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2. Partial Differentiation
2A. Functions and Partial Derivatives
2A-1 In the pictures below, not all of the level curves are labeled. In (c) and (d), the
picture is the same, but the labelings are different. In more detail:
b) the origin is the level curve 0; the other two unlabeled level curves are .5 and 1.5;
c) on the left, two level curves are labeled; the unlabeled ones are 2 and 3; the origin is a
the level curve 0;
d) on the right, two level curves are labeled; the unlabeled ones are −1 and −2; the origin
is the level curve 1;
The crude sketches of the graph in the first octant are at the right. b
2
-2
1
4
1 0
-3
2
1
-1
1
1
0
2
0
0
-1
c
3 -2
2 1
a b c, d e
2 2 3 1 x d
2A-2 a) fx = 3x y −3y ; fy = x −6xy +4y b) zx = y; zy = −y2
c) fx = 3cos(3x + 2y); fy = 2cos(3x + 2y)
2 2 2x x
d) fx = 2xyex y; fy = x2ex y e) zx = ln(2x + y) + ; zy =
2x + y 2x + y
3 2 2
f) fx = 2xz; fy = −2z ; fz = x − 6yz
2A-3 a) both sides are mnxm−1yn−1
b) fx = y ; fxy = (fx)y = x − y ; fy = −x ; fyx = −(y −x).
(x + y)2 (x + y)3 (x + y)2 (x + y)3
c) f = −2xsin(x2 + y); f =(f ) = −2xcos(x2 + y);
x 2 xy x y 2
fy = −sin(x + y); fyx = −cos(x + y) · 2x:
d) both sides are f′ (x)g ′ (y).
2A-4 (fx)y = ax+6y; (fy)x = 2x+6y; therefore fxy = fyx ⇔ a = 2. By inspection,
one sees that if a = 2, f(x;y) = x2y +3xy2 is a function with the given f and f .
x y
2A-5
a) wx = aeax sinay; wxx = a2eax sinay;
wy = eaxacosay; wyy = eaxa2(−sinay); therefore wyy = −wxx.
2x 2(y2 −x2)
b) We have wx = ; wxx = . If we interchange x and y, the function
x2 + y2 (x2 + y2)2
w = ln(x2 + y2) remains the same, while w gets turned into w ; since the interchange
xx yy
just changes the sign of the right hand side, it follows that wyy = −wxx.
2B. Tangent Plane; Linear Approximation
2
2B-1 a) zx = y ; zy = 2xy; therefore at (1,1,1), we get zx = 1; zy = 2, so that the
tangent plane is z = 1 + (x − 1) + 2(y − 1), or z = x + 2y − 2.
0
2. PARTIAL DIFFERENTIATION 1
2 2
b) wx = −y =x ; wy = 2y=x; therefore at (1,2,4), we get wx = −4; wy = 4, so
that the tangent plane is w = 4 − 4(x − 1) + 4(y − 2), or w = −4x + 4y.
x x y
2B-2 a) zx = � 2 2 = z; by symmetry (interchanging x and y), zy = z ; then the
x + y
x0 y0 x0 y0 2 2 2
tangent plane is z = z0 + z0 (x−x0)+ z0 (y−y0), or z = z0 x+ z0 y , since x0 +y0 = z0.
b) The line is x = x0t; y = y0t; z = z0t; substituting into the equations of the cone
and the tangent plane, both are satisfied for all values of t; this shows the line lies on both
the cone and tangent plane (this can also be seen geometrically).
2B-3 Letting x;y;z be respectively the lengths of the two legs and the hypotenuse, we
� 2 2
have z = x + y ; thus the calculation of partial derivatives is the same as in 2B-2, and
we get Δz ≈ 3Δx + 4 Δy. Taking Δx = Δy = :01, we get Δz ≈ 7 (:01) = :014:
5 5 5
2B-4 From the formula, we get R = R1R2 . From this we calculate
R1 + R2
� � � �
∂R R2 2 ∂R R1 2
∂R = R + R , and by symmetry, ∂R = R + R .
