MATH1120: Mathematics for Engineering, Science and Technology 2

Section 6.1 Directional Derivatives

Consider the function $z(x,y)=5-{\displaystyle \frac{x^2+y^2}{2}}\text{.}$ Figure 6.1.2 shows the graph of this function along with its the tangent plane at $(x,y)=(2,1)\text{.}$ Also shown on the diagram are the vectors $\mathbf{u}=⟨2,2⟩$ and $\mathbf{v}=⟨2,-1⟩$ drawn in the $xy$-plane with their tails at the point $(x,y)=(2,1)\text{.}$ Then the directional derivative $D_{\mathbf{u}}(f(2,1))$ will be the slope of the line joining the points $(2,1,z(2,1))$ and $(4,3,L(4,3))$ while the directional derivative $D_{\mathbf{v}}(f(2,1))$ will be the slope of the line joining the points $(2,1,z(2,1))$ and $(4,0,L(4,0))\text{.}$

1. $\mathrm{\nabla }f$ points in the direction in which the directional derivative takes on its largest value. To see this, note that

   $$\begin{array}{rl}{D}_{\mathbf{u}}f(x,y)& =\mathrm{\nabla }f(x,y)\cdot \hat{\mathbf{u}}\\ & =\Vert \mathrm{\nabla }f\Vert \Vert \hat{\mathbf{u}}\Vert \cos (\theta )\\ & =\Vert \mathrm{\nabla }f\Vert \cos (\theta )\end{array}$$

   At a given point $\Vert \mathrm{\nabla }f\Vert$ is fixed and so the largest value of $D_{\mathbf{u}}f(x,y)$ will occur when $\cos (\theta )=1\text{,}$ i.e. when $\theta =0$ or put another way, when $\hat{\mathbf{u}}$ is parallel to $\mathrm{\nabla }f\text{.}$ We can also see from this that the largest value that the directional derivative can take is $\Vert \mathrm{\nabla }f\Vert \text{.}$

   Similarly, the directional derivative takes on its smallest value in the direction of $-\mathrm{\nabla }f$ and has value $-\Vert \mathrm{\nabla }f\Vert \text{.}$

2. $\mathrm{\nabla }f(x_0,y_0)$ is orthogonal (i.e. at right angles) to the level curve passing through $(x_0,y_0)\text{.}$

   For the function $z=f(x,y)$ the level curve passing through the point $(x_0,y_0)$ is given by

   $$f(x,y)=f(x_0,y_0)\text{.}$$

   Now, as shown in Figure 6.1.8, a vector parallel to the tangent to this curve at the point $(x_0,y_0)$ will be $⟨1,{\displaystyle \frac{dy}{dx}}⟩\text{.}$

   

   Thus a vector normal to the curve at the point $(x_0,y_0)$ will be $⟨-{\displaystyle \frac{dy}{dx}},1⟩\text{.}$ We will see subsequently, via implicit differentiation, that for the curve $f(x,y)=k\text{,}$

   $$\frac{dy}{dx}=-\frac{f_x}{f_y}$$

   and so a vector normal to the curve at the point $(x_0,y_0)$ will be $⟨{\displaystyle \frac{f_x(x_0,y_0)}{f_y(x_0,y_0)}},1⟩\text{,}$ which is parallel to $\mathrm{\nabla }f(x_0,y_0)\text{.}$

   Notice that since $\mathrm{\nabla }f(x_0,y_0)$ is orthogonal to the level curve passing through the point $(x_0,y_0)$ and that $\mathrm{\nabla }f(x_0,y_0)$ is the direction in which the directional derivative takes on its largest value, the “path of steepest ascent” on any surface $z=f(x,y)$ is always at right angles to its contours. See the following diagram for an attempt to show this.

   

For the function $f(x,y)=x^2-y$ find the level curve, the tangent line and the gradient vector at the point $(-3,1)\text{.}$

Answer.

The level curve is $y=x^2-8\text{.}$

The tangent line is $y=-6x-17\text{.}$

The gradient vector is $\mathrm{\nabla }f(-3,1)=⟨-6,-1⟩$

Solution.

Since $f(-3,1)=8$ the level curve through the point $(-3,1)$ is $x^2-y=8$ or $y=x^2-8\text{.}$ We can find the equation of the tangent by standard calculus to obtain

$$y=-6x-17\text{.}$$

Next, the gradient vector is

$$\mathrm{\nabla }f(-3,1)=⟨f_x(-3,1),f_y(-3,1)⟩=⟨-6,-1⟩\text{.}$$

As can be seen in the diagram below, the gradient vector is orthogonal to the level curve.

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Exercises Example Tasks

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1.

Find the directional derivative for $f(x,y)=(x-2y{)}^2+5x^2$ at the point $(-3,1)$ in the direction of the point $(1,4)\text{.}$

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2.

Find the maximum value of the rate of change of $h(s,t)={\displaystyle \frac{1}{\sqrt{s^2+t^2}}}$ at $(3,4)\text{.}$

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3.

For the curve $e^x\ln (y)-xy=0$ use the gradient vector of a two variable function to find the tangent line and the normal line at the point $(2,e^2)\text{.}$

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4.

For the following contour plot for some unspecified function of two variables estimate the sign of the directional derivatives at:

1. The point $A$ and in the direction of $\mathbf{u}=⟨1,2⟩\text{.}$

2. The point $B$ and in the direction of $\mathbf{w}=⟨-1,-1⟩\text{.}$

3. The point $A$ and in the direction of the origin.

4. The point $B$ and in the direction of the origin.