140
CHAPTER 18
18.17
Approximate
Let
Then
and
By the
approximation principle,
(This is correct to two decimal places.)
18.18
Approximate'
Let f(x) = x
114 , x = Sl, A* = -l.
Then
and
By the approximation principle,
(This is correct to four decimal places.)
18.19
Estimate
Let
Then
and
By
the approximation principle,
0.0167 = 1.9833. (The correct answer to four decimals places is 1.9832.)
18.20
Estimate
Let f(x) = Vx, A-= 25, Ax = 0.4. Then
and
By
the approximation principle,
correct to two decimal places.)
So
(This is
18.21
Approximate
Let f(x) =\Sx, x = 27, Ax = -0.54. Then
and
By the approximation principle,
0.02 = 2.98. (This is correct to two decimal places.)
18.22
Estimate (26.5)
2 '
3 .
Let f(x) = x
213 , x = 27, x = -0.5. Then /'(*) = 2/3v^ and Ay = (26.5)
2 '
3 - (27)
2 " = (26.5)
2 '
3 - 9.
By the approximation principle, (26.5)
2 '
3 - 9 = (2/3v^)(-0.5) = | -(-0.5) = -§ = -0.1111.
9 - 0.1111 = 8.8889. (The correct answer to four decimal places is 8.8883.)
18.23
Estimate sin 60° 1'.
radians = it 110,800 radians. Let f(x) = sin x, x=ir/3, A* = Tr/10,800.
1 hen / (x) = cos x. and A>> = sin 60° 1' - sin 60° = sin 60° 1' - V3/2. By the approximation principle,
0.00015 = 0.86618. (The correct answer to five decimal places is 0.86617.)
18.24
Find the approximate change in the area of a square of side s caused by increasing the side by 1 percent.
By the approximation principle, A/4 = 25 • (0.01s) = 0.02s
2
.
18.25
Approximate cos 59°.
Let /(*) = cos AT, x = 7r/3, Ax = -ir/180. Then /'(*) = -sin x and Ay = cos 59° - cos 60°. By
the approximation principle, cos59° - | = -sinx • (-77/180) = (V3/2)- (Tr/180) =0.0151.
0.5151. (The actual answer is 0.5150 to four decimal places.)
So cos 59° =
18.26
Estimate tan 44°.
f(x)=&x, A: = 64, Ajc = -l.
So
So
f(x) =\/x, x = 8, AJC = -0.2.
So
So
So (26.5)
2/3 =
/l=5
2 . Then As=0.0li. D 5 /4=25.
So sin 60° 1' « V3/2 + 0.00015 = 0.86603 +
sin 60° 1' - V3/2 = cos x • (TT/10,800) = \ • (irl 10,800) =0.00015.
1'= <| b of a degree =