Question

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)?

Answer 1

`d = 0.0000142382`

Explanation:

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`