gilbert HW09 (55035)
This print-out should have 18 questions.
Multiple-choice questions may continue
on the next column or page find all
choices before answering.
001
10.0 points
Find an equation for the plane passing
through the origin that is parallel to the
tan-gent plane to the graph of
2
2
z = f (x, y) = 4x 2y 3x + y
at the point (1, 1, f (1, 1)).
1
keywords:
002
10.0 points
Find an equation for the tangent plane
to the graph of
p
2
2
f (x, y) =
5 x + 3y
at the point P (2, 1, f (2, 1)).
1. 3x + 2y 2z 5 = 0
1. z + 5x + 3y 8 = 0
2. 2x 3y + 2z 5 = 0 correct
2. z 5x + 3y = 0 correct
3 . 2 x 3y 2z + 3 = 0
3 . z 5 x 3y = 0
4. 3x 2y + 2z + 3 = 0
4 . z + 5x 3y 2 = 0
5. 2x + 3y + 2z 11 = 0
5. z + 5x + 3y = 0
6. 3x + 2y 2z 11 = 0
6 . z 5 x 3y + 8 = 0
Explanation:
Parallel planes have parallel normals.
On the other hand, the tangent plane to
the graph of z = f (x, y) at the point (a, b,
f (a, b)) has normal
n = hfx(a, b), fy(a, b), 1 O .
Explanation:
The equation of the tangent plane to
the graph of z = f (x, y) at the point
P (a, b, f (a, b))
is given by
z
But when
2
2
f (x, y) = 4x 2y 3x + y
= f (a, b)
+
f
x
(a, b)(x a) +
f
we see that
fx = 8x 3 ,
fy = 4y + 1 ,
z 5x + 3y = 0 .
(a, b)(y b) .
f (x, y) =
we see that
Thus an equation for the plane through
the origin with normal parallel to n is
hx, y, z On = hx, y, z Oh5, 3, 1 O = 0 ,
which after evaluation becomes
Now when
and so when a = 1, b = 1,
n = h5, 3, 1 O .
y
p
5 x 2 + 3 y2 ,
f
x
x
while
= p
f
y
5 x2 + 3y 2
3y
=p
5 x + 3y
2
.
2
,
HW09 gilbert (55035)
Thus at P ,
But the partial derivatives of
f (2, 1) = 2 ,
while
f
x
(2,
1)
2
f
= 1 ,
y
(2,
1)
f (x, y) = tan
=
are
3
2.
fx =
z = 2 (x 2) +
f
1
y
f
2(y 1) ,
,
x
(1,
0)
=2
(1,
0)
and
f
.
y
L(x, y) =
10.0 points
1
(x + 4y)
3. L(x, y) =
1
2x
+
4. L(x, y) = x + 2y
5. L(x, y) = x + 2y
6. L(x, y) =
1
004
1
4
1.
(a,b)
y
(a,b)
z 0.72 correct
4.
.
z 0.82
3.
L(x, y) = f (a, b)
f
f
+ (x a)
+(y b)
z 0.42
2.
Explanation:
The linearization of f = f (x, y) at a
point (a, b) is given by
10.0 points
as (x, y) changes from (2, 4) to (2.04, 4.03).
1
2x + 2y + 4 2 correct
x
.
2
2
z = 3x + y
+
1
1
2x + 2y + 4 2
Use Linear Approximation to estimate
the change, z, in
2
1
2y 2
1
+ 4 2
1
2
2. L(x, y) = x + 2y +
1
keywords:
Find the linearization, L(x, y), of the func-
f (x, y) = tan
at the point (1, 0) .
1 + (x + 4y)
Consequently,
keywords: tangent plane, partial derivative,
radical function, square root function,
tion
2.
= 2.
003
4
which after rearrangement becomes
2x 3y + 2z 5 = 0
=
2
1 + (x + 4y)
It follows that
So at P the tangent plane has equation
3
1
(x + 4y)
z 0.52
5. z 0.62
Explanation:
The Linear Approximation to z = f
(x, y) at a point (a, b) is
L(x, y) = f (a, b)
+ fx(a, b)(x a) + fy(a, b)(y b) ,
HW09 gilbert (55035)
and so the change,
z in z as (x, y), changes
from (a, b) to (a + x, b + y) can be estimated by
z = L(a +
x, b +
f
x (a, b)
=
y) f (a, b)
f
x + y (a, b)
=
4.
dw
f
in which case at (2, 4),
f
f
x (2, 4) = 12 ,
y (2,
dw
dt
4) = 8 .
4
100 ,
x=
w = xe
2
+ 8 100
3
= 0.72
.
dt
=
2t
006
dw
dt
when
y/z
dw
dt
dw
2.
dt
y = 1t,
=
t
=
t
x
z
+
x
z
4 xy y/z
z 2e .
x
4 xy
z z 2 ey/z
z = 1 + 4t .
xy y/z
ze
+
xy y/z
ze
.
