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Section 6.2 In Three Variables

The concepts of the directional derivative and the gradient vector extend to functions of more than two variables. In this section we will look at some examples for functions of three variables.

Find the rate of change of the function \(f(x,y,z) = xy+yz^2+xz^3\) at the point \((2,0,3)\) in the direction \(\hat{\mathbf{u}} = \left \langle -\dfrac{2}{3}, -\dfrac{1}{3}, \dfrac{2}{3} \right \rangle\text{.}\)

Answer.

\(D_{\hat{\mathbf{u}}} f(2,0,3) = \dfrac{43}{3}\text{.}\)

Solution.

The gradient vector for the given function is

\begin{align*} \nabla f \amp = \left \langle f_x, f_y, f_z \right \rangle\\ \amp = \left \langle y +z^3, x+z^2, 2yz+3xz^2 \right \rangle\text{.} \end{align*}

Thus

\begin{equation*} \nabla f (2,0,3) = \langle 27, 11, 54 \rangle \end{equation*}

and so the required directional derivative is

\begin{equation*} D_{\hat{\mathbf{u}}} f(2,0,3) = \langle 27, 11, 54 \rangle \cdot \left \langle -\dfrac{2}{3}, -\dfrac{1}{3}, \dfrac{2}{3} \right \rangle = \dfrac{43}{3}\text{.} \end{equation*}

The temperature at the point \((x,y,z)\) is given by the function

\begin{equation*} T(x,y,z) = 200e^{-x^2-3y^2-9z^2}\text{.} \end{equation*}

  1. Find the rate of change of temperature at the point \(P = (2,-1,2)\) in the direction \(\overrightarrow{PQ}\) where \(Q = (3,-3,3)\text{.}\)

  2. In which direction does the temperature increase the fastest at \(P\text{?}\)

  3. Find the maximum rate of increase at \(P\text{.}\)

Answer.

  1. \(D_{\hat{\mathbf{u}}} T(2,-1,2) = \dfrac{-10400}{\sqrt{6}} e^{-43}\text{.}\)

  2. \(\nabla T (2,-1,2) = 400e^{-43} \langle -2, 3, -18 \rangle\text{.}\)

  3. \(\| \nabla T (2,-1,2) \| = \dfrac{400 \sqrt{337}}{e^{43}}\text{.}\)

Solution.

The gradient vector for the given function is

\begin{align*} \nabla T \amp = \left \langle T_x, T_y, T_z \right \rangle\\ \amp = \left \langle -400xe^{-x^2-3y^2-9z^2}, -1200ye^{-x^2-3y^2-9z^2},-3600ze^{-x^2-3y^2-9z^2} \right \rangle\text{.} \end{align*}

Thus

\begin{equation*} \nabla T (2,-1,2) = 400e^{-43} \langle -2, 3, -18 \rangle\text{.} \end{equation*}

  1. Since \(\overrightarrow{PQ} = \langle 1, -2, 1 \rangle\text{,}\) the required rate of change is given by the directional derivative

    \begin{equation*} D_{\hat{\mathbf{u}}} T(2,-1,2) = 400 e^{-43} \langle -2, 3, -18 \rangle \cdot \dfrac{1}{\sqrt{6}} \langle 1, -2, 1 \rangle = \dfrac{-10400}{\sqrt{6}} e^{-43}\text{.} \end{equation*}

  2. The direction in which the temperature increase the fastest at \(P\) is

    \begin{equation*} \nabla T (2,-1,2) = 400e^{-43} \langle -2, 3, -18 \rangle\text{.} \end{equation*}

  3. The maximum rate of increase at \(P\) is the maximum value of \(D_{\mathbf{u}} T(2,-1,2)\) which is

    \begin{equation*} \| \nabla T (2,-1,2) \| = \dfrac{400 \sqrt{337}}{e^{43}}\text{.} \end{equation*}

Find the equation of the tangent plane to the level surface of \(f(x,y,z) = x^2+y^2 - z + \cos(z)\) at the point \((-1,1,0)\text{.}\)

Answer.

\(2x-2y+z = -4\text{.}\)

Solution.

Since \(f(-1,1,0) = 3\text{,}\) the level surface for this function satisfies the equation

\begin{equation*} x^2+y^2 - z + \cos(z) = 3\text{.} \end{equation*}

A normal to this surface at the point \((-1,1,0)\text{,}\) and hence to the tangent plane at this point, is given by \(\nabla f (-1,1,0)\text{.}\) Now,

\begin{equation*} \nabla f = \left \langle 2x, 2y, -1-\sin(z) \right \rangle\text{,} \end{equation*}

and so

\begin{equation*} \nabla f (-1,1,0) = \langle-2, 2, -1 \rangle\text{.} \end{equation*}

Thus the equation of the tangent plane is

\begin{equation*} \langle -2, 2, -1 \rangle \cdot \left( \langle x, y, z \rangle - \langle -1, 1, 0 \rangle \right) = 0 \end{equation*}

which simplifies to

\begin{equation*} 2x-2y+z = -4\text{.} \end{equation*}

Exercises Example Tasks

1.

Find the directional derivative of \(g(x,y,z) = \dfrac{z-x}{z+y}\) at \((1,0,-3)\) in the direction \(\mathbf{a} = -6\mathbf{i} + 3\mathbf{j} - 2\mathbf{k}\text{.}\)

2.

By thinking of level surfaces to a function of \(3\) variables show that the normal lines to a sphere pass through its centre.