφ ⇒ ρ sin φ = 2 ⇒ r = 2 or inside the cylinder of radius 2. 1 #7 Solution Boundary (13 pt.) Lagrange multiplier: 2x + y2 = 2 λ x λ λ 2y = 0 & . 2 2xy + 2y = 4 λ y => 2y (x + 1) = 2y (2 λ ) => 2y = 0 => y = 0 => 2x + (0)2 = 2 λ x So, λ =1 and x = . => ( => f ( 2 2 x + 1 = 2λ x + 1 = 2 λ => 2x + y = (x+1) x = x + x => y = x2 – x Apply y2 = x2 – x to g = x2 + 2y2 = 1. So, x2 + 2(x2 – x) = 1. 3x2 – 2x – 1 = 0 => (x – 1) (3x + 1) = 0 => x = 1 and x = . When x = Since f ( Interior (7 pt.) Let & . 2y = 0 & x+1=0 => 2y (x + 1) = 0 => ,y= => f ( f( . , maximum value is 1. . 2y = 0 => y = 0 => 2x + 02 = 0 => x = 0. => (0, 0) => f (0, 0) = 0. x + 1 = 0 => x = - 1 => 2 (- 1) + y2 = 0 => y = => (- 1, ) = > f ( - 1, ) = 1. Since the points ) are NOT in the elliptic region, our real maximum value is 1 at ( . ...
View Full Document

This note was uploaded on 12/25/2009 for the course MATH 1920 taught by Professor Pantano during the Spring '06 term at Cornell University (Engineering School).

Ask a homework question - tutors are online