10.0 points
z
u
when
and
r = uv ,
2
y/z
r
z = e cos
and
x=t,
ze
Use the Chain Rule to find
10.0 points
w = xe
x
Consequently,
keywords: dierentials, multi-variable
esti-mate, polynomial function
Use the Chain Rule to find
z = 1 + 4t ,
therefore,
dw
005
y/z
y = 1t,
dw
y/z
dt = 2te
4
4xy
When
Consequently,
z 12 100
+
4xy
z ey/z
z
dx
w d y
w d z
dt + y dt + z dt .
x
x=t,
3
100 .
x=
x
x
and
On the other hand, when (x, y)
changes from (2, 4) to (2.04, 4.03),
1.
w
=
+
2
2t
2t +
z
2
= 2y ,
y
x
ey/z dt z z
Explanation:
By the Chain Rule for Partial Di
erentia-tion,
we see that
= 6x ,
=
=
+
dt z z
dw
dt
6.
x
xy
y/z
z2 e
x
4xy
2t
ey/z
t
=
correct
5.
2
2
z = f (x, y) = 3x + y ,
f
dw
dt
dw
3.
y.
Now when
3
=
p
u2 + v2 .
z
sin
r
= e v cos 2 u2+ v2
1 . u
z
u sin
r
= e v cos u2 + v2 correct
2 . u
z
sin
2
2
3. u = ue v cos + u + v
r
HW09 gilbert (55035)
z
sin
r
2
2
4. u = e v cos + u + v
z
u sin
while
z
= e
Thus
sin ,
x dx
=
+
F
F dy
y dx
x
.
u2 + v2
z
u sin
r
= e v cos u2 + v2 .
u
+
F
y dx
F
dy
=v
u
u =
F dx
Thus
r
u
r
2y
sin(x 4y) + e
6.
=
2y
4xe 2 sin(x 4y)
dx
Explanation:
Applying the Chain Rule to both sides
of the equation F (x, y) = 0, we see that
dy
2 2
r
5. u = e v cos + 2 u + v
Explanation:
By the Chain Rule for Partial Di
erentia-tion,
z
z
r z
u = r u + u .
But
z
r
r = e cos ,
4
dx = F
x
y
007
10.0 points
Use partial di erentiation and the
Chain Rule applied to F (x, y) = 0 to
determine dy/dx when
F (x, y) = cos(x 4y) xe
2y
dx
1.
2y
sin(x 4y) + e
=
2y
2xe 4 sin(x 4y)
2y
= sin(x 4y) + 2xe
2y
4 sin(x 4y) e
dy
=
dy
dx
2. dy
3.
dx
4.
5.
= 0.
dy
dx
dy
dx
sin(x
4y) + e
F (x, y)
therefore,
4 sin(x 4y) 2xe
2y
sin(x 4y) 2e
=
2y
2 sin(x 4y) 4xe
2y
sin(x 4y) 2xe
=
2y
2 sin(x 4y) 4e
Fx
= Fy .
= cos(x 4y) xe
2y
sin(x 4y) e
dy =
4 sin( x
dx
4y)
= 0,
2y
2x e2y
.
Consequently,
2y
sin(x 4y) + e
=
2y .
4 sin(x 4y) 2xe
dy
dx
008
10.0 points
The temperature at a point (x, y) in the
plane is T (x, y) C. If a bug crawls on the
plane so that its position in the plane after
t minutes is given by (x(t), y(t)) where
x=
5+ t ,
y = 3+
3
4t ,
determine how fast the temperature is
rising on the bugs path at t = 4 when
2y
2y
= 0.
When
keywords: partial dierentiation, Chain Rule,
dy
correct
Tx(3, 6) = 18 ,
Ty(3, 6) = 4 .
1. rate = 6 C/min correct
2. rate = 4 C/min
3. rate = 5 C/min
HW09 gilbert (55035)
4. rate = 3 C/min
5. rate = 48 cu. ins/min
5. rate = 2 C/min
Explanation:
Explanation:
By the Chain Rule for partial dierentiation, the rate of change of temperature on
the bugs path at (x(t), y(t)) is given by
dT
dt
=
dT (x(t), y(t))
dt
=
5
T dx T dy
x dt + y dt .
The volume of a right circular cone having
base r radius and height h is given by
V
1
dy
dt
= 2 5 + t ,
dt =
dV =
dt
But when t = 4, the bug is at the point (3,
6), while
1 dy
1
dx
3
dt t=4 = 4 ,
t=4 = 25 + t t = 4 = 6 ,
dt
and we are told that
Tx(3, 6) = 18 ,
Ty(3, 6) = 4 .
