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Chapter 15 of Calculus by James Stewart
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14 PARTIAL DERIVATIVES14.1 Functions of Several Variables
2 12 (3) 3 (−2)2 (−1) 4
1. (a) ( ) = ⇒ (1 3) = =− (b) (−2 −1) = =
2 − 2 2(1) − 32 7 2(−2) − (−1)2 5
( + )2 2 3 2
(c) ( + ) = (d) ( ) = = =
2( + ) − 2 2 − 2 (2 − ) 2−
2. (a) ( ) = sin + sin ⇒ ( 0) = sin 0 + 0 sin = · 0 + 0 · 0 = 0
√ √
2 2+1
(b) = sin + sin = + (1) =
2 4 2 4 4 2 2 2 4 4
(c) (0 ) = 0 sin + sin 0 = 0 + · 0 = 0
(d) ( + ) = sin( + ) + ( + ) sin
3. (a) ( ) = 2 ln( + ) ⇒ (3 1) = 32 ln(3 + 1) = 9 ln 4
(b) ln( + ) is defined only when + 0 ⇒ −.
Thus, the domain of is {( ) | −}.
(c) The range of ln( + ) is , so the range of is .
√ √ √
−2 2
4. (a) ( ) = ⇒ (−2 5) = 5−(−2) = 1 =
(b) − 2 is defined only when − 2 ≥ 0 ⇒ ≥ 2 .
Thus, the domain of is {( ) | ≥ 2 }.
√
2
(c) We know − 2 ≥ 0 ⇒ − ≥ 1. Thus, the range of is [1 ∞].
√ √
5. (a) ( ) = − − 2 ⇒ (3 4 1) = 4 − 3 − 2(1) = 2 − 1 = 1
√ √
(b) is defined only when ≥ 0. − 2 is defined only when − 2 ≥ 0 ⇒ ≤ 12 . Thus, the domain is
{( ) | ≥ 2 ≥ 0}, which is the set of points on or below the plane = 12 and on or to the right of the plane.
√
6. (a) ( ) = ln − 2 + 2 ⇒ (4 −3 6) = ln 6 − 42 + (−3)2 = ln 6 − 25 = ln 1 = 0
(b) ln − 2 + 2 is defined only when − 2 + 2 0 ⇔ 2 + 2 ⇒ 2 2 + 2 . Thus, the
domain is {( ) | 2 + 2 }, which is the set of points inside the top half of the cone 2 = 2 + 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.
° 1341
,1342 ¤ CHAPTER 14 PARTIAL DERIVATIVES
√ √ √
7. ( ) = − 2 + − 1. − 2 is defined only when − 2 ≥ 0, or
√
≥ 2, and − 1 is defined only when − 1 ≥ 0, or ≥ 1. So the
domain of is {( ) | ≥ 2 ≥ 1}.
√ √
8. ( ) = 4
− 3. 4 − 3 is defined only when − 3 ≥ 0, or
≥ 3. So the domain of is {( ) | ≥ 3} or equivalently
( ) | ≤ 13 .
√ √
9. ( ) = + 4 − 42 − 2 . is defined only when ≥ 0.
4 − 42 − 2 is defined only when 4 − 42 − 2 ≥ 0 ⇔
1 ≥ 2 + 14 2 . So the domain of is ( ) | 2 + 14 2 ≤ 1 ≥ 0 .
10. ( ) = ln(2 + 2 − 9). ln(2 + 2 − 9) is defined only when
2 + 2 − 9 0 ⇔ 2 + 2 9. So the domain of is
{( ) | 2 + 2 9}.
−
11. ( ) = . is not defined if + = 0 ⇔ = − (and
+
is defined otherwise). Thus, the domain of is {( ) | 6= −}, the set of
all points in 2 that are not on the line = −.
ln(2 − )
12. ( ) = . ln(2 − ) is defined only when 2 − 0 ⇔
1 − 2 − 2
2. In addition, is not defined if 1 − 2 − 2 = 0 ⇔ 2 + 2 = 1.
Thus, the domain of is ( ) | 2, 2 + 2 6= 1 , the set of all
points to the left of the line = 2 and not on the unit circle.
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 14.1 FUNCTIONS OF SEVERAL VARIABLES ¤ 1343
13. ( ) = . is defined only when ≥ 0. Further, is defined
+1
only when + 1 6= 0 ⇔ 6= −1. So the domain of is
{( ) | ≥ 0, 6= −1}.
14. ( ) = sin−1 ( + ). sin−1 ( + ) is defined only when
−1 ≤ + ≤ 1 ⇔ −1 − ≤ ≤ 1 − . Thus, the domain of is
{( ) | −1 − ≤ ≤ 1 − }, which consists of those points on or
between the parallel lines = −1 − and = 1 − .
