My answers to the exercises

14.4 25, 27, 28, 35
25–30 Find the di¤erential of the function.
25. z = e 2x cos 2 t
dz = zx dx + zt dt = 2e 2x cos 2 t dx 2 e 2x sin 2 t dt =
2e 2x (cos 2 t dx + sin 2 t dt).
27. m = p5 q 3
dm = mp dp+mq dq = 5p4 q 3 dp+3p5 q 2 dq = p4 q 2 (5q dp + 3p dq).
v
28. T =
1 + uvw
v2w
1
dT = Tu du+Tv dv+Tw dw =
dv
2 du+
(1 + uvw)
(1 + uvw)2
v2u
dw.
(1 + uvw)2
35. Use di¤erentials to estimate the amount of tin in a closed tin can
with diameter 8 cm and height 12 cm if the tin is 0:04 cm thick.
The volume of a cylinder is V = r2 h, where r is the radius and h
is the height. So dV = 2 rh dr + r2 dh. If dt is the thickness of
the tin, then dr = dt and dh = 2dt, so dV = (2 rh + 2 r2 ) dt =
(96 + 32 ) 0:04 = 5:12 cubic centimeters of tin. Note that
V = 42 12 = 192 cubic centimeters. Dumb problem?
Exact computation where 8 is the outer diameter and 12 is the
outer height:
42 12
(4
0:04)2 (12
0:08) = 5:0753
Where 8 is the inner diameter and 12 is the inner height: (4 + 0:04)2 (12 + 0:08)
42 12 = 5:1649
Where 8 is the average diameter and 12 is the average height:
(4 + 0:02)2 (12 + 0:04)
(4
0:02)2 (12
0:04) = 5:12
14.5 1, 3, 5, 7, 9, 11, 13, 15, 21, 23, 27, 31, 33, 35, 43
1–6 Use the chain rule to …nd dz=dt or dw=dt. These are a bit tedious.
1. z = x2 + y 2 + xy, x = sin t, y = et
dz=dt = zx cos t + zy et = (2x + y) cos t + (2y + x) et =
(2 sin t + et ) cos t+(2et + sin t) et = sin 2t+2e2t +et (cos t + sin t)
1
p
3. z = 1 + x2 + y 2 , x = ln t, y = cos t
dz=dt = zx =t zy sin t = p 1 2 2 (x=t y sin t) =
1+x +y
p
1
1+ln2 t+cos2 t
y=z
((ln t) =t
cos t sin t)
5. w = xe , x = t2 , y = 1 t, z = 1 + 2t
dw=dt = wx 2t wy + 2wz =
ey=z (2t (x=z) 2 (xy=z 2 ))
7–12 Use the chain rule to …nd @z=@s and @z=@t. These are a bit
tedious.
7. z = x2 y 3 , x = s cos t, y = s sin t
@z=@s = zx cos t + zy sin t = 2xy 3 cos t + 3x2 y 2 sin t =
2s4 cos2 t sin3 t + 3s4 cos2 t sin3 t = 5s4 cos2 t sin3 t
@z=@t = zx s sin t + zy s cos t = 2xy 3 s sin t + 3x2 y 2 s cos t =
2s5 cos t sin4 t+3s5 cos3 t sin2 t = 3 cos2 t 2 sin2 t s5 cos t sin2 t
9. z = sin cos ,
= st2 ,
= s2 t
@z=@s = z t2 + z 2st = t2 cos cos
2st sin sin =
2
2
2
2
2
t cos st cos s t 2st sin st sin s t
@z=@t = z 2st + z s2 = 2st cos cos
s2 sin sin =
2
2
2
2
2
2st cos st cos s t s sin st sin s t
p
11. z = er cos , r = st,
= s2 + t2
p
p s
@z=@s = zr t + z s= s2 + t2 = ter cos
er sin =
s2 +t2
p
p
s
est t cos s2 + t2 p
sin s2 + t2
2
2
p s +t
p t
@z=@t = zr s + z t= s2 + t2 = ser cos
er sin =
s2 +t2
p
p
t
est s cos s2 + t2 p
sin s2 + t2
s2 + t2
13. If z = f (x; y) where f is di¤erentiable, and
x = g (t)
g (3) = 2
g 0 (3) = 5
fx (2; 7) = 6
y = h (t)
h (3) = 7
h0 (3) = 4
fy (2; 7) = 8
…nd dz=dt when t = 3.
