MATHEMATICAL PROBLEM PAGE Directed by EDWARD L. HAENISCH Montana State College, Bozeman, Montana
I
F u is a function of two .or more variables, i. e., u = f( x,y,z . . . ), ( a u l b x ) , , . . is obtained by diierentiating u with respect to x keeping y,e, . constant. The total differential d u is:
..
It is sometimes convenient to write this statement in the form:
2
=
.
+
2
dz +(
&
) , y . .
-
).(
The higher partials are independent of the order of differentiation, i. e.,
-2
REFERENCES
+
PROBLEMS
1. Find du if u is (a)
sin (x
+ y)
(h) 9
(c) yez
+
+
+
0.526) Z(0.833 0.714 0.625 0.555)lO.l = 0.6930. The In 2 is 0.6932. 3. A count of the squares under the curve as shown in Figure 2 gives 0.217 for A S for one gram or 32 X 0.217 = 9.23 for m e mole. Care must be taken to apply the proper conversion factor from logarithms to the base 10 to natural logarithms (or the plot may be made directly with In T). Wben nvves are plotted on a scale such as has been used here sometimes an error is made by not including the total area down to the .. n of t h c.. ~ axi.i.
DANIELS,p. 174 LEWISAND RANDALL,p. 27 MELLOR,p. 68
+ xeu
2. Show that (C) is true for (a)
raya
+
+ 4xy'
(h) ei sin y
3. If x8y y2x - xy = 0 , finddyldx. (Hint: Use (A) but recall that here du = 0.)
C*
4.
=
rg)*+ 9 (g)* rnd
and show that fo. a perfect sume that U = f(T, gas Cp - Ca = R. A perfect gas is one which obeys
pV
=
s+o
..
nRT and for which SOLUTIONS TO NOVEMBER PROBLEMS
! 25 ; :
+
+ +
+
Z '
+ + + +
+ +
4. d L
1
+
+
+
+
:
;,*I
:
:
;
:
I
:
:
;
;
35,--,
FIGWEE 2 dL2 = -
2 19.64
( d ) ~ l x ~ d ~ = - ] -= l6 ---=-= l5 3.75 4 , 4 4 4 In order of increasing accuracy the rank is trapezoidal rule, counting squares and Simpson's Rule. (Figure 1.) 2. Application of Simpson's Rule with intervals of & = 0.1 gives: '/a[(l.OOO 0.500) 4(0.909 0.769 0.667 0.588
+
:
fog
1. (a) A count of the squares gives a value of 3.65 for the area. (b) [I/%(-1) (-0.125) 0 0.125 1 3.375 '/.(8)] 0.5 = 3.94 ( c ) I/d(-l 8) 4(-0.125 0.125 3.375) I ) ] 0.5 = 3.75 2(0
+ +
;
+ 592
= -
m,Z
dZ;
= 0wehaveL =
the area under a m
-
:
40
1
:
9.64
;
:
I
:
45
N,
d z . Since at
84
ddZI. We must determine
e plotted with % values as abscissas and
5 values as ordinates. N'
I
Let us make the assumption that
- = 154 can be neglected. This. of the area past a value of NL N2
top c-5 A count of the squares gives = -654 (the value of La determined by more exact means is -605).
Since Y, is the value of y a t x = a t x = Ax we have y, = ao - a 1 A x + & A x X YI = ao ya = ao alAx aazAxl
+
+
and we have a. = y2 and Q = Yt
- Az, y,
at x = 0 and ya
+ Y, - 2y2 2Axa
By the use of these values of a0 and ar in (A) we have Ax Area for fust two = (Y, 4y2 ys) intervals If we apply the same process to the next two intervals (or for ordinates y,, y,, and yJ we have Area for az 3rdand4tb = 7 (ya 4y, yd intervals Or, by the addition of the &ea for the four intervals.
7
+ +
+ +
course, introduces a distinct error. The cross-hatched area (Figure 3) must be subtracted from the total area under the
AREA =
A%
3
(YS
+ 4yn + 2% + 4 ~ 4+ yd
