Multivariable Calculus 9th Edition by James Stewart solution manual

we first graph () =  and () = 1 on the
same coordinate axes. Then create the graph of
 by plotting the largest ­value of  and  for
every value of .
(b)
(c)
On the TI­84 Plus, max is found under LIST, then under MATH. To graph () = max
  2 2 + 2 −   , use
Y = max( 2 max(2 + 2 − )).
10. (a) If max{2} = 1, then either  = 1 and 2 ≤ 1   ≤ 1 and
2 = 1. Thus, we obtain the set of points such that  = 1 and
 ≤
1
2
[a vertical line with highest point (1
1
2 )

  ≤ 1 and
 =
1
2
 a horizontal line with rightmost point (1
1
2 )
 .
c ° 2021 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
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CHAPTER 1 PRINCIPLESOF PROBLEMSOLVING
(b) The graph of max{2} = 1 is shown in part (a), and the graph of
max{2} = −1 can be found in a similar manner. The
inequalities in −1 ≤ max{2} ≤ 1 give us all the points on or
inside the boundaries.
(c) max{ 2 } = 1 ⇔  = 1 and  2 ≤ 1 [−1 ≤  ≤ 1]
  ≤ 1 and  2 = 1 [ = ±1].
11.
1
log 2 
+
1
log 3 
+
1
log 5 
=
1
log
log2
+
1
log
log3
+
1
log
log5
[Change of Base formula]
=
log2
log
+
log3
log
+
log5
log
=
log2 + log3 + log5
log
=
log(2 · 3 · 5)
log
[Law 1 of Lograithms]
=
log30
log
=
1
log
log30
=
1
log 30 
[Change of Base formula]
12. We note that −1 ≤ sin ≤ 1 for all . Thus, any solution of sin = 100 will have −1 ≤ 100 ≤ 1, or
−100 ≤  ≤ 100. We next observe that the period of sin is 2, and sin takes on each value in its range, except for −1
and 1, twice each cycle. We observe that  = 0 is a solution. Finally, we note that because sin and 100 are both odd
functions, every solution on 0 ≤  ≤ 100 gives us a corresponding solution on −100 ≤  ≤ 0.
1002 ≈ 159, so there 15 full cycles of sin on [0100]Each of the 15 intervals [02], [24], , [2830] must
contain two solutions of sin = 100, as the graph of sin will intersect the graph of 100 twice each cycle. We must
be careful with the next (16th) interval [3032], because 100 is contained in the interval. A graph of  1 = sin and
 2 = 100 over this interval reveals that two intersections occur within the interval with  ≤ 100
Thus, there are 16 · 2 = 32 solutions of sin = 100 on [0100]. There are also 32 solutions of the equation on
[−1000]. Being careful to not count the solution  = 0 twice, we find that there are 32 + 32 − 1 = 63 solutions of the
equation sin = 100.
c ° 2021 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
CHAPTER 1 PRINCIPLESOF PROBLEMSOLVING
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71
13. By rearranging terms, we write the given expression as

sin

100
+ sin
199
100

+

sin
2
100
+ sin
198
100

+ ··· +

sin
99
100
+ sin
101
100

+ sin
100
100
+ sin
200
100
Each grouped sum is of the form sin + sin with  +  = 2 so that
 + 
2
=
2
2
= . We now derive a useful identity
from the product­to­sum identity sin cos =
1
2
[sin( + ) + sin( − )]. If in this identity we replace  with
 + 
2
and
 with
 − 
2
, we have
sin
  + 
2

cos
  − 
2

=
1
2

sin
  + 
2
+
 − 
2

+ sin
  + 
2
−
 − 
2

=
1
2 (sin + sin)
Multiplication of the left and right members of this equality by 2 gives the sum­to­product identity
sin + sin = 2sin
  + 
2