Let $f(x,y,z) = x^2 -2y + 3z^2$
I need to find the direction derivative of f at $(0,1,0)$ in the direction $v=[1,1,1]$.
The formula given is
f(a + tv) - f(a) / t|v| = f( (0,1,0) + t[1,1,1] ) - f(a) / tv
= f(t, 1+t, t) - f(0,1,0) / tv
I'm not sure how (0,1,0) + t[1,1,1] = t, 1+t, t? If expanded, we get (0,1,0) + [t,t,t]. How was this part done (since it is a point + a vector)? Is it just [0 + t, 1 + t, 0 + t]?
Best Answer
Given $f(x,y)=x^2-2y+3z^2$
The unit vector in the direction of $\vec v=\dfrac{\vec v}{||\vec v||}=\left<\dfrac{1}{\sqrt{3}},\dfrac{1}{\sqrt{3}},\dfrac{1}{\sqrt{3}}\right>$
The partial derivatives are $$f_x=2x,f_y=-2,f_z=6z$$ $$f_x(0,1,0)=0,f_y(0,1,0)=-2,f_z(0,1,0)=0$$
Then the directional derivative is$$D_{\vec u}f(0,1,0)=0\cdot\left(\dfrac{1}{\sqrt{3}}\right)-2\left(\dfrac{1}{\sqrt{3}}\right)+0\cdot\left(\dfrac{1}{\sqrt{3}}\right)=-\dfrac{2}{\sqrt{3}}$$