1 1 2 2 1 2
Substituting R = 1; R = 2 the approximation formula then gives ΔR = 4ΔR + 1 ΔR :
1 2 9 1 9 2
By hypothesis, |ΔR | ≤ :1; for i = 1;2, so that |ΔR| ≤ 4(:1) + 1(:1) = 5 (:1) ≈ :06; thus
i 9 9 9
R = 2 = :67 ± :06.
3
2
2B-5 a) We have f(x;y) = (x+y+2) ; fx = 2(x+y+2); fy = 2(x+y+2). Therefore
at (0;0), fx(0;0) = fy(0;0) = 4; f(0;0) = 4; linearization is 4 + 4x + 4y;
at (1;2), fx(1;2) = fy(1;2) = 10; f(1;2) = 25;
linearization is 10(x − 1) + 10(y − 2) + 25, or 10x + 10y − 5.
b) f = ex cosy; fx = ex cosy; fy = −ex siny .
linearization at (0;0): 1+x; linearization at (0;π=2): −(y − π=2)
2B-6 We have V = πr2h, ∂V = 2πrh; ∂V = πr2; ΔV ≈ � ∂V � Δr + � ∂V � Δh.
∂r ∂h ∂r 0 ∂h 0
Evaluating the partials at r = 2; h = 3, we get
ΔV ≈ 12πΔr +4πΔh:
Assuming the same accuracy |Δr| ≤ ǫ; |Δh| ≤ ǫ for both measurements, we get
|ΔV| ≤ 12πǫ+4πǫ = 16πǫ; which is <:1 if ǫ < 1 <:002 :
160π
� 2 2 −1 y ∂r x ∂r y
2B-7 We have r = x + y ; θ = tan x ; ∂x = r; ∂y = r :
Therefore at (3;4); r = 5, and Δr ≈ 3Δx + 4Δy: If |Δx| and |Δy| are both ≤ :01, then
5 5
3 4 7
|Δr| ≤ |Δx|+ |Δy| = (:01) = :014 (or .02).
5 5 5
Similarly, ∂θ = −y ; ∂θ = x , so at the point (3;4),
∂x x2 + y2 ∂y x2 + y2
2 S. 18.02 SOLUTIONS TO EXERCISES
|Δθ| ≤ |−4Δx|+ | 3 Δy| ≤ 7 (:01) = :0028 (or .003).
25 25 25
Since at (3;4) we have |r | > |r |, r is more sensitive there to changes in y; by analogous
y x
reasoning, θ is more sensitive there to x.
2B-9 a) w = x2(y +1); w = 2x(y +1) = 2 at (1;0), and w = x2 = 1 at (1;0); therefore
x y
w is more sensitive to changes in x around this point.
b) To first order approximation, Δw ≈ 2Δx + Δy, using the above values of the
partial derivatives.
If we want Δw = 0, then by the above, 2Δx + Δy = 0, or Δy=Δx = −2 .
2C. Differentials; Approximations
2C-1 a) dw = dx + dy + dz b) dw = 3x2y2zdx +2x3yzdy + x3y2dz
x y z
2ydx−2xdy tdu − udt
c) dz = (x + y)2 d) dw = t√t2 −u2
2C-2 The volume is V = xyz; so dV = yz dx+xzdy +xydz. For x = 5; y = 10; z = 20;
ΔV ≈ dV = 200dx+100dy +50dz;
from which we see that |ΔV | ≤ 350(:1); therefore V = 1000 ± 35.
2C-3 a) A = 1absinθ. Therefore, dA = 1(bsinθda + asinθdb + abcosθdθ).
2 2 √ √
b) dA = 1(2 · 1 da + 1 · 1 db + 1 · 2 · 1 3dθ) = 1(da+ 1 db+ 3dθ);
2 2 2 2 2 2
therefore most sensitive to θ, least senstitive to b, since dθ and db have respectively the
largest and smallest coefficients.
