Directional Derivatives and the Gradient
The rate of change in an arbitrary direction is the dot product of the gradient with a unit vector, so the gradient points the steepest way uphill and sits perpendicular to level curves.
What a learner can do afterwards
- Compute a directional derivative from the gradient and a unit direction
- State the direction of fastest increase and its rate
- Show that the gradient is perpendicular to the level curve through a point
1 · Read
The gradient packs both partials into one vector. For f = 3x + 4y it is (3, 4) everywhere. Dotting it with any unit direction gives the slope there: for x squared + y squared at (1, 2), the gradient (2, 4) dotted with (1, 0) gives 2.
The gradient points toward the steepest uphill climb, and its length is that fastest rate. For (3, 4) the fastest rate is 5. In the unit direction (3/5, 4/5) the slope is 9/5 + 16/5 = 5, the full maximum.
A zero dot means you walk along a level curve, so the gradient sits perpendicular to it. For x squared + y squared = 25 at (3, 4), the gradient (6, 8) meets the curve at a right angle. Watch arrows meet curves squarely on any contour map.
The cosine view unites everything: slope equals gradient length times cosine. Cosine 1 gives the fastest increase along the gradient, minus 1 the fastest decrease opposite it, and 0 gives zero change along the level curve.
Dot the gradient with a unit direction for any slope, follow it for the fastest climb, and expect right angles with level curves.
2 · Watch
Take it off screen
Where it sits
8 questions wait behind this lesson, each with its answer explained. Every answer feeds the sky: stars light as they are learned, and dim when it is time to come back.