- #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...