Find linear approximations and approximations of the function. 1. Find the linea

Question

Find linear approximations and approximations of the function.

1. Find the linear approximation of the function below at the indicated point. f(x, y) = square root 26-x^2-4y^2 at (1, 2) f(x, y) is approximately equal to Use this approximation to find f(0.92, 2.05). (Round the answer to three decimal places.) f(0.92, 2.05) is approximately equal to 2. Find the linear approximations of the function below and at the indicated point. F(x, y) = in(x ? 3y) at (4, 1) f(x, y) is approximately equal to Use the approximation to find f(3.94, 1.08). (Round the answer to two decimal places.) f(3.94, 1.08) is approximately equal to

Explanation / Answer

1.

Linear approximation is given by

f(X, Y) = f(xo, yo) + f_x(xo, yo) (x - xo) + f_y(xo, yo) (y - yo).............(1)

f(x, y) = (26 - x2 - 4y2)1/2
xo = 1, y0 = 2
f_x = 1/2(26 - x2 - 4y2)-1/2(-2x)
f_y = 1/2(26 - x2 - 4y2)-1/2(-2y)
f(1, 2) = 3
f_x(1, 2) = 1/3
f_y(1, 2) = 2/3

putting these values in (1)
f(X, Y) = 3 + 1/3(x - 1) + 2/3(y - 2)
f(X, Y) = 3 + x/3 - 1/3 + 2y/3 - 4/3
f(X, Y) = x/3 + 2y/3 + 4/3................(2)
f(0.92, 2.05) = 0.92/3 + 2(2.05)/3 + 4/3
f(0.92, 2.05) = 3.0066

2.

f(X, Y) = f(xo, yo) + f_x(xo, yo) (x - xo) + f_y(xo, yo) (y - yo).............(1)
f(x, y) = ln(x - 3y)
xo = 4, yo = 1
f_x = 1/(x - 3y)
f_y = -3/(x - 3y)
f(4, 1) = ln1 = 0
f_x(4, 1) = 1
f_y(4, 1) = -3
Putting these values in (1)
f(X, Y) = 0 + 1(x - 4) - 3(y - 1)
f(X, Y) = x - 4 - 3y + 3
f(X, Y) = x - 3y - 1
f(3.94, 1.08) = 3.94 - 3.24 - 1
f(3.94, 1.08) = -0.3

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