 #1
Poetria
 267
 42
 Homework Statement:

##f(x,y) = x^3+x*y^2+y^3##
Then
f(1,1)=3
##f_x (1,1) = 4##
##f_y (1,1) = 5##
##\nabla f(1,1) = (4,5)##
Find the closest point to (1,1) at f(x,y) = 3.01, i.e. ##(1+\Delta x, 1+\Delta y)##
 Relevant Equations:
 Linear approximation
I tried to use a Taylor series expanded at 3 and set to 3.01:
https://www.wolframalpha.com/input/?i=27+++9+(3+++x)^2+++(3+++x)^3+++3+y^2+++y^3+++(3+++x)+(27+++y^2)=3.01
I got the vector ## (\Delta x, \Delta y)= (0.37887, 0.54038)##
It does give a desired result but it is marked as wrong.
There is also a question about what statement is correct:
##(1+\Delta x, 1 + \Delta y)## is nearly parallel or perpendicular to ##(4,5)##
Or
##(\Delta x, \Delta y)## is nearly parallel or perpendicular to ##(4,5)##
I thought the vector (0.37887, 0.54038) was nearly perpendicular. Well...
https://www.wolframalpha.com/input/?i=27+++9+(3+++x)^2+++(3+++x)^3+++3+y^2+++y^3+++(3+++x)+(27+++y^2)=3.01
I got the vector ## (\Delta x, \Delta y)= (0.37887, 0.54038)##
It does give a desired result but it is marked as wrong.
There is also a question about what statement is correct:
##(1+\Delta x, 1 + \Delta y)## is nearly parallel or perpendicular to ##(4,5)##
Or
##(\Delta x, \Delta y)## is nearly parallel or perpendicular to ##(4,5)##
I thought the vector (0.37887, 0.54038) was nearly perpendicular. Well...