How to Find the Steepest Slope Without Coordinates

For a surface written as $z=f(x,y)$, the steepest slope depends on where you are on the surface. At a specified point, it is found from the gradient. If no coordinates are given, you can often find a formula for the steepest slope as a function of $x$ and $y$, but you cannot evaluate it to one number.

Here, “slope” means the rate at which the height $z$ changes per unit horizontal distance in a chosen direction. This is the usual directional derivative interpretation in multivariable calculus.

The gradient gives the steepest direction

At a point $(x,y)$ where $f$ is differentiable, its gradient is

$$\nabla f(x,y)=\left(f_x(x,y),f_y(x,y)\right).$$

If you move in a horizontal direction given by a unit vector $\mathbf{u}=(u_1,u_2)$, the directional slope is

$$D_{\mathbf{u}}f=\nabla f\cdot\mathbf{u}.$$

Because $\mathbf{u}$ has length $1$, the largest possible value of this dot product is $\|\nabla f\|$. When the gradient is nonzero, this maximum occurs when $\mathbf{u}$ points in the same direction as the gradient. If the gradient is zero, every unit direction gives a directional derivative of zero, so there is no unique steepest direction. Thus

$$\text{steepest uphill slope}=\|\nabla f(x,y)\|=\sqrt{f_x(x,y)^2+f_y(x,y)^2}.$$

When the gradient is nonzero, the steepest downhill direction is the unit vector $-\nabla f/\|\nabla f\|$. Its directional derivative is $-\|\nabla f\|$. If you want the downhill slope as a signed rate of change, retain the minus sign; if you want its magnitude, use $\|\nabla f\|$.

What to do when coordinates are missing

First check what information the problem does give:

• A function but no point: calculate the gradient and leave the result in terms of the coordinates. This gives the steepest slope at each point.
• A point but no function: there is not enough information to calculate a slope. You need the surface or its derivatives there.
• Neither a function nor a point: no particular numerical answer can be determined. The gradient rule is the general method.

For example, if $f(x,y)=x^2+3xy+y^2$ and the point is not specified, then

$$\nabla f(x,y)=(2x+3y,3x+2y),$$

so the steepest uphill slope at $(x,y)$ is

$$\sqrt{(2x+3y)^2+(3x+2y)^2} =\sqrt{13x^2+24xy+13y^2}.$$

This is a formula, not one fixed number: its value changes with the point. The uphill direction is the gradient normalized to have length $1$:

$$\frac{(2x+3y,3x+2y)}{\sqrt{(2x+3y)^2+(3x+2y)^2}},$$

provided the gradient is nonzero.

To see how coordinates turn that into a number, evaluate at $(1,2)$. The gradient is $(8,7)$, so the steepest slope is

$$\sqrt{8^2+7^2}=\sqrt{113}\approx 10.63.$$

The direction of steepest ascent is $(8,7)/\sqrt{113}$. For the surface in this example, $f(1,2)=11$, so this point on the graph is $(1,2,11)$.

Common mistakes

• Reporting a direction instead of a slope: The gradient points uphill, but its magnitude gives the steepest slope.
• Forgetting to normalize a direction: The directional derivative formula uses a unit direction vector. The gradient itself is generally not a unit vector.
• Treating a symbolic result as a numerical one: If the point is absent, a formula in $x$ and $y$ may be the complete answer; do not substitute arbitrary coordinates.
• Assuming zero gradient means a local maximum or minimum: If $\nabla f=0$, every first-order directional derivative is zero, but the point could be a maximum, minimum, or saddle point.
• Ignoring restrictions on motion: The gradient gives the steepest direction when you may move in any horizontal direction. If motion is restricted to a curve or boundary, maximize the directional derivative only over the allowed directions.

If “slope” means the angle of the tangent plane above the horizontal rather than rise per unit horizontal distance, that angle is $\arctan(\|\nabla f\|)$. The gradient magnitude is still the quantity to calculate first.