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# 14.5: Directional Derivatives

• • Contributed by David Guichard
• Professor (Mathematics) at Whitman College
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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 $$f(x,y)$$, 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 14.3.1, 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 $$xy$$ 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 $$t$$; then we can write $$z$$ in terms of $$t$$ since we know $$z$$ in terms of $$x$$ and $$y$$; 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 $${\bf u}$$ 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 $${\bf v}(t)=\langle u_1t+x_0,u_2t+y_0\rangle$$. The height of the surface above the point $$(u_1t+x_0,u_2t+y_0)$$ is $$g(t)=f(u_1t+x_0,u_2t+y_0)$$. Because $$\bf u$$ is a unit vector, the value of $$t$$ is precisely the distance along the line from $$(x_0,y_0)$$ to $$(u_1t+x_0,u_2t+y_0)$$; this means that the line is effectively a $$t$$ axis, with origin at the point $$(x_0,y_0)$$, so the slope we seek is

\eqalign{ g'(0)&=\langle f_x(x_0,y_0),f_y(x_0,y_0)\rangle\cdot \langle u_1,u_2\rangle\cr &=\langle f_x,f_y\rangle\cdot{\bf u}\cr &=\nabla f\cdot {\bf u}.\cr }

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

Example $$\PageIndex{1}$$

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

Solution

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

$\langle 2,4\rangle\cdot\langle 3/5,4/5\rangle=6/5+16/5=22/5.\nonumber$

Example $$\PageIndex{2}$$

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: $$\langle 3/5,4/5,22/5\rangle$$. 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

$\langle f_x(1,2),f_y(1,2),-1\rangle = \langle 2,4,-1\rangle, \nonumber$

and the dot product is

$\langle 2,4,-1\rangle\cdot\langle 3/5,4/5,22/5\rangle=6/5+16/5-22/5=0 \nonumber$

so the two vectors are perpendicular. (Note that the vector normal to the surface, namely $$\langle f_x,f_y,-1\rangle$$, 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 $$\bf u$$ is called the directional derivative of $$f$$, written $$D_{\bf u}f$$. The directional derivative immediately provides us with some additional information. We know that

$D_{\bf u}f=\nabla f\cdot {\bf u}=|\nabla f||{\bf u}|\cos\theta= |\nabla f|\cos\theta$

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

Example $$\PageIndex{3}$$

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

Solution

The gradient is $$\langle 2x,2y\rangle=2\langle x,y\rangle$$; this is a vector parallel to the vector $$\langle x,y\rangle$$, so the direction of steepest ascent is directly away from the origin, starting at the point $$(x,y)$$. The direction of steepest descent is thus directly toward the origin from $$(x,y)$$. Note that at $$(0,0)$$ the gradient vector is $$\langle 0,0\rangle$$, 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 $$\bf u$$, $$D_{\bf u}f=|\nabla f|\cos(\pi/2)=0$$, since $$\cos(\pi/2)=0$$. This means that in either of the two directions perpendicular to $$\nabla f$$, 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 $$\nabla f=\langle f_x,f_y\rangle$$, 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 $$\nabla f\cdot{\bf u}$$, where now $$\nabla f=\langle f_x,f_y,f_z\rangle$$ and $${\bf u}=\langle u_1,u_2,u_3\rangle$$ 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.

Example $$\PageIndex{4}$$

Suppose the temperature at a point in space is given by $$T(x,y,z)=T_0/(1+x^2+y^2+z^2)$$; at the origin the temperature in Kelvin is $$T_0>0$$, 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

\eqalign{ \nabla T&=\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\cr &={-2T_0\over (1+x^2+y^2+z^2)^2}\langle x,y,z\rangle.\cr } \nonumber

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.

Example $$\PageIndex{5}$$

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 $$3x-y+3z=1$$.

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 $$\langle 3,-1,3\rangle$$. Let $$f=x^2+2y^2+3z^2$$; 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 $$\langle 3,-1,3\rangle$$. The gradient is $$\langle 2x,4y,6z\rangle$$; if it is parallel or anti-parallel to $$\langle 3,-1,3\rangle$$, then

$\langle 2x,4y,6z\rangle=k\langle 3,-1,3\rangle \nonumber$

for some $$k$$. This means we need a solution to the equations

$2x=3k\qquad 4y=-k\qquad 6z=3k\nonumber$

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 $$x^2+2y^2+3z^2=1$$; this is the fourth equation. If we solve the first three equations for $$x$$, $$y$$, and $$z$$ and substitute into the fourth equation we get

\eqalign{ 1&=\left({3k\over2}\right)^2+2\left({-k\over4}\right)^2+3\left({3k\over6}\right)^2\cr &=\left({9\over4}+{2\over16}+{3\over4}\right)k^2\cr &={25\over8}k^2\cr } \nonumber

so $$k=\pm{2\sqrt2\over 5}$$. The desired points are $$\left({3\sqrt2\over5},-{\sqrt2\over10},{\sqrt2\over 5}\right)$$ and $$\left(-{3\sqrt2\over5},{\sqrt2\over10},-{\sqrt2\over 5}\right)$$. Here are the original plane and the two tangent planes, shown with the ellipsoid.

## Contributors

• Integrated by Justin Marshall.