Question

1 If f(x, y) = tan-1(xy), find an approximate value of f(1.1,0.8) using the Taylor's
series linear approximation​

Answer 1

Given :

f(x,y) = tan⁻¹(xy)

f(a,b)=f(1.1,0.8)

To find :

L(x,y)

Solution :

L(x,y) = f(a,b) + fₓ(a,b)(x-a) + fy(a,b)(y-b)

f(a,b) = tan⁻¹(1.1 * 0.8)

f(a,b) = tan⁻¹(0.88)

$fx = \frac{1}{1 + x^{2}y^{2} } *y\\fx = \frac{y}{1 + x^{2} y^{2} }$

fₓ(a,b) = 0.451

$fy = \frac{1}{1+x^{2} y^{2} } *x\\fy = \frac{x}{1+x^{2} y^{2} }$

fy(a,b) = 0.62

L(x,y) = f(a,b) +fₓ (a,b)(x-a) + fy(a,b)(y-b)

L(x,y) = tan⁻¹(0.88) + 0.451(x - 1.1) + 0.62(y - 0.8)

L(x,y) = tan⁻¹(0.88) + 0.451x - 0.496 + 0.62y - 0.496

L(x,y) = tan⁻¹(0.88) + 0.451x + 0.62y