1 1
c) dA = 2(:02 + :01 + 1:73(:02) ≈ 2(:065) ≈ :03
2C-4 a) P = kT ; therefore dP = k dT − kT dV
V V V 2
b) V dP + P dV = kdT; therefore dP = kdT − P dV .
V
c) Substituting P = kT=V into (b) turns it into (a).
dw dt du dv � dt du dv �
2C-5 a) − = − − − ; therefore dw = w2 + + .
w2 t2 u2 v2 t2 u2 v2
b) 2udu +4vdv +6wdw = 0; therefore dw = − udu +2vdv .
3w
2D. Gradient; Directional Derivative
� √
2 2 df � i − j 3 2
2D-1 a) ∇f = 3x i +6y j; (∇f)P = 3i +6j; � =(3i +6j)· √ =−
ds � u 2 2
�
y x xy dw � i +2j −2k 1
b) ∇w = i + j− k; (∇w)P = −i+2j+2k; � =(∇w)P· = −
z z z2 ds � u 3 3
c) ∇z = (siny −ysinx)i +(xcosy +cosx)j; (∇z)P = i + j;
�
dz � −3i +4j 1
� =(i + j) · =
ds � u 5 5
2. PARTIAL DIFFERENTIATION 3
�
2i + 3j dw � 4i −3j 1
d) ∇w = ; (∇w)P = 2i +3j; � =(2i +3j)· =−
2t + 3u ds � u 5 5
e) ∇f = 2(u + 2v + 3w)(i + 2j + 3k); (∇f)P = 4(i +2j +3k)
�
df � −2i +2j − k 4
� =4(i +2j +3k)· = −
ds � u 3 3
2D-2 a) ∇w = 4i −3j ; (∇w)P = 4i −3j
� 4x − 3y
dw � 4i − 3j
� =(4i −3j)· u has maximum 5, in the direction u = ,
ds � u 5
and minimum −5 in the opposite direction.
�
dw � 3i + 4j
� =0inthe directions ± .
ds � u 5
b) ∇w = �y + z;x + z;x + y�; (∇w)P = �1;3;0�;
� √ � √
dw � i +3j dw � i +3j
max � = 10, direction √ ; min � = − 10, direction − √ ;
ds � u 10 ds � u 10
�
dw � −3i + j + ck
� =0inthe directions u = ± √ (for all c)
ds � u 10+c2
c) ∇z = 2sin(t − u)cos(t − u)(i − j) = sin2(t − u)(i − j); (∇z)P = i − j;
� √ � √
dz � i − j dz � −i + j
max � = 2, direction √ ; min � = − 2, direction − √ ;
ds � u 2 ds � u 2
�
dz � i + j
� =0inthe directions ± √
ds � u 2 3 3 2 2 2
2D-3 a) ∇f = �y z ;2xyz ;3xy z �; (∇f)P = �4;12;36�; normal at P: �1;3;9�;
tangent plane at P: x + 3y + 9z = 18
b) ∇f = �2x;8y;18z�; normal at P: �1;4;9�, tangent plane: x + 4y + 9z = 14.
c) (∇w)P = �2x0;2y0;−2z0�; tangent plane: x0(x − x0) + y0(y − y0) − z0(z − z0) = 0;
or x x + y y − z z = 0; since x2 + y2 − z2 = 0.
0 0 0 0 0 0
2D-4 a) ∇T = 2xi +2yj ; (∇T) = 2i +4j ;
x2 + y2 P 5
i +2j
T is increasing at P most rapidly in the direction of (∇T)P, which is √ :
5
2 i +2j
b) |∇T| = √ = rate of increase in direction √ : Call the distance to go Δs, then
5 5
√ √
2 :2 5 5
√5Δs = :20 ⇒ Δs = 2 = 10 ≈ :22.
�
dT � 2i + 4j i + j 6
c) � =(∇T)P · u = · √ = √ ;
ds � u 5 2 5 2
√ √
6 5 2
√ Δs = :12 ⇒ Δs = 6 (:12) ≈ (:10)( 2) ≈ :14
5 2
2i − j
d) In the directions orthogonal to the gradient: ± √5
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