1

The directional derivative of the scalar function $\log(x^2+y^2+z^2)$ at point P(1,1,1) in direction of line joining p to $p_0(3,2,1)$ is

parametric equation of line joining p to $p_0$ is $\overrightarrow{v}=(2t+1)i+(t+1)j+k$

unit vector in direction of $\overrightarrow{v}$=$\frac{(2t+1)i+(t+1)j+k}{\sqrt{5t^2+6t+3}}$

$\nabla{f} at (1,1,1)=\frac{2}{3}(i+j+k)$

${D^f}_\overrightarrow{v}$=$\nabla{f}.\overrightarrow{v}$

but how to take dot product at $(1,1,1)$?

Kenta S
  • 16,151
  • 15
  • 26
  • 53
dipali mali
  • 555
  • $\vec v$ doesn’t point in the direction of the line. That should be a constant! – amd Apr 27 '18 at 17:50

1 Answers1

1

The directional derivative for $f(x,y,z)$ in the direction $\vec v$ is computed as

$$ \left <\nabla f, \vec v \right > $$

with $\vec v = (v_x,y_y,v_z)$ Here

$$ \nabla f = 2\left(\frac{(x,y,z)}{x^2+y^2+z^2}\right) $$

and

$$ \left <\nabla f, \vec v \right > = 2\left(\frac{x v_x+y v_y+z v_z}{x^2+y^2+z^2}\right) $$

and at point $(x_0,y_0,y_0)$

$$ 2\left(\frac{x_0 v_x+y_0 v_y+z_0 v_z}{x^2_0+y^2_0+z^2_0}\right) = 4 $$

Cesareo
  • 33,252