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#