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# Directional 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 $$f(x,y)$$, 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 14.5.1

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

Example 14.5.2

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 $\langle f_x(1,2),f_y(1,2),-1\rangle = \langle 2,4,-1\rangle,$ and the dot product is $$\langle 2,4,-1\rangle\cdot\langle 3/5,4/5,22/5\rangle=6/5+16/5-22/5=0$$, 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 14.5.3

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 14.5.4

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 }

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 14.5.5

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

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

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

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

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 }

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.

## Exercises 14.5

Ex 14.5.1 Find $$D_{\bf u} f$$ for $$f=x^2+xy+y^2$$ in the direction of $${\bf u}=\langle 2,1\rangle$$ at the point $$(1,1)$$. (answer)

Ex 14.5.2 Find $$D_{\bf u} f$$ for $$f=\sin(xy)$$ in the direction of $${\bf u}=\langle -1,1\rangle$$ at the point $$(3,1)$$. (answer)

Ex 14.5.3 Find $$D_{\bf u} f$$ for $$f=e^x\cos(y)$$ in the direction 30 degrees from the positive $$x$$ axis at the point $$(1,\pi/4)$$. (answer)

Ex 14.5.4 The temperature of a thin plate in the $$x$$-\)y\) plane is $$T=x^2+y^2$$. How fast does temperature change at the point $$(1,5)$$ moving in a direction 30 degrees from the positive $$x$$ axis? (answer)

Ex 14.5.5 Suppose the density of a thin plate at $$(x,y)$$ is $$1/\sqrt{x^2+y^2+1}$$. Find the rate of change of the density at $$(2,1)$$ in a direction $$\pi/3$$ radians from the positive $$x$$ axis. (answer)

Ex 14.5.6 Suppose the electric potential at $$(x,y)$$ is $$\ln\sqrt{x^2+y^2}$$. 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)

Ex 14.5.7 A plane perpendicular to the $$x$$-\)y\) plane contains the point $$(2,1,8)$$ on the paraboloid $$z=x^2+4y^2$$. The cross-section of the paraboloid created by this plane has slope 0 at this point. Find an equation of the plane. (answer)

Ex 14.5.8 A plane perpendicular to the $$x$$-\)y\) plane contains the point $$(3,2,2)$$ on the paraboloid $$36z=4x^2+9y^2$$. The cross-section of the paraboloid created by this plane has slope 0 at this point. Find an equation of the plane. (answer)

Ex 14.5.9 Suppose the temperature at $$(x,y,z)$$ is given by $$T=xy+\sin(yz)$$. 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)

Ex 14.5.10 Suppose the temperature at $$(x,y,z)$$ is given by $$T=xyz$$. In what direction can you go from the point $$(1,1,1)$$ to maintain the same temperature? (answer)

Ex 14.5.11 Find an equation for the plane tangent to $$x^2-3y^2+z^2=7$$ at $$(1,1,3)$$. (answer)

Ex 14.5.12 Find an equation for the plane tangent to $$xyz=6$$ at $$(1,2,3)$$. (answer)

Ex 14.5.13 Find an equation for the line normal to $$x^2+2y^2+4z^2=26$$ at $$(2,-3,-1)$$. (answer)

Ex 14.5.14 Find an equation for the line normal to $$x^2+y^2+9z^2=56$$ at $$(4,2,-2)$$. (answer)

Ex 14.5.15 Find an equation for the line normal to $$x^2+5y^2-z^2=0$$ at $$(4,2,6)$$. (answer)

Ex 14.5.16 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)

Ex 14.5.17 Show that the curve $${\bf r}(t) = \langle\ln(t),t\ln(t),t\rangle$$ is tangent to the surface $$xz^2-yz+\cos(xy) = 1$$ at the point $$(0,0,1)$$.

Ex 14.5.18 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 $$T(x,y)=100-x^2-3y^3$$.

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

b. If the bug is at the point $$(1,3)$$, in what direction should it move in order to maintain its temperature? (answer)

Ex 14.5.19 The elevation on a portion of a hill is given by $$f(x,y) = 100 -4x^2 - 2y$$. From the location above $$(2,1)$$, in which direction will water run? (answer)

Ex 14.5.20 Suppose that $$g(x,y)=y-x^2$$. Find the gradient at the point $$(-1, 3)$$. Sketch the level curve to the graph of $$g$$ when $$g(x,y)=2$$, and plot both the tangent line and the gradient vector at the point $$(-1,3)$$. (Make your sketch large). What do you notice, geometrically? (answer)

Ex 14.5.21 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.

• a. $$\nabla(fg)=f\nabla(g)+g\nabla(f)$$ b. $$\nabla(f/g)=(g\nabla f - f \nabla g)/g^2$$ c. $$\nabla((f(x,y))^n)=nf(x,y)^{n-1}\nabla f$$