When t = 3 we have x = g (3) = 2 and y = h (3) = 7. So
dz=dt = fx (2; 7) g 0 (3) + fy (2; 7) h0 (3) = 6 5
2
8 ( 4) = 62
15. Suppose f is a di¤erentiable function of x and y, and g (u; v) =
f (eu + sin v; eu + cos v). Use the table of values to calculate gu (0; 0)
and gv (0; 0).
f g f x fy
(0; 0) 3 6 4 8
(1; 2) 6 3 2 5
gu (u; v) = fx (eu + sin v; eu + cos v) eu +fy (eu + sin v; eu + cos v) eu
so gu (0; 0) = fx (1; 2) + fy (1; 2) = 7.
gv (u; v) = fx (eu + sin v; eu + cos v) cos v fy (eu + sin v; eu + cos v) sin v
so gv (0; 0) = fx (1; 2) + 0 = 2.
21–26 Use the chain rule to …nd the indicated partial derivatives.
21. z = x4 + x2 y, x = s + 2t u, y = stu2 ;
@z @z @z
,
,
when s = 4, t = 2, u = 1
@s @t @u
For those values of s; t; u we have x = 7 and y = 8, so
@z
= zx +zy tu2 = 4x3 +2xy+x2 tu2 = 4 73 +112+49 2 = 1582
@s
@z
= zx 2 + zy su2 = 8x3 + 4xy + x2 su2 = 8 73 + 224 + 49 4 =
@t
3164
@z
= zx + zy 2stu = 4x3 2xy + 2x2 stu = 4 73 112 +
@s
2 49 8 = 700
23. w = xy + yz + zx; x = r cos , y = r sin , z = r ;
@w @w
,
when r = 2, = =2
@r @
For those values of r and we have x = 0, y = 2, and z = ,
so
@w
= wx cos + wy sin + wz = 0 + (x + z) + (x + y) =2 =
@r
0+ + =2
@w
= wx r sin +wy r cos +wz r = (y + z) r+0+(x + y) r =
@
(x z) r = 2
27–30 Use Equation 6 to …nd dy=dx.
27. y cos x = x2 + y 2
Without using equation 6 we just di¤erentiate with respect to
x treating y as a function to get y 0 cos x y sin x = 2x + 2yy 0 ,
so y 0 = (2x + y sin x) = (cos x 2y).
3
Using equation 6 we set F (x; y) = y cos x x2 y 2 , so dy=dx =
Fx =Fy = (2x + y sin x) = (cos x 2y).
31–34 Use Equations 7 to …nd @z=@x and @z=@y.
31. x2 + 2y 2 + 3z 2 = 1
Set F (x; y; z) = x2 + 2y 2 + 3z 2 1. Then @z=@x = Fx =Fz =
2x= (6z) = x= (3z), and @z=@y = Fy =Fz = 4y= (6z) =
2y= (3z).
33. ez = xyz
Set F (x; y; z) = ez xyz. Then Then @z=@x = Fx =Fz =
yz= (ez xy), and @z=@y = Fy =Fz = xz= (ez xy).
35. The temperature at a point (x; y) is T (x; y), measured in degrees
Celsius. A
p bug crawls so that its position after t seconds is given
by x = 1 + t, y = 2 + t=3, where x and y are measured in
centimeters. The temperature function satis…es Tx (2; 3) = 4 and
Ty (2; 3) = 3. How fast is the temperature rising on the bug’s path
after 3 seconds?
p
Note that x0 (t) = 1=2 1 + t and y 0 (t) = 1=3, so x0 (3) = 1=4
andpy 0 (3) = 1=3. We are asked to compute f 0 (3) where f (t) =
1 + t; 2 + t=3 . By the chain rule,
T
f 0 (3) = Tx (2; 3) x0 (3) + Ty (2; 3) y 0 (3) = 4 (1=4) + 3 (1=3) = 2
43. One side of a triangle is increasing at a rate of 3 cm/s and a
second side is decreasing at a rate of 2 cm/s. If the area of the
triangle remains constant, at what rate does the angle between the
two sides change when the …rst side is 20 cm long, the second side
is 30 cm, and the angle is =6?
If the length of the two sides are denoted by a and b, and the
angle between them by , then the area of the triangle is 21 ab sin .
We know that a0 = 3 and b0 = 2. We want to determine 0 . So
di¤erentiate the constant ab sin to get
a0 b sin + ab0 sin + ab 0 cos = 0
Plugging in a = 20, b = 30, and
= =6 gives
3 30 (1=2) + 20 ( 2) (1=2) + 600
4
0
p
3=2 = 0
so
25 + 300
whence
0
=
0
p
3=0
p
1= 12 3 . Is it reasonable that
5
0
is negative?

Download Report