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## Section7.5Directional Derivatives

We still have not answered one of our first questions about the steepness of a surface: starting at a point on a surface given by $f(x,y)\text{,}$ and walking in a particular direction, how steep is the surface? We are now ready to answer the question.

We already know roughly what has to be done: as shown in Figure 7.3, we extend a line in the $x$-$y$-plane to a vertical plane, and we then compute the slope of the curve that is the cross-section of the surface in that plane. The major stumbling block is that what appears in this plane to be the horizontal axis, namely the line in the $x$-$y$-plane, is not an actual axis—we know nothing about the “units” along the axis. Our goal is to make this line into a $t$ axis; then we need formulas to write $x$ and $y$ in terms of this new variable $t\text{;}$ then we can write $z$ in terms of $t$ since we know $z$ in terms of $x$ and $y\text{;}$ and finally we can simply take the derivative.

So we need to somehow “mark off” units on the line, and we need a convenient way to refer to the line in calculations. It turns out that we can accomplish both by using the vector form of a line. Suppose that $\vect{u}$ is a unit vector $\langle u_1,u_2\rangle$ in the direction of interest. A vector equation for the line through $(x_0,y_0)$ in this direction is $\vect{v}(t)=\langle u_1t+x_0,u_2t+y_0\rangle\text{.}$ The height of the surface above the point $(u_1t+x_0,u_2t+y_0)$ is $g(t)=f(u_1t+x_0,u_2t+y_0)\text{.}$ Because $\vect{u}$ is a unit vector, the value of $t$ is precisely the distance along the line from $(x_0,y_0)$ to $(u_1t+x_0,u_2t+y_0)\text{;}$ this means that the line is effectively a $t$ axis, with origin at the point $(x_0,y_0)\text{,}$ so the slope we seek is

\begin{align*} g'(0)\amp =\langle f_x(x_0,y_0),f_y(x_0,y_0)\rangle\cdot \langle u_1,u_2\rangle\\ \amp =\langle f_x,f_y\rangle\cdot\vect{u}\\ \amp =\nabla f\cdot \vect{u} \end{align*}

Here we have used the Chain Rule and the derivatives ${d\over dt}(u_1t+x_0)=u_1$ and ${d\over dt}(u_2t+y_0)=u_2\text{.}$ The vector $\langle f_x,f_y\rangle$ is very useful, so it has its own symbol, $\nabla f\text{,}$ pronounced “del f” ; it is also called the gradient of $f\text{.}$

###### Example7.37. Slope.

Find the slope of $z=x^2+y^2$ at $(1,2)$ in the direction of the vector $\langle 3,4\rangle\text{.}$

Solution

We first compute the gradient at $(1,2)\text{:}$ $\nabla f=\langle 2x,2y\rangle\text{,}$ which is $\langle 2,4\rangle$ at $(1,2)\text{.}$ A unit vector in the desired direction is $\langle 3/5,4/5\rangle\text{,}$ and the desired slope is then $\langle 2,4\rangle\cdot\langle 3/5,4/5\rangle=6/5+16/5=22/5\text{.}$

###### Example7.38. Tangent Vector.

Find a tangent vector to $z=x^2+y^2$ at $(1,2)$ in the direction of the vector $\langle 3,4\rangle$ and show that it is parallel to the tangent plane at that point.

Solution

Since $\langle 3/5,4/5\rangle$ is a unit vector in the desired direction, we can easily expand it to a tangent vector simply by adding the third coordinate computed in the previous example: $\langle 3/5,4/5,22/5\rangle\text{.}$ To see that this vector is parallel to the tangent plane, we can compute its dot product with a normal to the plane. We know that a normal to the tangent plane is

\begin{equation*} \langle f_x(1,2),f_y(1,2),-1\rangle = \langle 2,4,-1\rangle\text{,} \end{equation*}

and the dot product is $\langle 2,4,-1\rangle\cdot\langle 3/5,4/5,22/5\rangle=6/5+16/5-22/5=0\text{,}$ so the two vectors are perpendicular. (Note that the vector normal to the surface, namely $\langle f_x,f_y,-1\rangle\text{,}$ is simply the gradient with a $-1$ tacked on as the third component.)

