Directional Derivative Of A Tangent Plane

A tangent plane is defined like this:

$$ z = D_x (x-a) + D_y (y-b) + c $$

where the tangent plane crosses the point \((a,b,c)\), and has derivative \(D_x\) in the \(x\) direction and \(D_y\) in the \(y\) direction.

Let \(\textbf{u}\) be a unit vector in the \(xy\) plane. We need to find the derivative at \((a,b)\) in the direction of \(\textbf{u}\) (let's call it \(D_{\textbf{u}}\)). Let components of \(\textbf{u}\) be \(\langle \cos(\theta) \textbf{i} + \sin(\theta)\textbf{j} \rangle\), where \(\theta\) is the angle from the \(x\)-axis in the \(xy\) plane.

When going from \((a,b)\) to \((a + \cos(\theta),b + \sin(\theta))\), what happens to the \(z\) value? Initially it's \(c\), but it changes to:

$$\begin{gather} z_{new} = D_x ((a+\cos(\theta))-a) + D_y ((b+\sin(\theta))-b) + c \\ z_{new} = D_x (\cos(\theta)) + D_y (\sin(\theta)) + c \end{gather}$$

This means the change in \(z\) is:

$$\Delta z = z_{new} - c = D_x (\cos(\theta)) + D_y (\sin(\theta))$$

This means, going one unit to the direction of our interest which is at angle \(\theta\), the increase is \(D_x (\cos(\theta)) + D_y (\sin(\theta))\). Since this is a tangent plane, the the derivative in a specific direction is constant; this means the directional derivative at direction \(\textbf{u}\) is:

$$D_x (\cos(\theta)) + D_y (\sin(\theta))$$

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