 f  f '  3 x  6  0.02  0.12
a3
x  0.02
1
1
1
f '  3    f    0.02  
 0.002
9
9
450
f  x  x  x
3
1/3
1 2/3
1
 f ' x  x 
3
3 3x
 
2
a  27
x  0.2
1
1
 f  f '  27   0.02 
 0.2 
 0.0074074
27
135
f  x  x  x
1/2
a  25
x  1
1 1/2
1
 f ' x  x 
2
2 x
1
 f  f '  25  1 
 0.1
10
The exact change is f  f  a  x   f  a  
The error in this estimate is 0.09901955136  0.1 
1
1 3/2
1
1/2
f  x 
 x  f ' x   x

2
x
2 x
 
3
a  100
x  1
1
 f  f ' 100  1  
 0.0005
2000
The exact change is f  f  a  x   f  a  
The error in this estimate is 0.000496280979  0.0005 
f  x  x
1/3
1 2/3
1
 f ' x  x 
3
3 3x
 
2
a 8
x  1
1
 f  f '  8  1 
 0.083
12
The exact change is f  f  a  x   f  a  
The error in this estimate is 0.000496280979  0.0005 
x  4.03, a  4  f  4.03  f '  4  0.03  f  4 
2 1
1
f  4   2, f '  4     f  4.03   0.03  2  2.01
6 3
3
F  F '  35   s  4.88 1  4.88'
F  F '  55   s  7.04 1  7.04 '
35. Newton’s Law of Gravitation shows that if a person weighs w pounds on the surface of the
earth, then his or her weight at distance x from the center of the earth is
where R = 3,960 miles is the radius of the earth.
(a) Show that the weight lost at altitude h miles above the earth’s
surface is approximately ΔW ≈ −(0.0005w)h.
Hint: Use the Linear Approximation with dx = h.
dx  h  a  3960
w & R are constant
2
2
wR
W  x   wR 2 x 2  W '  x    3
x
2 wR 2
2 wh
W  W '  a  x  
h
 0.0005wh
3
R
R
(b)
Estimate the weight lost by a 200-lb football player flying in a jet at an altitude of 7 miles.
w  200, h  7  W  0.0005  200  7   0.7 lbs
39. A player located 18.1 ft from the basket launches a successful jump
shot from a height of 10 ft (level with the rim of the basket), at an angle
θ = 34° and initial velocity v = 25 ft/s.)
(a) Show that Δs ≈ 0.255Δθ ft for a small change of Δθ.
(b) Is it likely that the shot would have been successful if the angle
had been off by 2°?
1 2
a) s  v0 sin 2 ft (given) *** and  must be in radians
32
625

17
s
sin 2 ft,   34 

32
180
90
  1250
 
17


 s  s '     

cos
2




,





180 
32
180 
90

 




 0.255


180 

b)   2  s  0.255  2 
 s 
1250
 17
cos 
32
 45
*** The

has already beed accounted for in our factor of 0.255
180
 s  6"  the shot was likely missed