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Multivariable Calculus 9th Edition by James Stewart solution manual

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(b) The power model in part (a) is approximately  =  15 . Squaring both sides gives us 
2
=  3 , so the model matches
Kepler’s Third Law, 
2
=  3 .
35. (a) If  = 60, then  = 07 03  239, so you would expect to find 2 species of bats in that cave.
(b)  = 4 ⇒ 4 = 07 03 ⇒
40
7
=  310 ⇒  =
 40
7
 103
 3336, so we estimate the surface area of the cave
to be 334 m 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.
30
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CHAPTER 1 FUNCTIONSAND MODELS
36. (a) Using a computing device, we obtain a power function  =   , where   31046 and   0308.
(b) If  = 291, then  =    178, so you would expect to find 18 species of reptiles and amphibians on Dominica.
37. We have  =

4 2
=


4

1
 2

=
(4)
 2
. Thus,  =

 2
with  =

4
.
1.3 New Functions from Old Functions
1. (a) If the graph of  is shifted 3 units upward, its equation becomes  = () + 3.
(b) If the graph of  is shifted 3 units downward, its equation becomes  = () − 3.
(c) If the graph of  is shifted 3 units to the right, its equation becomes  = ( − 3).
(d) If the graph of  is shifted 3 units to the left, its equation becomes  = ( + 3).
(e) If the graph of  is reflected about the ­axis, its equation becomes  = −().
(f) If the graph of  is reflected about the ­axis, its equation becomes  = (−).
(g) If the graph of  is stretched vertically by a factor of 3, its equation becomes  = 3().
(h) If the graph of  is shrunk vertically by a factor of 3, its equation becomes  =
1
3 ().
2. (a) To obtain the graph of  = () + 8 from the graph of  = (), shift the graph 8 units upward.
(b) To obtain the graph of  = ( + 8) from the graph of  = (), shift the graph 8 units to the left.
(c) To obtain the graph of  = 8() from the graph of  = (), stretch the graph vertically by a factor of 8.
(d) To obtain the graph of  = (8) from the graph of  = (), shrink the graph horizontally by a factor of 8.
(e) To obtain the graph of  = −() − 1 from the graph of  = (), first reflect the graph about the ­axis, and then shift it
1 unit downward.
(f) To obtain the graph of  = 8( 1
8 ) from the graph of  = (), stretch the graph horizontally and vertically by a factor
of 8.
3. (a) Graph 3: The graph of  is shifted 4 units to the right and has equation  = ( − 4).
(b) Graph 1: The graph of  is shifted 3 units upward and has equation  = () + 3.
(c) Graph 4: The graph of  is shrunk vertically by a factor of 3 and has equation  =
1
3 ().
(d) Graph 5: The graph of  is shifted 4 units to the left and reflected about the ­axis. Its equation is  = −( + 4).
(e) Graph 2: The graph of  is shifted 6 units to the left and stretched vertically by a factor of 2. Its equation is
 = 2( + 6).
4. (a)  = () − 3: Shift the graph of  3 units down. (b)  = ( + 1): Shift the graph of  1 unit to the left.
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.
SECTION 1.3 NEW FUNCTIONSFROM OLD FUNCTIONS
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31
(c)  =
1
2 (): Shrink the graph of  vertically by a
factor of 2.
(d)  = −(): Reflect the graph of  about the ­axis.
5. (a) To graph  = (2) we shrink the graph of 
horizontally by a factor of 2.
The point (4−1) on the graph of  corresponds to the
point
 1
2
· 4−1  = (2−1).
(b) To graph  =   1
2 

we stretch the graph of 
horizontally by a factor of 2.
The point (4−1) on the graph of  corresponds to the
point (2 · 4−1) = (8−1).
(c) To graph  = (−) we reflect the graph of  about
the ­axis.
The point (4−1) on the graph of  corresponds to the
point (−1 · 4−1) = (−4−1).
(d) To graph  = −(−) we reflect the graph of  about

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