# Find the directions in which the directional derivative has the value 1

Can anyone show me how to adjust my work below so that it is a correct answer? This is question number 14.6.28 in the 7th edition of Stewart Calculus.

Find the directions in which the directional derivative of $f(x,y)=ye^{-xy}$ at the point $(0,2)$ has the value 1.

My work is:

$\nabla f(x,y) = <-y^2 e^{-xy}, e^{-xy}(1-xy)>$
$\nabla f(0,2) = <-4,1>$
$|\nabla f(0,2)| = \sqrt{17}$
$D_v f(x,y)=|\nabla f|\cos{\theta}=\sqrt{17}\cos{\theta}$
$\sqrt{17} \cos{\theta}=1$ when $\cos{\theta}={{\sqrt{17}}/17}$
$\theta =\arccos{{\sqrt{17}}/17}\approx +1.326$ and $-1.326$

EDIT:
I tried the following, based on suggestions below:
$\vec{u}=<\cos{\theta},\sin{\theta}>$
$D_u f(x,y)=\nabla f(x,y)\cdot \vec{u}$
$D_u f(0,2)=\nabla f(0,2)\cdot \vec{u}=-4\cos{\theta}+\sin{\theta}=1$
I then plugged this into a spreadsheet and found that $-4\cos{\theta}+\sin{\theta}=1$ when $\theta = \pi/2 , 5\pi/2 , 9\pi/2 , ...$ and when $\theta = 4\pi/3 , 10\pi/3 , 16\pi/3 , ...$
Can anyone check the correctness of this approach?
Also, I found this result experimentally. If it is correct, I would rather be able to find it using calculus.

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1
Directional derivative is "del f dot u" where u is a unit vector in the given direction. If $u=(x,y)$ then in your case the "del f dot u" is -4x+y, which you set to 1, along with $x^2+y^2=1$ to make u a unit vector. I don't follow your method... – coffeemath Oct 25 '12 at 22:14
@TMS: check your math. – robjohn♦ Oct 25 '12 at 22:21
@coffeemath Can you elaborate? I am trying to work on this now. If you post it as an answer, and if I can follow/check it, I will mark it as the answer. – CodeMed Oct 26 '12 at 20:46
Just entered an "answer"-- turns out there are two different directions! – coffeemath Oct 27 '12 at 2:14
But $-4\cos(4\pi/3)+\sin(4\pi/3)=2-\sqrt{3}/2$ which is not 1. – coffeemath Oct 27 '12 at 9:50

## 2 Answers

The directional derivative is $\nabla f \bullet u$, where $u$ is a unit vector which points in the direction desired. What you want is the unit vector $u=(x,y)$; your del $f$ is $(-4,1)$ as you say, and then $\nabla f \bullet u$ is simply $-4x+1y$. Since it should be 1 you know that $-4x+y=1$, i.e. $y=1+4x$. Since $(x,y)$ is a unit vector you also know that $x^2+y^2=1$.

So plugging in we have $x^2+(1+4x)^2=1$, which when you move the 1 over and expand gives the equation $17x^2+8x=0$. This factors as $x(17x+8)=0$. So either $x=0$ or else $x=-8/17$. Then plugging these into $y=1+4x$ gives the two unit vectors $(0,1)$ and $(-8/17,-15/17)$.

[note the question said "find the directions" rather than "direction"; I think in general for a desired value of the directional derivative strictly between the gradient and the negative of the gradient, one usually has two directions. If you're on a hill not pointing straight up, and you find one way to walk so you're going up at a lesser rate than straight up, there should be another such direction...]

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Thanks.  +1 for giving a working answer.  Sorry for the delay in response, but it took me this long to have the time to sit down to really think through what you wrote.  Eric Angle also seems to have given the start of an answer.  In his case, x and y are both functions of $\theta$.  I just did not know how to get from $<\cos{\theta}, \sin{\theta}>$ to the answer, so I am marking yours as the answer.  In theory, we should be able to use both methods to check each other. – CodeMed Oct 28 '12 at 19:48
Actually, from $cos^2\theta+\sin^2\theta=1$, the other approach is equivalent. However I don't myself immediately see how the formula for the cosine of the sum could help in this case. The two angles $\pi/2$ and $\arctan(15/8)+\pi$ [one is in quadrant 3] don't seem related enough for addition trig formulas to be of use... – coffeemath Oct 28 '12 at 20:15

Let $u$ be the unit vector along which you would like to take the directional derivative of $f$. With $\theta$ the angle that $u$ makes with the $x$ axis, $$u = \left(\cos\theta,\sin\theta\right).$$ You can check that $u$ is a unit vector.

Now, the directional derivative of $f$ in the direction of $u$, at the point $\left(0,2\right)$, is $$\nabla f \Big|_{\left(0,2\right)} \bullet u = \left(-4,1\right) \bullet \left(\cos\theta,\sin\theta\right) = -4 \cos \theta + \sin \theta.$$ The problem states that this should be 1, so $$-4 \cos \theta + \sin \theta = 1.$$ Now it remains to solve for $\theta$. You can probably do this with $$- \cos\left(a+b\right) = - \cos a \cos b + \sin a \sin b.$$

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Can you please elaborate?  I am not sure that I understand. – CodeMed Oct 26 '12 at 0:09
I tried your approach in a revision to my post above.  Not sure if it is correct.  Can you elaborate with something that I can verify? – CodeMed Oct 26 '12 at 22:21
What do I use for a and b? – CodeMed Oct 26 '12 at 23:23
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