14.4 | Gradients and Directional Derivatives in the Plane


If \(f\) is a function of two variables \(x\) and \(y\), then the gradient of \(f\) is the vector \(\nabla f\) defined by
$$
\begin{aligned}
\nabla f(x,y) & = \left \langle f_x(x,y), f_y(x,y) \right \rangle \\
& = \frac{\partial f}{\partial x} \mathbf{i} + \frac{\partial f}{\partial y} \mathbf{j}
\end{aligned}
$$
The directional derivative of \(f\) at \((x_0,y_0)\) in the direction of a unit vector \(\mathbf{u}= \left \langle a, b \right \rangle\) is

$$
D_u f(x_0,y_0) = \lim_{h \rightarrow 0} \frac{f(x_0 + ha, y_0 + hb) – f(x_0, y_0)}{h}
$$
$$
\begin{aligned}
D_u & f(x_0,y_0) \\
&= \lim_{h \rightarrow 0} \frac{f(x_0 + ha, y_0 + hb) – f(x_0, y_0)}{h}
\end{aligned}
$$

if this limit exists.

If \(f\) is a differentiable function of \(x\) and \(y\), then \(f\) has a directional derivative in the direction of any unit vector \(\mathbf{u} = \left \langle a,b \right \rangle\) and
$$
\begin{aligned}
D_u f(x,y) & = f_x(x,y) a + f_y (x,y) b \\
& = \left \langle f_x(x,y), f_y(x,y) \right \rangle \cdot \left \langle a,b \right \rangle
\end{aligned}
$$
If \(\mathbf{u} = \left \langle a,b \right \rangle\) makes an angle \(\theta\) with the positive \(x\)-axis then

$$
D_u f(x,y) = \left \langle f_x(x,y), f_y(x,y) \right \rangle \cdot \left \langle \cos \theta, \sin \theta \right \rangle
$$
$$
\begin{aligned}
D_u & f(x,y) \\
&= \left \langle f_x(x,y), f_y(x,y) \right \rangle \cdot \left \langle \cos \theta, \sin \theta \right \rangle
\end{aligned}
$$
Suppose \(f\) is a differentiable function of two or three variables. The maximum value of the directional derivative \(D_u f(\mathbf{x})\) is \(\left| \nabla f(\mathbf{x})\right|\) and it occurs when \(\mathbf{u}\) has the same direction as the gradient vector \(\nabla f(\mathbf{x})\).
An equation for the tangent plane to the surface \(z = f(x,y)\) at \((x_0, y_0, z_0)\) is given by
$$
\nabla F (x_0,y_0,z_0) \cdot \left \langle x-x_0, y-y_0, z-z_0 \right \rangle = 0
$$
where \(F(x,y,z) = f(x,y) – z = 0 \).
Let \(f(x,y) = x^3 y^2 – y^4\). Find the directional derivative of \(f\) at \((2,1)\) in the of a vector \(\mathbf{u}\) that makes an angle of \(\pi/6\) with the positive \(x\)-axis.


$$
\begin{aligned}
D_u f(x,y) & = \left \langle f_x(x,y), f_y(x,y) \right \rangle \cdot \left \langle \cos \theta, \sin \theta \right \rangle \\
& = \left \langle 3x^2 y^2, 2x^3y – 4y^3 \right \rangle \cdot \left \langle \cos (\pi/6), \sin (\pi/6) \right \rangle \\
& = \left \langle 3x^2 y^2, 2x^3y – 4y^3 \right \rangle \cdot \left \langle \sqrt{3}/2, 1/2 \right \rangle \\
\end{aligned}
$$
$$
\begin{aligned}
& D_u f(x,y) \\
& = \left \langle f_x(x,y), f_y(x,y) \right \rangle \cdot \left \langle \cos \theta, \sin \theta \right \rangle \\
& = \left \langle 3x^2 y^2, 2x^3y – 4y^3 \right \rangle \cdot \left \langle \cos (\pi/6), \sin (\pi/6) \right \rangle \\
& = \left \langle 3x^2 y^2, 2x^3y – 4y^3 \right \rangle \cdot \left \langle \sqrt{3}/2, 1/2 \right \rangle \\
\end{aligned}
$$

So,
$$
\begin{aligned}
D_u f(2,1) & = \left \langle 12, 12 \right \rangle \cdot \left \langle \sqrt{3}/2, 1/2 \right \rangle \\
& = 6\sqrt{3} + 6
\end{aligned}
$$

HH144EX2
  1. Let \(f(x,y) = 9 \cos x \sin y\) and let \(S\) be the surface \(z = f(x,y)\). Find a unit vector \(\vec{u}\) that is normal to the surface \(S\) at the point \((0, \pi/2, 9)\).
  2. What is an equation of the tangent plane to the surface \(S\) at the point \((0, \pi/2, 9)\)?

  1. Let \(F(x,y,z) = 9 \cos x \sin y – z\), then \(\nabla F = \langle -9 \sin x \sin y, 9 \cos x \cos y, -1 \rangle\). A normal to the surface at \((0, \pi/2, 9)\) is given by \(\nabla F(0, \pi/2, 9) = \langle 0,0,1 \rangle\), which is already unit.
  2. An equation of the tangent plane to the surface \(S\) at the point \((0, \pi/2, 9)\) is given by
    $$
    \nabla F (0, \pi/2, 9) \cdot \left \langle x, y-\pi/2, z-9 \right \rangle = 0.
    $$
    So, an equation for the tangent plane is \(z = 9\).

E 14.4 Exercises