The slope of a surface given by $z=f(x,y)$ in the direction of a (two-dimensional) vector $\vect{u}$ is called the directional derivative of $f\text{,}$ written $D_{\vect{u}}f\text{.}$ The directional derivative immediately provides us with some additional information. We know that

\begin{equation*} D_{\vect{u}}f=\nabla f\cdot \vect{u}=|\nabla f||\vect{u}|\cos\theta= |\nabla f|\cos\theta \end{equation*}

if $\vect{u}$ is a unit vector; $\theta$ is the angle between $\nabla f$ and $\vect{u}\text{.}$ This tells us immediately that the largest value of $D_{\vect{u}}f$ occurs when $\cos\theta=1\text{,}$ namely, when $\theta=0\text{,}$ so $\nabla f$ is parallel to $\vect{u}\text{.}$ In other words, the gradient $\nabla f$ points in the direction of steepest ascent of the surface, and $|\nabla f|$ is the slope in that direction. Likewise, the smallest value of $D_{\vect{u}}f$ occurs when $\cos\theta=-1\text{,}$ namely, when $\theta=\pi\text{,}$ so $\nabla f$ is anti-parallel to $\vect{u}\text{.}$ In other words, $-\nabla f$ points in the direction of steepest descent of the surface, and $-|\nabla f|$ is the slope in that direction.

###### Example7.39. Direction of Steepest Ascent and Descent.

Investigate the direction of steepest ascent and descent for $z=x^2+y^2\text{.}$

Solution

The gradient is $\langle 2x,2y\rangle=2\langle x,y\rangle\text{;}$ this is a vector parallel to the vector $\langle x,y\rangle\text{,}$ so the direction of steepest ascent is directly away from the origin, starting at the point $(x,y)\text{.}$ The direction of steepest descent is thus directly toward the origin from $(x,y)\text{.}$ Note that at $(0,0)$ the gradient vector is $\langle 0,0\rangle\text{,}$ which has no direction, and it is clear from the plot of this surface that there is a minimum point at the origin, and tangent vectors in all directions are parallel to the $x$-$y$-plane.

If $\nabla f$ is perpendicular to $\vect{u}\text{,}$ $D_{\vect{u}}f=|\nabla f|\cos(\pi/2)=0\text{,}$ since $\cos(\pi/2)=0\text{.}$ This means that in either of the two directions perpendicular to $\nabla f\text{,}$ the slope of the surface is 0; this implies that a vector in either of these directions is tangent to the level curve at that point. Starting with $\nabla f=\langle f_x,f_y\rangle\text{,}$ it is easy to find a vector perpendicular to it: either $\langle f_y,-f_x\rangle$ or $\langle -f_y,f_x\rangle$ will work.

If $f(x,y,z)$ is a function of three variables, all the calculations proceed in essentially the same way. The rate at which $f$ changes in a particular direction is $\nabla f\cdot\vect{u}\text{,}$ where now $\nabla f=\langle f_x,f_y,f_z\rangle$ and $\vect{u}=\langle u_1,u_2,u_3\rangle$ is a unit vector. Again $\nabla f$ points in the direction of maximum rate of increase, $-\nabla f$ points in the direction of maximum rate of decrease, and any vector perpendicular to $\nabla f$ is tangent to the level surface $f(x,y,z)=k$ at the point in question. Of course there are no longer just two such vectors; the vectors perpendicular to $\nabla f$ describe the tangent plane to the level surface, or in other words $\nabla f$ is a normal to the tangent plane.

###### Example7.40. Gradient.

Suppose the temperature at a point in space is given by $T(x,y,z)=T_0/(1+x^2+y^2+z^2)\text{;}$ at the origin the temperature in Kelvin is $T_0>0\text{,}$ and it decreases in every direction from there. It might be, for example, that there is a source of heat at the origin, and as we get farther from the source, the temperature decreases. The gradient is

\begin{align*} \nabla T\amp =\langle {-2T_0x\over (1+x^2+y^2+z^2)^2}+ {-2T_0x\over (1+x^2+y^2+z^2)^2}+{-2T_0x\over (1+x^2+y^2+z^2)^2}\rangle\\ \amp ={-2T_0\over (1+x^2+y^2+z^2)^2}\langle x,y,z\rangle\text{.} \end{align*}

The gradient points directly at the origin from the point $(x,y,z)$—by moving directly toward the heat source, we increase the temperature as quickly as possible.

