Gradients and Directional Derivatives

Combine coordinate rates into one vector

For differentiable $f$ , the gradient is $\nabla f=(f_x,f_y)$ , or $(f_x,f_y,f_z)$ in space. The derivative per unit distance in a unit direction $u$ is $D_uf=\nabla f\cdot u$ . A vector describing a direction must be normalized before using this as a rate per unit distance.

Worked example: normalize before measuring

For $f=x^2+y^2$ at $(1,2)$ , the gradient is $(2,4)$ . Direction $v=(3,4)$ has length five, so $u=(3/5,4/5)$ . The directional derivative is $6/5+16/5=22/5$ .

Using $v$ directly would give $22$ , the derivative along the path $(1,2)+tv$ per unit $t$ , whose speed is five. Both numbers have meanings, but they answer different rate questions.

Explore

Direction and rate of change

Try this. At (1,1), compare directions 45°, 135° and 225°: the rate is positive, zero and negative. Move to (0,0); every directional derivative is zero.

Point x1Point y1Direction (degrees)0

f(x,y) = x² + y². At (1, 1), f = 2, gradient = (2, 2). Directional rate = 2; maximum rate = 2.828. Orange: chosen unit direction. Green: unit direction of greatest increase. Rings: equal heights. Arrows show direction, not gradient magnitude.

The explorer uses $f=x^2+y^2$ . Its green arrow shows the unit direction of greatest increase, not the gradient’s magnitude. Orange shows the chosen unit direction; the readout gives the resulting rate.

Worked example: perpendicular to a level curve

For $f=x^2+y^2$ , the level curve through $(1,2)$ is $x^2+y^2=5$ . A tangent direction there is $(-2,1)$ , whose dot product with $(2,4)$ is zero. Moving tangentially causes no first-order height change.

More generally, differentiating $f(r(t))=c$ gives $\nabla f\cdot r'=0$ . At regular points, the gradient is normal to the level set. A zero gradient cannot supply a normal direction and can mark a crossing, cusp or other exceptional behavior.

Practice

Find the greatest directional rate of $f=3x-4y$ .
Find its derivative in direction $(1,1)$ .
What are all first-order directional derivatives of $x^2+y^2$ at zero?

Show worked solutions

$\|(3,-4)\|=5$ .
Normalize to $(1,1)/\sqrt2$ , giving $-1/\sqrt2$ .
All are zero because the gradient vanishes, although the function increases quadratically away from zero.

Further study

MIT OpenCourseWare: Multivariable Calculus provides a full university course with additional lectures and exercises.