Consequently, after 4 minutes the temperature on the bugs path is changing at a
rate =
1
3
18 6 + 4 4
009
V dr + V
r dt
3
4.
= 6 C/min .
10.0 points
The radius of the base of a right circular
cone is increasing at a rate of 4 ins/min while
its height is increasing at a rate of 2 ins/min.
2
h.
3r
When h and r are changing with t, then
by the Chain Rule the rate of change of
V with respect to t is given by
Now
dx
1
=
=
2
h dt
3rh
But
dr
dt = 4,
in which +
1 case
dV
dh
dr
dt 2 dh
3r dt .
dh
dt = 2 ,
8
2
2
dt = 3rh + 3 r .
Consequently, when r = 3 and h = 5,
dV
dlt (r=3, h=5)
= 46 cu.in./min.
.
010 10.0 points
The length and width w of the closed
box shown in
At what rate is the volume of the cone
changing when the radius is 3 inches
and the height is 5 inches?
h
1. rate = 46 cu. ins/min correct
2. rate = 50 cu. ins/min
w
3. rate = 47 cu. ins/min
4. rate = 49 cu. ins/min
are increasing at a rate of 4 ft/min while its
height h is decreasing at a rate of 6 ft/min.
HW09 gilbert (55035)
6
Find the rate at which the volume of the box
is increasing when
= 2,
011 10.0 points
w = h = 3 feet .
Find the directional derivative, fV, of
p
f (x, y) = 4x + 2y at
1. rate = 30 cu ft/min
2. rate = 26 cu ft/min
the point (3, 3) in the direction
3. volume is decreasing, not increasing
v = i+j.
4. rate = 28 cu ft/min
5. rate = 24 cu ft/min correct
1 . fV
V = w h .
Thus by the Chain Rule for partial di
eren-tiation, the rate at which the
surface area is changing is given by
V
t
V
V
w
V
h
t +
w
t +
h
t .
=
= 16
2 . fV
Explanation:
The box has volume
= 13
f
3.
V
1
4. fV =
f
Now
V
V
= wh ,
w
= h ,
V
5.
h = w ,
and so we see that
=0
V
2
correct
= 23
Explanation:
For an arbitrary vector v,
V
w
h
t = wh t + h t + w t .
fV = f , v
||
When
= 2,
w = h = 3,
and
v
where we have normalized the direction
vector so that it has unit length.
Now the partial derivatives of
t =
therefore,
V
t
w
t
p
h
= 4 , t = 6 ,
f (x, y) =
4x + 2y
are given by
=334
f
x = 4x + 2y ,
+ (3 2) 4 (3 2) 6 = 24 > 0 .
Consequently, the volume is increasing
at a rate = 24 cu ft/min .
2
and
f
1
y = 4x + 2y .
HW09 gilbert (55035)
Thus
7
we see that
f (x, y) =
f
x i +
f
2
2
3
f = y + 9x y, 2xy + 3x .
y j
2 1 j , = 4x + 2y i
+
keywords:
4x + 2y
013
and so
Determine the gradient of
2
f (x, y) = xy
at the point P (3, 2).
1 2 1 f (3, 3) =
2 3 i + 3 j .
On the other hand,
1
v
=
v = i+j
But then
f
v
v
= 21
=
|v|
2
1
3 i+3 j
1. f |P = 4 i 12 j correct
2 (i + j) .
2. f |P = 12 j 4 k
(i + j) .
3. f |P = 12 i + 4 k
4. f |P = 4 i + 12 j
||
Consequently,
f
V
=2
1
5. f |P = 4 i 12 k
2
3
+
3
1
=
2
1
.
keywords:
012
10.0 points
6. f |P = 12 j + 4 k
Explanation:
Since
2
f = fx i + fy j = y i + 2xy j ,
and P = (3, 2), we see that
f |P = 4 i 12 j
Find the gradient of
2
3
f (x, y) = xy + 3x y .
1. f
2. f
3. f
4. f
=
10.0 points
y2 + 9x2y,
014
3
2xy + 3x correct
2
2
= 9x y y ,
3
= 3x 2xy,
2
2
= y + 9x y,
3
= 2xy + 3x ,
2xy + 3x
3
f (x, y) =
The contour map given below for a
function f shows also a path r(t) traversed
counter-clockwise as indicated.
3x
2xy
2
9x y y
5. f
32
2
6. f
= 2xy + 3x , y + 9x y
Explanation:
Since
f
10.0 points
2
2
y + 9x y
3
2
2
1
0
3
Q
-3
-2
P
R
x
f
, y ,
.
-1
0
HW09 gilbert (55035)
Which of the following properties does
the derivative
d
dt
f (r(t))
II FALSE: at Q we are ascending - the
con-tours are increasing in the counterclockwise direction.