###### Example7.41. Tangent Plane.

Find the points on the surface defined by $x^2+2y^2+3z^2=1$ where the tangent plane is parallel to the plane defined by $3x-y+3z=1\text{.}$

Solution

Two planes are parallel if their normals are parallel or anti-parallel, so we want to find the points on the surface with normal parallel or anti-parallel to $\langle 3,-1,3\rangle\text{.}$ Let $f=x^2+2y^2+3z^2\text{;}$ the gradient of $f$ is normal to the level surface at every point, so we are looking for a gradient parallel or anti-parallel to $\langle 3,-1,3\rangle\text{.}$ The gradient is $\langle 2x,4y,6z\rangle\text{;}$ if it is parallel or anti-parallel to $\langle 3,-1,3\rangle\text{,}$ then

\begin{equation*} \langle 2x,4y,6z\rangle=k\langle 3,-1,3\rangle \end{equation*}

for some $k\text{.}$ This means we need a solution to the equations

\begin{equation*} 2x=3k\qquad 4y=-k\qquad 6z=3k \end{equation*}

but this is three equations in four unknowns—we need another equation. What we haven't used so far is that the points we seek are on the surface $x^2+2y^2+3z^2=1\text{;}$ this is the fourth equation. If we solve the first three equations for $x\text{,}$ $y\text{,}$ and $z$ and substitute into the fourth equation we get

\begin{align*} 1\amp =\left({3k\over2}\right)^2+2\left({-k\over4}\right)^2+3\left({3k\over6}\right)^2\\ \amp =\left({9\over4}+{2\over16}+{3\over4}\right)k^2\\ \amp ={25\over8}k^2 \end{align*}

so $\ds k=\pm{2\sqrt2\over 5}\text{.}$ The desired points are $\ds\left({3\sqrt2\over5},-{\sqrt2\over10},{\sqrt2\over 5}\right)$ and $\ds\left(-{3\sqrt2\over5},{\sqrt2\over10},-{\sqrt2\over 5}\right)\text{.}$

##### Exercises for Section 7.5.

Find $D_{\vect{u}} f$ for $\ds f=x^2+xy+y^2$ in the direction of $\vect{u}=\langle 2,1\rangle$ at the point $(1,1)\text{.}$

Answer

$9\sqrt5/5$

Find $D_{\vect{u}} f$ for $\ds f=\sin(xy)$ in the direction of $\vect{u}=\langle -1,1\rangle$ at the point $(3,1)\text{.}$

Answer

$\sqrt2\cos3$

Find $D_{\vect{u}} f$ for $\ds f=e^x\cos(y)$ in the direction 30 degrees from the positive $x$ axis at the point $(1,\pi/4)\text{.}$

Answer

$e\sqrt2(\sqrt3-1)/4$

The temperature of a thin plate in the $x$-$y$-plane is $\ds T=x^2+y^2\text{.}$ How fast does temperature change at the point $(1,5)$ moving in a direction 30 degrees from the positive $x$ axis?

Answer

$\sqrt3+5$

Suppose the density of a thin plate at $(x,y)$ is $\ds 1/\sqrt{x^2+y^2+1}\text{.}$ Find the rate of change of the density at $(2,1)$ in a direction $\pi/3$ radians from the positive $x$ axis.

Answer

$-\sqrt6(2+\sqrt3)/72$

Suppose the electric potential at $(x,y)$ is $\ds\ln\sqrt{x^2+y^2}\text{.}$ Find the rate of change of the potential at $(3,4)$ toward the origin and also in a direction at a right angle to the direction toward the origin.

Answer

$-1/5\text{,}$ $0$

A plane perpendicular to the $x$-$y$-plane contains the point $(2,1,8)$ on the paraboloid $z=x^2+4y^2\text{.}$ The cross-section of the paraboloid created by this plane has slope 0 at this point. Find an equation of the plane.

