Directional Derivatives | What's the slope in any direction?

Main ideas and lessons

• Directional derivatives generalize partial derivatives by measuring the slope of a multivariable function in any chosen direction, not just along the coordinate axes.
• Partial derivatives correspond to slopes along:
  • the x-direction (holding (y) fixed, giving a 1D curve (f(x, y_0)))
  • the y-direction (holding (x) fixed, giving a 1D curve (f(x_0, y)))
• To get the slope in an arbitrary direction, use:
  • a unit direction vector (\mathbf{u}=\langle u_1,u_2\rangle) (unit length so the result doesn’t depend on how far you move)
  • the idea that moving along that direction traces a curve on the surface (z=f(x,y)), whose slope at the point is the directional derivative.

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Method / formula to compute directional derivatives

Step-by-step definition (geometric limit / secant slope)

Given a point ((x_0,y_0)) and a direction vector (\mathbf{u}) with (|\mathbf{u}|=1):

• Parametrize movement along the direction:

  • (x(s)=x_0+s u_1)
  • (y(s)=y_0+s u_2)

• Directional derivative definition: [ D_{\mathbf{u}}f(x_0,y_0) =\lim_{s\to 0}\frac{f(x_0+s u_1,\;y_0+s u_2)-f(x_0,y_0)}{s} ]

Interpretation: This is the limit of slopes of secant lines between two nearby points on the graph, as the points approach each other along direction (\mathbf{u}).

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Step-by-step computation using the chain rule (leads to a dot product)

In (f(x(s),y(s))), both (x) and (y) depend on (s). Apply the chain rule:

[ \frac{d}{ds}f(x(s),y(s)) = f_x(x_0,y_0)\cdot\frac{dx}{ds}+f_y(x_0,y_0)\cdot\frac{dy}{ds} ]

Since:

• (\frac{dx}{ds}=u_1)
• (\frac{dy}{ds}=u_2)

the directional derivative becomes:

[ D_{\mathbf{u}}f(x_0,y_0)= f_x(x_0,y_0)u_1 + f_y(x_0,y_0)u_2 ]

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Key shorthand: gradient and dot product

Define the gradient:

[ \nabla f(x,y)=\langle f_x(x,y),\; f_y(x,y)\rangle ]

Then:

[ D_{\mathbf{u}}f(x_0,y_0)= \nabla f(x_0,y_0)\cdot \mathbf{u} ]

Core lesson: Directional derivatives are computed as a dot product between the gradient (encoding the partial slopes) and the unit direction vector.

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Example from the video (computed directional derivative)

• Function: [ f(x,y)=2-x^2-y^2 ]

• Point: [ (x_0,y_0)=\left(\frac12,-\frac12\right) ]

• Direction vector: [ \mathbf{u}=\left(\frac{1}{\sqrt2},\frac{1}{\sqrt2}\right) ] (Normalized to unit length.)

• Partial derivatives: [ f_x=-2x,\quad f_y=-2y ]

• Gradient at the point: [ \nabla f\left(\frac12,-\frac12\right)=\left(-2\cdot\frac12,\; -2\cdot\left(-\frac12\right)\right)=(-1,1) ]

• Dot product: [ D_{\mathbf{u}}f(x_0,y_0)=(-1,1)\cdot\left(\frac{1}{\sqrt2},\frac{1}{\sqrt2}\right)=0 ]

• Visualization conclusion: A directional derivative of 0 corresponds to a horizontal tangent in that direction at the point.

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Speakers / sources featured

• No specific named speakers or external sources are identified in the provided subtitles (the narrator/teacher is unnamed).