The actual change of z is delta z = delta f = f(x, y) - f(x_0, y_0) approximatel

Question

The actual change of z is delta z = delta f = f(x, y) - f(x_0, y_0) approximatelyequalto dz. That is, the change in z is approximated by the differential, dz dz = f_x(x_0, y_0)(x - x_0) + f_y(x_0, y_0)(y - y_0) = f)x(x_0, y_0) dx + f_y(x_0, y_0) dy Given the function f(x, y) = squareroot 20 - x^2 - 7y^2 at the point (2, 1). Use differentials to approximate f(1.95, 1.08). (a) f(2, 1) = ____ (b) f_s(2, 1) = ____ (c) f_y(2, 1) = ____ (d) dx = ____ What is the sign of dx? ____ (e) dy = ____ (f) Use the results of (a) through (e) to evaluate df = ____ (g) Calculate the actual change in f, delta f = f(x, y) - f(x_0, y_0) (h) Compare the actual change and the differential as an approximation.

Explanation / Answer

Given f(x,y) = ( 20 - x2-7y2) 1/2

f(2,1) = ( 20- 4-7)1/2 =3

f( 1.95 , 1.08 ) = f(2,1) +dz = f(2,1)+ fx(2,1) dx+ fy(2,1) dy ----(1)

fx= 1/2 [ 20-x2-7y2]-1/2 x-2x at (2,1) fx= -2/3   

fy at (2,1) = -7/3 , substituting thses values in (1)

f ( 1.95, 1.08) = 3 -(2/3) ( 1.95-2) -(7/3) ( 1.08 -1)

= 3+0.1/3 - 0.56/3

=3 - 0.46/ 3 = 3 - 0. 15= 2.85

b. f(2,1) = (20 - 4-7)1/2 = (9)1/2=3

c . f x = - 2 ( 20-4-7) -1/2 = - 2/3

d . f y = - 7 ( 20 -4-7)-1/2 = -7/3

e dx = 1.95 - 2= 0.05 , dy = 1.08 -1 =0.08

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