MATH1120: Mathematics for Engineering, Science and Technology 2

Section 6.1 Directional Derivatives

Firstly, note that \(2D\) vectors are a convenient way to specify directions in the \(xy\)-plane. For example, we could say the slope of the function in the direction of the vector \(\mathbf{i} = \langle 1, 0 \rangle\) is \(f_x(x_0,y_0)\) while in the direction of the vector \(\mathbf{j} = \langle 0,1 \rangle\) it is \(f_y(x_0,y_0)\text{.}\) Thus the problem we are looking at is that of finding the slope of the function at the point \((x,y) = (x_0,y_0)\) in the direction given by some vector \(\mathbf{u} = \langle u_1, u_2 \rangle\text{.}\) Mathematically, we would say that we are trying to find the directional derivative of the function \(f(x,y)\) at the point \((x_0,y_0)\) in the direction \(\mathbf{u}\text{.}\) The notation that we use to denote this directional derivative is

One way to approach the problem of finding the directional derivative \(D_{\mathbf{u}}f(x_0,y_0)\) is to use the tangent plane to the function at the point \((x_0,y_0)\text{,}\) i.e.

Then the slope of the function \(f(x,y)\) in the direction of \(\mathbf{u}\) is the slope of \(L(x,y)\) in that direction. If \(\hat{\mathbf{u}} = \langle u_1, u_2 \rangle\) is a unit vector in the direction of \(\mathbf{u}\) then the required slope is the amount by which the value of \(L\) changes as the independent variables change from \((x_0,y_0)\) to \((x_0+u_1,y_0+u_2)\text{,}\) i.e.

Consider the function \(z(x,y) = 5 - \dfrac{x^2+y^2}{2}\text{.}\) 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 \(\mathbf{u} = \langle 2,2 \rangle\) and \(\mathbf{v} = \langle 2,-1 \rangle\) 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 \((4,0,L(4,0))\text{.}\)

To summarise:

Note that the directional derivative \(D_\mathbf{u} f(x_0,y_0)\) can be expressed in the terms of the scalar product if we use the following definition.

With this definition the directional derivative can be written as:

The gradient vector has some interesting facts associated with it. Note in the following we are assuming that \(\nabla f \neq \langle 0, 0 \rangle\text{.}\)

1. \(\nabla f\) points in the direction in which the directional derivative takes on its largest value. To see this, note that

   \begin{align*} D_{\mathbf{u}} f(x,y) \amp = \nabla f(x,y) \cdot \hat{\mathbf{u}}\\ \amp = \| \nabla f \| \| \hat{\mathbf{u}} \| \cos(\theta)\\ \amp = \| \nabla f \| \cos(\theta) \end{align*}

   At a given point \(\| \nabla f \|\) 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 \(\nabla f\text{.}\) We can also see from this that the largest value that the directional derivative can take is \(\| \nabla f \|\text{.}\)

   Similarly, the directional derivative takes on its smallest value in the direction of \(-\nabla f\) and has value \(- \| \nabla f \|\text{.}\)

2. \(\nabla f(x_0,y_0)\) 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

   \begin{equation*} f(x,y) = f(x_0,y_0)\text{.} \end{equation*}

   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 \(\left \langle 1, \dfrac{dy}{dx} \right \rangle\text{.}\)

   

   Thus a vector normal to the curve at the point \((x_0,y_0)\) will be \(\left\langle -\dfrac{dy}{dx},1\right\rangle\text{.}\) We will see subsequently, via implicit differentiation, that for the curve \(f(x,y)=k\text{,}\)

   \begin{equation*} \frac{dy}{dx}=-\frac{f_x}{f_y} \end{equation*}

   and so a vector normal to the curve at the point \((x_0,y_0)\) will be \(\left \langle \dfrac{f_x(x_0,y_0)}{f_y(x_0,y_0)}, 1 \right \rangle\text{,}\) which is parallel to \(\nabla f(x_0,y_0)\text{.}\)

   Notice that since \(\nabla f(x_0,y_0)\) is orthogonal to the level curve passing through the point \((x_0,y_0)\) and that \(\nabla f (x_0,y_0)\) 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 \(\nabla f(-3,1) = \langle -6, -1 \rangle\)

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

\begin{equation*} y=-6x-17\text{.} \end{equation*}

Next, the gradient vector is

\begin{equation*} \nabla f(-3,1) = \langle f_x(-3,1), f_y(-3,1) \rangle = \langle -6, -1 \rangle\text{.} \end{equation*}

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

Exercises Example Tasks

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 \((1,4)\text{.}\)

2.

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

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 \((2,e^2)\text{.}\)

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 \(\mathbf{u} = \langle 1,2 \rangle\text{.}\)

2. The point \(B\) and in the direction of \(\mathbf{w} = \langle -1,-1 \rangle\text{.}\)

3. The point \(A\) and in the direction of the origin.

4. The point \(B\) and in the direction of the origin.