Question

(3 points) The directional derivative of f(x, y) at (2, 1) in the direction going from (2, 1) toward the point (1, 3) is −2/ √ 5, and the directional derivative at (2, 1) in the direction going from (2, 1) toward the point (5, 5) is 1. Compute fx(2, 1) and fy(2, 1).

Answer 1

The vector pointing from (2, 1) to (1, 3) points in the same direction as the vector
`\vec u=(1,3)-(2,1)=(-1,2)`. The derivative of
`f` at (2, 1) in the direction of
`\vec u` is

`D_(\vec u)f(2,1)=\\abla f(2,1)\cdot(\vec u)/(\|\vec u\|)`

We have

`\|\vec u\|=√((-1)^2+2^2)=\sqrt5`

Then

`D_(\vec u)f(2,1)=(f_x(2,1),f_y(2,1))\cdot((-1,2))/(\sqrt5)=(-f_x(2,1)+2f_y(2,1))/(\sqrt5)=-\frac2{\sqrt5}`

`\implies f_x(2,1)-2f_y(2,1)=2`

The vector pointing from (2, 1) to (5, 5) has the same direction as the vector
`\vec v=(5,5)-(2,1)=(3,4)`. The derivative of
`f` at (2, 1) in the direction of
`\vec v` is

`D_(\vec v)f(2,1)=\\abla f(2,1)\cdot(\vec v)/(\|\vec v\|)`

`\|\vec v\|=√(3^2+4^2)=5`

so that

`(f_x(2,1),f_y(2,1))\cdot\frac{(3,4)}5=1`

`\implies3f_x(2,1)+4f_y(2,1)=5`

Solving the remaining system gives
`f_x(2,1)=\frac95` and
`f_y(2,1)=-\frac1{10}`.