Thread: 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)

gradient (1,0,0)=i+j+0k

direction derivative= gradient(u)

Thank you very much

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)

gradient (1,0,0)=i+j+0k

direction derivative= gradient(u)

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

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

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

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