Let g : R 2 → R be totally differentiable in the point a ∈ R 2 Let v 1 := 1 2 ( 1 1 ) and v 2 := 1 5 ( 4 − 3 ) be normalized direction vectors in R 2 It is ∂ g ∂ v 1 ( a ) = 5 2 and ∂ g ∂ v 2 ( a ) = 1With the info given above, how can one find out ∇ g ( a )?

Question

Let   g  :            R      2        →      R   be totally differentiable in the point   a  ∈            R      2      Let       v    1    :=      1          2            (                            1                                      1                      )   and       v    2    :=      1    5        (                            4                                      −          3                      )   be normalized direction vectors in             R      2      It is             ∂      g              ∂              v        1              (  a  )  =  5      2   and             ∂      g              ∂              v        2              (  a  )  =  1With the info given above, how can one find out   ∇  g  (  a  )?

Gauge Howard · Accepted Answer

Given that       v    1   and       v    2   are unit vectors, the the direction derivative gives      ∇                  v        2              g  =  ∇  g  ⋅      v    2  and      ∇                  v        2              g  =  ∇  g  ⋅      v    2  Use these to build a linear system of 2 equations in the 2 components of   ∇  g, and solve. So, if  ∇  g  =  [  x  ,  y      ]    T  , then            1              2              x  +            1              2              y  =  5      2  and            4      5        x  −            3      5        y  =  1After multiplying the first equation through by       2   and second equation by 5, they becomes  x  +  y  =  10  4  x  −  3  y  =  5It is trivial to solve to get x=5 , y=5Therefore,   ∇  g  =      [                            5                                      5                      ]

Gradient with v_1:=1/(\sqrt(2))((1),(1)) and v_2:=1/(5)((4),(-3)) given

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