Answer

$4(x-2)+8(y-1)=0$

A plane perpendicular to the $x$-$y$-plane contains the point $(3,2,2)$ on the paraboloid $36z=4x^2+9y^2\text{.}$ The cross-section of the paraboloid created by this plane has slope 0 at this point. Find an equation of the plane.

Answer

$2(x-3)+3(y-2)=0$

Suppose the temperature at $(x,y,z)$ is given by $\ds T=xy+\sin(yz)\text{.}$ In what direction should you go from the point $(1,1,1)$ to decrease the temperature as quickly as possible? What is the rate of change of temperature in this direction?

Answer

$\langle -1,-1-\cos1,-\cos1\rangle\text{,}$ $-\sqrt{2+2\cos1+2\cos^21}$

Suppose the temperature at $(x,y,z)$ is given by $\ds T=xyz\text{.}$ In what direction can you go from the point $(1,1,1)$ to maintain the same temperature?

Answer

Any direction perpendicular to $\nabla T=\langle 1,1,1\rangle\text{,}$ for example, $\langle -1,1,0\rangle$

Find an equation for the plane tangent to $\ds x^2-3y^2+z^2=7$ at $(1,1,3)\text{.}$

Answer

$2(x-1)-6(y-1)+6(z-3)=0$

Find an equation for the plane tangent to $\ds xyz=6$ at $(1,2,3)\text{.}$

Answer

$6(x-1)+3(y-2)+2(z-3)=0$

Find an equation for the line normal to $\ds x^2+2y^2+4z^2=26$ at $(2,-3,-1)\text{.}$

Answer

$\langle 2+4t,-3-12t,-1-8t\rangle$

Find an equation for the line normal to $\ds x^2+y^2+9z^2=56$ at $(4,2,-2)\text{.}$

Answer

$\langle 4+8t,2+4t,-2-36t\rangle$

Find an equation for the line normal to $\ds x^2+5y^2-z^2=0$ at $(4,2,6)\text{.}$

Answer

$\langle 4+8t,2+20t,6-12t\rangle$

Find the directions in which the directional derivative of $f(x,y)=x^2+\sin(xy)$ at the point $(1,0)$ has the value 1.

Answer

$\langle 0,1\rangle\text{,}$ $\langle 4/5,-3/5\rangle$

Show that the curve $\vect{r}(t) = \langle\ln(t),t\ln(t),t\rangle$ is tangent to the surface $xz^2-yz+\cos(xy) = 1$ at the point $(0,0,1)\text{.}$

A bug is crawling on the surface of a hot plate, the temperature of which at the point $x$ units to the right of the lower left corner and $y$ units up from the lower left corner is given by $T(x,y)=100-x^2-3y^3\text{.}$

1. If the bug is at the point $(2,1)\text{,}$ in what direction should it move to cool off the fastest? How fast will the temperature drop in this direction?

Answer
$\langle 4,9\rangle$
2. If the bug is at the point $(1,3)\text{,}$ in what direction should it move in order to maintain its temperature?

Answer
$\langle -81,2\rangle$ or $\langle 81,-2\rangle$

The elevation on a portion of a hill is given by $f(x,y) = 100 -4x^2 - 2y\text{.}$ From the location above $(2,1)\text{,}$ in which direction will water run?

Answer

in the direction of $\langle 8,1\rangle$

Suppose that $g(x,y)=y-x^2\text{.}$ Find the gradient at the point $(-1, 3)\text{.}$ Sketch the level curve to the graph of $g$ when $g(x,y)=2\text{,}$ and plot both the tangent line and the gradient vector at the point $(-1,3)\text{.}$ (Make your sketch large). What do you notice, geometrically?

Answer

$\ds \nabla g(-1,3)=\langle 2,1\rangle$

The gradient $\nabla f$ is a vector valued function of two variables. Prove the following gradient rules. Assume $f(x,y)$ and $g(x,y)$ are differentiable functions.

1. $\nabla(fg)=f\nabla(g)+g\nabla(f)$

2. $\nabla(f/g)=(g\nabla f - f \nabla g)/g^2$

3. $\nabla((f(x,y))^n)=nf(x,y)^{n-1}\nabla f$