SUNY Geneseo Department of Mathematics

Directional Derivatives Part 2

Tuesday, March 28

Math 223
Spring 2023
Prof. Doug Baldwin

Anything You Want to Talk About?

(No.)

Directional Derivatives

Consider finding the derivative of g(x, y) = (sin x)(cos y) in direction v = ⟨3, 4⟩. Yesterday we realized that simply finding a unit vector in the same direction as v will give us the cosine and sine we need as the x and y components of the unit vector, respectively.

So let’s do that in order to finish finding the directional derivative of f.

There were 3 preliminary steps, namely finding the unit vector and the two partial derivatives of g:

G of X and Y equals sine X times cosine Y and its partial derivatives and a unit vector in direction 3 comma 4

Then those pieces of information just need to get combined according to the formula for a directional derivative (which can be rephrased with the components of the unit direction vector replacing the cosine and sine):

Directional derivative is D F D X times X component of U plus D F D Y times Y component of U

Field Trip

To “see” multivariable functions, level curves, partial derivatives, directional derivatives, and gradients.

(Standing on a hillside, your elevation is a function of your north-south and east-west position. Rates at which the hill rises or drops in these directions are the partial derivatives of elevation, and the rate at which the hill rises or drops in an arbitrary direction is a directional derivative. If you walk along a path on which your elevation doesn’t change, you’re walking along a level curve — or technically walking above the level curve, because level curves are curves in the xy plane. Finally, if you face in the direction that the hill rises fastest, you’re facing in the direction given by the so-called “gradient” of the elevation; this direction happens to always be perpendicular to the level curve you’re standing on.)

The Gradient

Having been introduced to gradients on the field trip, let’s talk a bit more about them:

The gradient is written as an upside-down triangle, or “nabla” symbol, often in boldface or with a vector arrow over it, to indicate that it is a vector. It’s defined simply as the vector of a function’s partial derivatives:

Gradient of F of X and Y is vector of derivative of F with respect to X and derivative with respect to Y

For example, what’s the gradient of f(x, y, z) = 3xy - yz + e^xyz?

Making a vector of partial derivatives to find the gradient of a function

Notice that the definition of gradient gives us a new way to think about directional derivatives, one with no explicit mention of angles, and thus one that can be applied in any number of dimensions:

Derivative of F in direction U is gradient of F dotted with U

We can also get an informal sense of how gradients are perpendicular to level curves. For example, in the paraboloid z = x² + y², level curves are always circles with equations of the form x² + y² = c for some constant c. The gradient is also easy to find:

Z equals X squared plus Y squared, gradient of Z is vector 2 X comma 2 Y

Now consider a specific level curve, say x² + y² = 1, and the gradient at a specific point on that curve. The gradient is a vector that, if placed in “standard position” (i.e., starting at the origin), runs out along a radius of the circle and a little past it. Now any radius of a circle is perpendicular to the circle’s edge where the radius line meets that edge, showing that at least in this case the gradient vectors really are perpendicular to the level curves.

X Y plot with circular level curve and gradient vector projecting perpendicularly through it from origin

More about gradients.

Please read “Gradient” and “Gradients and Level Curves” in section 3.6 of the textbook.

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