III TRUE: at P we are descending - the
con-tours are decreasing in the counterclockwise direction.
have?
I zero at R,
II negative at Q,
III
8
keywords: contour map, contours,
slope, curve on surface, tangent, Chain
Rule, multi-variable Chain Rule,
negative at P .
1. none of them
015
2. all of them
Find the directional derivative, fV, of
the function
3. II and III only
f (x, y) = 3 + x y
4. I only
5. I and III only correct
6. III only
at the point P (1, 4) in the direction of the
vector
v = h 3, 4 O .
f
7. I and II only
1.
=
Explanation:
By the multi-variable Chain Rule,
d
f (r(t)) = (f )(r(t))r (t) .
2. fV
10
3.
d
dt
will be the sign of the slope of the surface in
the direction of the tangent to the curve r(t),
and we have to know which way the curve is
being traversed to know the direction the
tangent points. In other words, if we think of
the curve r(t) as defining a path on the graph
of f , then we need to know the slope of the
path as we travel around that path - are we
going uphill, downhill, or on the level. That
will depend on which way we are walking!
From the contour map we see that
I TRUE: at R we are on the level - we
are following the contour.
11
10
=
5
V
f
=
7
10
V
5 . fV
f (r(t))
=
4
f
4.
Thus the sign of
9
V
8. II only
dt
10.0 points
= 1 correct
Explanation:
Now for an arbitrary vector v,
fV
= f v
v
,
||
where we have normalized so that the
direc-tion vector has unit length. But when
f (x, y) = 3 + x y ,
then
1 x
f = ( y) i + 2
y
j.
HW09 gilbert (55035)
At P (1, 4), therefore,
f P = 2 i +
1
4
9
so that at P (3, 3),
1
1
(f )(3, 3) = 6 i + 6 j
j.
=
D 1, 1 E .
6
6
Consequently, when v = h 3, 4O ,
On the other hand, v =
fV(1, 4) = 2, 41 vv = 1 .
||
h
4, 3O which as a
vector of unit length becomes
43
v
=D5, 5 E.
v
||
keywords:
Consequently,
016
10.0 points
f |P =
Find the directional derivative, fV, of
f (x, y) =
y
017
x
P Q when Q = (7, 6).
2. f V
1
= 10
3. f V
1
= 30 correct
1
6
2. max slope = 5
3. max slope = 1
2
4. max slope = 20
= 15
1
= 15
5. max slope = 20 correct
6. max slope =
The directional derivative of f (x, y) at P
in the direction of v = P Q is given by the
dot product
fV|P = f |P |v
v
|
f (x, y) =
xy
1/2
8. max slope = 5
,
||
we see that
2
7. max slope = 4
Explanation:
At P (0, 5, 0) the slope in the direction
of v is given by
f (0,5) vv .
.
But when
=
10.0 points
1. max slope = 4
Explanation:
f
x
But when
1 y 1/2
x3
1
= 30 .
at the point P (0, 5).
=
5 . fV
5
3E
f (x, y) = 4 sin(xy)
1. f V
V
5,
Find the maximum slope on the graph of
4.
4
1/2
at P = (3, 3) in the direction of the vector
f
D 1, 1 E.D
66
,
f
y
=
2
1
1 1/2
xy
,
f (x, y) = 4 sin(xy) ,
HW09 gilbert (55035)
of the partial derivative fy at P = (1, 2).
the gradient of f is
f (x, y) = 4y cos(xy) i + 4x cos(xy) j ,
so at P (0, 5)
f (0,5)
10
On the other hand, the slope at P in
the direction of any vector
v = ai+bj
= 20 i .
Consequently, the slope at P will be
maxi-mized when v = i in which case
is given by
f (1,2)
max slope = 20 .
=
keywords: slope, gradient, trig function,
max-imum slope
018
10.0 points
If the graph of z = f (x, y) at P = (1, 2)
has slope = 2 in the x-direction and
slope = 3 in the y-direction, find the
slope at P in the direction of the vector
v = 4 i + 3 j .
1
1. slope = 5 correct
2
2. slope =
5
3. slope = 1
4
4. slope =
5
3
5. slope = 5
Explanation:
The slope of the graph of z = f (x, y) in
the x-direction at P = (1, 2) is the value
f
fx(1, 2) = x (1, 2) = 2
of the partial derivative fx at P , while the
slope of the graph of z = f (x, y) in the ydirection at P is the value
f
fy(1, 2) = y (1, 2) = 3
vv
||
1
|v | fx(1, 2) i + fy(1, 2) j (a i + b j) .
When
v = 4 i + 3 j ,
therefore, the graph of f has
slope =
1
5(2 i 3 j)(4 i + 3 j) =
at P in the direction of
v. keywords:
1
5