How do you use the linear approximation to $f(x, y)=(5x^2)/(y^2+12)$ at (4 ,10) to estimate f(4.1, 9.8)?

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Answer 1 · 2016-07-15T20:57:05

Tangent plane approximation for $f(4.1,9.8) = 0.77551$

The tangent plane to $f(x,y)-z=0$ in $p_0 = {x_0,y_0,f(x_0,y_0}$ can be obtained by doing

$Pi_0 = << p-p_0,vec n_0>> = 0$

where $p = {x,y,z}in Pi_0$ is a generic point and $vec n_0$ is the normal vector to the surface $f(x,y)-z=0$ at point $p_0$

but $vec n_0 = grad (f(x,y)-z)_0 = {f_x,f_y,-1}_0$

so

$vec n_0 = {(10 x_0)/(12 + y_0^2), -(10 x_0^2 y_0)/(12 + y_0^2)^2,-1}$

or

$vec n_0 = {5/14, -25/196,-1}$

given that $p_0 = {4,10,5/7}$ the tangent plane is

$55/98 + (5 x)/14 - (25 y)/196 - z=0$

Calculating the approximation for $f(4.1,9.8)$ gives

$z=55/98 + (5 xx4.1)/14 - (25xx 9.8)/196 =0.77551$

and

$f(4.1,9.8) = 0.777953$

How do you use the linear approximation to f(x, y)=(5x^2)/(y^2+12)f(x, y)=(5x^2)/(y^2+12) at (4 ,10) to estimate f(4.1, 9.8)?