How do you find the linear approximation L to f at the designated point P. compare the error in approximating f by L at the specified point Q with the distance between P and Q given #f(x,y) = 1/sqrt(x^2+y^2)#, P(4,3) and Q(3.92, 3.01)?

Question

Luke Alvarez · Accepted Answer

#d = 0.0000142382#

The tangent plane #Pi# to a Surface #S# in #p_0 in S# is obtained as

#Pi->(p-p_0).vec n_0 =0#

where

#p = (x,y,z)# and #vec n_0# is the normal vector to #S# at #p_0#.

The normal vector to #S# is computed as

#vec n = grad S = ((partial S)/(partial x),(partial S)/(partial y),(partial S)/(partial z))#.

Giving

#S(x,y,z)=z-1/sqrt(x^2+y^2) = 0# then
#vec n = (x/(x^2 + y^2)^(3/2), y/(x^2 + y^2)^(3/2), 1)#
At #p_0 = (4,3,1/sqrt(4^2+3^2))# gives
#vec n_0 = (4/125, 3/125, 1)#

so the tangent plane reads

#Pi->4/125 (x-4) + 3/125 ( y-3) + z-1/5 = 0#

Given a point #q_S = (3.92, 3.01, 0.202334) in S#
and a point #q_{Pi} = (3.92, 3.01, 0.20232) in Pi#
Their distance is #norm(p_S-p_{Pi}) = 0.0000142382#

Cameron Alexander · Answer

undefined To find the linear approximation $ L $ to $ f $ at the designated point $ P $, first, compute the partial derivatives $ f_x $ and $ f_y $ of $ f $ at $ P $. Then, use the formula:

$$ L(x,y) = f(P) + f_x(P)(x-x_P) + f_y(P)(y-y_P) $$

where $ (x_P, y_P) $ are the coordinates of point $ P $.

The error in approximating $ f $ by $ L $ at the specified point $ Q $ can be approximated by the absolute difference between $ f(Q) $ and $ L(Q) $.

To compare the error in approximating $ f $ by $ L $ at $ Q $ with the distance between $ P $ and $ Q $, compute the distance between $ P $ and $ Q $ using the distance formula:

$$ \text{Distance} = \sqrt{(x_Q-x_P)^2 + (y_Q-y_P)^2} $$

Then, compare this distance with the error obtained above.