Higher_Mathematics_Part_2
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lim (x2 + y2 ) arctg |
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y |
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x→0, |
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y→0 |
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4. Does lim |
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2xy |
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exist ? |
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x2 |
+ xy + y 2 |
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x→0, |
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y→0 |
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Solution. Let the point (x, y) approach (0; 0) along the line y = kx (fig. 1.7). Then
lim |
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2xy |
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lim |
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2x2 k |
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= lim |
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2k |
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2k |
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x→0, x2 |
+ xy + y 2 |
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x→0 x2 + x2 k + x2 k2 |
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x→0 1+ k + k2 |
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+ k + k2 |
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y→0 |
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The values of an expression |
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2k |
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are if change the slope k change. |
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1 + k + k 2 |
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Hence, there are infinitely many diverse straight lines along which the point |
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(x, y) approaches (0; 0). Thus the limit does not exist. |
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y=k2x |
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y=k1x |
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2 х |
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–2 |
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–2 |
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Рис. 1.7 |
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Рис. 1.6 |
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5. Calculate |
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lim |
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e x4 |
− e y4 |
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x 2 |
− y 2 |
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x→0, |
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y→0 |
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Solution. Here |
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lim |
e x4 − e y4 |
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= lim |
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e y4 (e x4 − y4 |
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− 1) |
= |
lim |
e x4 − y4 |
− |
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x2 − y 2 |
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x2 − y 2 |
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x2 |
− y 2 |
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x→0, |
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x |
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x→0, |
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y |
→0 |
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y |
→0 |
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y→0 |
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x 4 − y 4 → 0 |
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= lim |
x4 |
− y 4 |
= lim (x2 + y 2 ) = 0 . |
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e x4 − y4 − 1 ≈ x 4 − y 4 |
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x2 |
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x→0, |
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x + y |
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y→0 |
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6. Let f(x, y) = |
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x − y ≠ 0, is f continuous at (0, 0) ? |
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x − y |
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0, |
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if |
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x − y = 0. |
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11 |
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Solution. Let’s consider |
limit of f |
along the y-axis. Since x = 0 and |
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lim |
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= −1 the limit at (0, 0) is not equal to the value of function at (0, 0): |
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x − y |
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y→0 |
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x=0, |
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–1 ≠ 0. Thus the limit is not continuous at (0, 0). |
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7. Investigate for continuity. |
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sin(x2 + y2 ) |
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f(x, y) = |
x2 + y2 |
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1, |
if x = y = 0. |
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Solution. The given function is determined at all points of the xy-plane. Another,
lim |
sin(x2 + y2 ) |
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sin(x2 |
+ y2 ) ~ x2 + y2 |
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= lim |
x2 + y2 |
= 1 , |
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x2 + y2 |
x2 + y2 |
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x→0, |
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x→0, |
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y→0 |
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y |
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sin(x2 |
+ y2 ) |
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sin(x2 |
+ y2 ) |
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x2 |
+ y2 ≠ 0 . |
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lim |
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= |
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0 |
for |
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x→ x0 , x2 + y2 |
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x02 + y02 |
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y→ y0 |
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As (1.1) shows, function f (x, y) is continuity at any point of xy-plane.
Т.1 Self-test and class assignments
Find and graph the domain on the xy-plane of the following functions.
1. |
z = ln(y − x2 |
+ 3x − 2) . |
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z = |
(x − 1)( y − x2 ) . |
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3. |
z = arcsin(x2 |
+ 4y 2 ) . |
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z = tg |
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πx |
ctg πy . |
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z = (ln(5 − x2 |
− y 2 ))−1/ 2 . |
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z = |
x2 + y 2 − 1 + |
1 |
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x2 − y 2 |
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Find and graph the domain in the xyz-space of the following functions.
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u = |
4 − x2 − y2 − z2 + x . |
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u = 4 (x2 + y 2 − 1)(z 2 − 1) . |
9.u = ln(1− x2 ) + ln(4 − y2 ) + ln(9 − z2 ) .
10.u = arccos x + arccos 2 y + arccos z 2 . Do the following limits exist? If so, evaluate them.
