1. ## directional derivative

Hello everyone,

Could you please tell me if this is correct?

FInd the directional derivative using f(x, y, z)=xy+z^2 as you leave the point (1,1,0) heading in the direction of the point (0,1,1)

u=-i+0j+1k

fx=y (0,1.1)--> 1

fy=x --> 1

fz=2z-->0

u=v/magnitude= -i+1k/square root (-1^2+1^2)= -1i+k/square root (2)

directional derivative=i+j(-1/square root of 2i+1/square root of 2k)

(i+j)(-1/square root 2i+1/square root of 2k)

Thank you very much

2. Originally Posted by Chocolatelover2
Hello everyone,

Could you please tell me if this is correct?

FInd the directional derivative using f(x, y, z)=xy+z^2 as you leave the point (1,1,0) heading in the direction of the point (0,1,1)

u=-i+0j+1k

fx=y (0,1.1)--> 1

fy=x --> 1

fz=2z-->0

u=v/magnitude= -i+1k/square root (-1^2+1^2)= -1i+k/square root (2)

directional derivative=i+j(-1/square root of 2i+1/square root of 2k)

(i+j)(-1/square root 2i+1/square root of 2k)

Thank you very much
no

the directional derivative of a differentiable function $\displaystyle f(x,y,z)$ at the point $\displaystyle (x_0,y_0,z_0)$ in the direction of a unit vector $\displaystyle \bold{u}$ is given by

$\displaystyle D_uf(x_0,y_0,z_0) = \nabla f \cdot \bold {u}$

3. Thank you very much

Do you see where I went wrong? The answer is -1/square root of 2 and I got -1/square root of 2+1/square root of 2.

Thank you

4. Originally Posted by Chocolatelover2
Thank you very much

Do you see where I went wrong? The answer is -1/square root of 2 and I got -1/square root of 2+1/square root of 2.

Thank you
i believe i see where you went wrong, but i might not be sure. what you did wasn't very coherent and confused me.

you have $\displaystyle \bold{u} = \left< \frac {-1}{\sqrt{2}}~,~ 0~ ,~ \frac 1{\sqrt{2}}\right>$

and $\displaystyle \nabla f (1,1,0) = \left< 1,1,0 \right>$

now just take the dot product