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I was working on the following question about directional derivatives.

The directional derivative of a function $\phi(x, y, z)$ at the point $(2, 0, 3)$ in the direction towards $(3, -2, 3)$ is $1.789$, in the direction towards $(2, 4, 4)$ it is $0.243$, whilst in the direction towards $(4, -1, 2)$ it is zero. By showing that the three first order partial derivatives of $\phi$ with respect to $x,$ $y$ and $z$ are $2.0$, $-1.0$ and $5.0$ respectively, verify that the value of the directional derivative of $\phi$ at $(2, 0, 3)$ in the direction towards $(0, 2, 14)$ is $4.314$.

Now I used the following approach. We know that the directional derivative of a scalar field $\phi$ in the direction of a unit vector $\hat{\boldsymbol{\mathrm{s}}}$ is given by

$$\frac{\partial \phi}{\partial s}=\hat{\boldsymbol{\mathrm{s}}}\cdot\nabla\phi$$

Let the first direction given be $\boldsymbol{\mathrm{s}}_1=(3,-2,3)$. Then $\hat{\boldsymbol{\mathrm{s}}}=\frac{1}{\sqrt {22}}(3,-2,3)$. Substituting into the result above gives

$$\frac{\partial \phi}{\partial s_1}=\frac{1}{\sqrt{22}}\begin{pmatrix}3\\-2\\3\end{pmatrix}\cdot\nabla\phi=\frac{3}{\sqrt{22}}\frac{\partial \phi}{\partial x}-\frac{2}{\sqrt{22}}\frac{\partial \phi}{\partial y}+\frac{3}{\sqrt{22}}\frac{\partial \phi}{\partial z}=1.789$$

However if we substitute the given results $\partial\phi/\partial x=2$, $\partial\phi/\partial y=-1$ and $\partial\phi/\partial z=5$, equality does not hold; nor does it hold if we repeat the procedure for the other two directional derivatives.

Is there a flaw in my approach? Perhaps the results in the question are mistaken? I'd appreciate any feedback.

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    $\begingroup$ I think that $\mathbf{s}_1$ is not $(3,-2,3)$, but the displacement vector from $(2,0,3)$ to $(3,-2,3)$, namely $\left(1,-2,0\right)$. $\endgroup$ – Matthew Leingang Jan 4 '17 at 0:00
  • $\begingroup$ Ohhhh... so direction towards meaning, towards the point, not the direction given by the vector. That was a bit confusing. $\endgroup$ – Luke Collins Jan 4 '17 at 0:01
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The procedure is correct, but the vectors that need to be taken are the displacement vectors, i.e. the vector from $(2,0,3)$ to $(3,-2,3)$. In other words, $\hat{\boldsymbol{\mathbb s}}_1=(3,-2,3)-(2,0,3)=(1,-2,0)$.

This gives

$$\frac{\partial \phi}{\partial s_1}=\frac{1}{\sqrt{5}}\begin{pmatrix}1\\-2\\0\end{pmatrix}\cdot\nabla\phi=\frac{1}{\sqrt{5}}\frac{\partial \phi}{\partial x}-\frac{2}{\sqrt{5}}\frac{\partial \phi}{\partial y}$$ which $=1.798$ when substituting the given partial derivatives.

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