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11. |
lim |
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x + 2 y |
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12. |
lim |
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4 + y |
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y→0 x + 6y |
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y→0 |
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+ y |
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x→0, |
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x→0, |
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14. |
lim |
arctg(xy) |
. 15. |
lim |
tg(xy) |
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x2 + y 2 |
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x→∞, |
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x→0, |
xy2 |
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y→∞ |
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y→2 |
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xy |
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lim xe−(x2 + y2 ) . |
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17. |
lim e |
x2 + y2 |
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18. |
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x→∞, |
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x→∞, |
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y→1 |
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y→∞ |
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Investigate for continuity.
13. |
lim |
sin(x + y) |
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x→0, |
x2 + y 2 |
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y→0 |
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16. |
lim |
sin(x + 2y) |
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x→∞, |
2xy − 3 |
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y→1 |
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19. |
f (x, y) = |
x2 |
+ y2 |
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20. f (x, y) = |
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x2 + y 2 |
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x − y |
x2 |
− xy + 2y 2 |
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sin x +sin y |
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x + y |
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21. f(x, y)= |
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1, if x + y = 0. |
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x y |
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22. f(x, y) = x |
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0, if x = x = 0. |
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Find the points or lines where z is not continuous.
23. z = |
x + y |
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24. z = |
x − y |
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x3 + y3 |
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x2 + y2 |
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25. z = sin(x + y) . sin( y − x)
Answers
1. Fig. 1.8. |
2. Fig. 1.9. 3. Fig. 1.10. 11. Does not exist. 12. 0. 13. ∞ . |
14. 0. |
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15. |
1/ 2 . |
16. 0. 17. 1. 18. 0. 19. Continuous at |
all points, except |
points |
on line |
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у = |
х. 20. Continuous at all points, except (0; 0). 21. Continuous |
at all |
points. |
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22. |
Continuous |
at all points. 23. (0; 0) is not continuous. 24. On line у = –х is not |
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continuous. 25. |
There are the plural of straight lines |
y = x + πn,n Z |
where z is not |
continuous.
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у |
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у |
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у=х2–3х+2 |
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у=х |
2 |
1/2 |
х2+ 4у2=1 |
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1 х |
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2 |
х |
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х=1 |
х |
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Fig. 1.9 |
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Т.1 |
Individual test problems |
1.1. Find and graph the domain on the x,y-plane of the following functions.
1.1.1. z = ln(x − y) arccos(x2 + y2 ) . 1.1.2. z = x2 − y 2 ln x .
1.1.3. z = |
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1.1.11. z = ln sin x + ln cos y .
1.1.13. z = arcsin |
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1.1.15. z = y − − x + ln(4− y2 ) .
1.1.17. z = 9 − x2 − y 2 arccos y .
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1.1.12.z = ln(sin x sin y) .
1.1.14.. z = 4 x2 + y2 −1 . 16−x2 − y2
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1.1.22. |
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14
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1.1.23. z = ln((x + y) ln( y − x)) . |
1.1.24. z = |
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1.1.27. z = ctg πx +arccos |
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1.1.29. z = ((x 2 + y 2 − 1)(x + y))1/ 2 . |
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Topic 2. Partial derivatives and differentials of a function of several variables
The partial and total changes and the partial derivatives of a function of two variables. The differential of a function of several variables. Applications of the differentials. The partial derivatives and the differentials of a function of more than two variables.
Literature: [2, section 1, § 1.2], [3, ch . 6, § 2], [4, section 6, § 16], [6, section 6, ch. 6.1], [7, section 8, §§5––12], [9, § 44].
Т.1 |
Main concepts |
2.1. The partial and total changes of a function of several variables
Let z = f(x, y) by a function and M (x, y) D by a point on the region D.
The problem is to estimate the difference between f(x + x, y + y) and f(x, y), the lengths of the vertical segments shown in fig. 1.2. (For convenience, assume
that both function values are positive.) Denote the difference by f or z and is called the total change:
f = z = f(x + x, y + y) – f(x, y).
If only the x coordinate changes from x to x+ x and y is constant the difference of f, namely x, changes and is denoted by xf or xz:
x z = f (x + x, y) − f (x, y) . Likewise, if only the y coordinate changes:
y z = f (x, y + y) − f (x, y) .
15
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The total change |
z is a function of two variables x |
and y , x z is a |
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x ; y z is a function of one variable |
y . Therefore, |
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2.2. Partial derivatives
Let f be a function of x and y. The graph of z = f(x, y) is a surface. Consider a point (x0, y0) in the xy plane. The graph of z = f(x, y) for (x, y) near (x0, y0) may look like the surface in fig 1.11. Let (x0, y0, f(x0, y0)) be a point on the surface directly above (x0, y0). The plane P through (x0, y0) perpendicular to the y axis meets the surface in a curve z = f(x, y0).