√ √
15. ( ) = 4 − 2 + 9 − 2 + 1 − 2 . is defined only
when 4 − 2 ≥ 0 ⇔ −2 ≤ ≤ 2, and 9 − 2 ≥ 0 ⇔
−3 ≤ ≤ 3, and 1 − 2 ≥ 0 ⇔ −1 ≤ ≤ 1. Thus, the domain
of is {( ) | −2 ≤ ≤ 2 −3 ≤ ≤ 3 −1 ≤ ≤ 1},
which is a solid rectangular box with vertices (±2 ±3 ±1)
(all 8 combinations).
16. ( ) = ln(16 − 42 − 4 2 − 2 ). is defined only when
2 2 2
16 − 42 − 4 2 − 2 0 ⇒ + + 1. Thus,
4 4 16
2
2 2
= ( ) + + 1 , that is, the points inside the
4 4 16
2 2 2
ellipsoid + + = 1.
4 4 16
17. (a) (160 70) = 01091(160)0425 (70)0725 ≈ 205, which means that the surface area of a person 70 inches (5 feet 10
inches) tall who weighs 160 pounds is approximately 20.5 square feet.
(b) Answers will vary depending on the height and weight of the reader.
18. (120 20) = 147(120)065 (20)035 ≈ 942, so when the manufacturer invests $20 million in capital and 120,000 hours of
labor are completed yearly, the monetary value of the production is about $94.2 million.
19. (a) From Table 1, (−15 40) = −27, which means that if the temperature is −15◦ C and the wind speed is 40 kmh, then the
air would feel equivalent to approximately −27◦ C without wind.
(b) The question is asking: when the temperature is −20◦ C, what wind speed gives a windchill index of −30◦ C? From
Table 1, the speed is 20 kmh.
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.
°
, 1344 ¤ CHAPTER 14 PARTIAL DERIVATIVES
(c) The question is asking: when the wind speed is 20 kmh, what temperature gives a windchill index of −49◦ C? From
Table 1, the temperature is −35◦ C.
(d) The function = (−5 ) means that we fix at −5 and allow to vary, resulting in a function of one variable. In
other words, the function gives windchill index values for different wind speeds when the temperature is −5◦ C. From
Table 1 (look at the row corresponding to = −5), the function decreases and appears to approach a constant value as
increases.
(e) The function = ( 50) means that we fix at 50 and allow to vary, again giving a function of one variable. In
other words, the function gives windchill index values for different temperatures when the wind speed is 50 kmh . From
Table 1 (look at the column corresponding to = 50), the function increases almost linearly as increases.
20. (a) From Table 3, (95 70) = 124, which means that when the actual temperature is 95◦ F and the relative humidity is 70%,
the perceived air temperature is approximately 124◦ F.
(b) Looking at the row corresponding to = 90, we see that (90 ) = 100 when = 60.
(c) Looking at the column corresponding to = 50, we see that ( 50) = 88 when = 85.
(d) = (80 ) means that is fixed at 80 and is allowed to vary, resulting in a function of that gives the humidex values
for different relative humidities when the actual temperature is 80◦ F. Similarly, = (100 ) is a function of one
variable that gives the humidex values for different relative humidities when the actual temperature is 100◦ F. Looking at
the rows of the table corresponding to = 80 and = 100 we see that (80 ) increases at a relatively constant rate of
approximately 1◦ F per 10% relative humidity, while (100 ) increases more quickly (at first with an average rate of
change of 5◦ F per 10% relative humidity) and at an increasing rate (approximately 12◦ F per 10% relative humidity for
larger values of ).
21. (a) According to Table 4, (40 15) = 25, which means that if a 40knot wind has been blowing in the open sea for 15 hours,
it will create waves with estimated heights of 25 feet.
(b) = (30 ) means we fix at 30 and allow to vary, resulting in a function of one variable. Thus here, = (30 )
gives the wave heights produced by 30knot winds blowing for hours. From the table (look at the row corresponding to
= 30), the function increases but at a declining rate as increases. In fact, the function values appear to be approaching a
limiting value of approximately 19, which suggests that 30knot winds cannot produce waves higher than about 19 feet.
(c) = ( 30) means we fix at 30, again giving a function of one variable. So, = ( 30) gives the wave heights
produced by winds of speed blowing for 30 hours. From the table (look at the column corresponding to = 30), the
function appears to increase at an increasing rate, with no apparent limiting value. This suggests that faster winds (lasting
30 hours) always create higher waves.
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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