Thomas Calculus 13th [Solutions]

Chapter 1 Practice Exercises 43
80.
81.
t t t
ln(2)
500(1.0475) 1000 1.0475 2 ln(1.0475 ) ln(2) t ln(1.0475) ln(2) t
14.936
ln(1.0475)
It will take about 14.936 years. (If the interest is paid at the end of each year, it will take 15 years.)
t t 8 t
375,000(1.0225) 1,000,000 1.0225 ln(1.0225 ) ln 8 t ln(1.0225) ln 8
3 3 3
8
3
ln
t
44.081
ln(1.0225)
It will take about 44.081 years.
82.
0.18t
y y0e represents the decay equation; solving
0.18t
ln(0.9)
(0.9) y0 y0e t 0.585 days
0.18
CHAPTER 1 PRACTICE EXERCISES
2
2 2
1. The area is A r and the circumference is C 2 r . Thus, r C A C C .
2 2 4
2
1/2
2. The surface area is 4 S
3
S r r . The volume is V 4 r r 3 3V
4
3 4 . Substitution into the formula
2
2/3
for surface area gives S 4 r 4 3V
.
4
3. The coordinates of a point on the parabola are (x, x 2 ). The angle of inclination joining this point to the origin
2
satisfies the equation tan x
2
2
x . Thus the point has coordinates ( x, x ) (tan , tan ).
x
4. tan rise h h 500 tan ft.
run 500
5.
6.
Symmetric about the origin.
Symmetric about the y-axis.
7.
8.
Neither
Symmetric about the y-axis.
9.
2 2
y( x) ( x) 1 x 1 y( x ). Even.
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