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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?

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"Substituting the values of P in the parametric equation of the curve, you get (1,1,1)"

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, $(0, 5, 4)$.

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  • $\begingroup$ thanks for clearing it up.. what about the part "in the direction in which $z$ is growing?" $\endgroup$ – Victor S. Apr 15 at 0:01
  • $\begingroup$ 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). $\endgroup$ – Peter Foreman Apr 15 at 0:31
  • $\begingroup$ Well the official answer is $13\sqrt{6}/6$. This is why I think I must be wrong. $\endgroup$ – Victor S. Apr 15 at 0:40

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