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Fig. 1.11 |
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is the partial derivative of f with respect to y at (x0, y0).
A quantity may depend on more than two variables. For instence, the volume of a box dependens on three variables: the length s, width n, and heigth h, V = snh.
16
http://vk.com/studentu_tk, http://studentu.tk/
Procedure to Find ∂∂xz and ∂∂yz
To find ∂∂xz , treat y as a constant and differentiate f with respect to x in the usual way.
To find ∂∂yz , treat x as a constant and differentiate f with respect to y in the usual way
The notions and notations of partial derivatives carry over to functions of more than two variables. If u = f (x1 , x2 ,…, xm ) , there are m first order partial
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moreover |
∂F ≠ 0 |
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∂u |
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, i = 1, 2, …, m . |
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For instance, if F(x, y, z(x, y)) = 0 , hence |
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∂z |
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Let z = f(x, y) have continuous partial derivatives |
z′ |
and |
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z′ |
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for all points within some disk with center at the point |
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(x, y). Then z, which is the change f(x + |
x, y + |
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x + ε2 |
y, |
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functions of the four variables x, y, |
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2.3. The differential |
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When |
x and y are small, the quantities ε1 |
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y, being the products |
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of small quantities, are usually negligible when compared with |
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x and |
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y |
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y is often a good |
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estimate of |
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If z is a function of two variables, and x, y, |
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is called the differential of z at x, y, |
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dz = |
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http://vk.com/studentu_tk, http://studentu.tk/
In the case of a function of one variable the differential dy represents the change in y along a tangent line to the graph of the function. In the case of a function of two variables it turns out that the differential dz represents the change in z along a “tangent plane” to the graph of a function.
By the formula (1.4):
z = dz + ε1 x + ε2 y.
Hence
z ≈ dz
or
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f (x + x, y + |
y) ≈ f (x, y) + |
∂f (x, y) |
x + |
∂f (x, y) |
y . |
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Let u = f (x1 , x2 ,…, xm ) is a |
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differential is |
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du = |
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dx1 + |
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dx2 |
+ …+ |
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dxm . |
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The chain rules. The theorem in Sec. 2.2 is the basis for the chain rules for differentiating composite functions of more than one variable. Theorem 1 of this section concerns the case in which z is a function of x and y, and x and y are functions of one variable. Theorem 2 concerns the case in which z is a function of x and y, and x and y are, in turn, functions of two variables.
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Chain rule. Let z = f(x, y) have continuous partial |
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derivatives z |
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dz = |
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Theorem 2 |
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∂x + |
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The differential |
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du + dx dv , dy = |
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http://vk.com/studentu_tk, http://studentu.tk/
2.4. Higher partials and differentials |
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If z = f(x, y), then not only is z a function of x and y, but also |
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and z′ |
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second-order partial derivative of z. Symbolically, |
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′′ |
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(zx ) x , |
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They are also denoted |
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or |
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and so on. |
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∂x 2 |
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Note that to find |
z′′ |
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∂ 2 z |
first differentiate z with respect to x. |
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xy |
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∂x∂y |
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For most functions met in practice the two “mixed partials” z′′ |
and |
z′′ |
are |
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xy |
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equal. In view of the importance of this remark it is stated as a theorem. |
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Mixed |
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partial. If z = f(x, y) has continuous partial |
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Theorem |
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derivatives |
z′ , |
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, z′′ |
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x |
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yx |
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z′′ |
= z′′ . |
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xy |
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For a function |
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u = f (x1 , x2 ,…, xm ) |
exist |
m2 second-order |
partial |
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∂ 2u |
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∂ |
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∂z |
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derivatives: |
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= |
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i, k = 1, 2, …, m . |
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∂xk ∂xi |
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∂xk |
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∂xi |
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We can extend our notation beyond second-order partial derivatives. |
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′′′ |
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∂ 3 z |
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(or |
∂x2∂y ) is a third-order partial derivative of z. It is the |
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For example, zxxy |
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′′ |
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∂ 2 z |
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partial derivative of |
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(or ∂x2 |
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zxx |
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Second-order differential d 2 z |
of a function |
z = f (x, y) at point |
M (x, y) is |
called the differential of the first-order differential dz , thus d 2 z = d (dz) .
20
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