Exercicios resolvidos James Stewart vol. 2 7ª ed - ingles

SECTION 14.6 DIRECTIONAL DERIVATIVES AND THE GRADIENT VECTOR 0 213
(b) Differentiating both sides of f(tx, ty) = t" f(x, y) with 'respect tot using the Chain Rule, we get
8 a
8 t f(tx, ty) = t [t" f(x,y)] ¢:>
8
8 8(tx) 8 8(ty) 8 a -1
8 (tx) f(tx, ty) · {j't + 8 (ty) f(tx, ty) · at = x a(tx ) f(tx, ty) + y 8
(ty) f(tx, ty) = nt" f(x , y).
8
8
Setting t = 1: x f(x , y) 8
+ y f (x, y) 8
= nf(x, y).
X . y .
57. Differentiating both sides of f(tx, ty) = tn f(x, y) with respect to x using the Chain Rule, we get
8 8
ax f(tx,ty) = 8x [t"'f(x, y)] ¢:>
8 8(tx) 8 8(ty) n 8
a (tx) f.(tx , ty) . a;;- + 8 (ty) f(tx , ty). a;:- = t 8x f(x, y) ¢:>
tfx(tx, ty) = tn f,.(x, y).
Thus f ,.(tx, ty) = t"·- 1 J,.(x, y).
59. Given a function defined implicitly by F(x, y) = 0, where F is differentiable and F 11 =J: 0, we know that ~~ = - F,. . Let
. ~ ~
G(x, y) = - F,. so ddy = G(x, y). Differentiating both sides with respect to x and using the Chain Rule gives
Fy X
d 2 y = 8G dx + 8G dy where 8G = ,E_ ( - Fx) = _ F11Fxx - F,.F11., , 8G = ,E_ (-F.,) =
dx2 8x dx ay dx 8x 8x Fy FJ ay 8y Fy
T~us
d 2 y = ( FuF:.,,. - F,.Fyx) (l)
2
+ (- FyFxll - FxF,m) (-Fa;)
dx 2 F
2
11
F 11
F 11
FxxF; - FyxFxFy - F,.uF11Fx + FyyF;
FJ
But F has continuous second derivatives, so by Clauraut's Theorem, F 11 ,
= F., 11
and we have
d211 F.,.,F; - 2F., 11 F, Fu + FuuF; d . ed
dx 2 = - F. 3 as es1r .
II
14.6 Directional Derivatives and the Gradient Vector
1. We can approximate the directional derivative of the pressure function at K in the direction of S by the average rate of change
of pressure between the points where the red line intersects the contour lines closest to K (extend the red line slightly at the
left). In the direction ofS, the pressure changes from 1000 millibars to 996 millibars and we estimate the distance between
these two points to be approximately 50 km (using the fact that U1e distance from K to S is 300 km). Then the rate of change of
pressure in the direction given is approximately 996 ; 0
1000
= -0.08 millibar /km.
3. Du /{ -20, 30) = 'V j { -20, 30) · U = /T{ -20, 30) ( 7z) + /v{ - 20, 30) ( ?z).
/T( - 20, 30) ~ lim !( - 20 + h, 30 )-!( - 20 • 30 ), so we can approximate /T( -20, 30) by considering h = ±5 and
h -+0
t
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