# Directional Derivative in the direction in which $z$ is growing

Find the directional derivative of $$f(x, y, z) = xy + 2xz - y^2 + z^2$$, at the point $$P = (1, -2, 1)$$, passing through the curve $$x = t, y = t -3, z = t^2$$, in the direction in which $$z$$ is growing.

The work I've done:

$$\nabla f = (y + 2z, x - 2y, 2x + 2z) \Rightarrow \nabla f \rvert _P = (0, 5, 4)$$

Now I'm not sure of what I'm doing. Substituting the values of $$P$$ in the parametric equation of the curve, you get $$(1, 1, 1)$$. What I mean is,

$$1 = t, -2 = t - 3, 1 = t^2$$

When you solve for $$t$$ in each equation you get $$(1, 1, 1)$$. So my guess is that

$$\vec{v} = (1, 1, 1)$$

Therefore the directional derivative is

$$\nabla f \rvert _P \cdot \frac{\vec{v}}{|\vec{v}|} = \frac{4 + 5}{\sqrt{3}} = 3\sqrt{3}$$

Is that correct?

No, you get $$t= 1$$, not three different values. Each value of $$t$$ gives a point on the curve. $$t= 1$$ gives $$x= 1, y= 1- 3= -2, z= 1^2$$ or the point $$(1, -2, 1)$$ that you were given to begin with. The "directional derivative", also called a "tangent vector" is the function you got, $$(0, 5, 4)$$.
• thanks for clearing it up.. what about the part "in the direction in which $z$ is growing?" – Victor S. Apr 15 at 0:01
• The tangent vector could also be given as $(0,-5,-4)$ but that would not be in the correct direction it would be $z$ decreasing (pointing downwards in the 3d plane). – Peter Foreman Apr 15 at 0:31
• Well the official answer is $13\sqrt{6}/6$. This is why I think I must be wrong. – Victor S. Apr 15 at 0:40