Find the unit vector given the directional derivative

Question

Given the equation $$ f(x)=x^2 -xy+y^2-y $$

I am asked to find the unit vector $\hat n$, such that $D_{\hat n} f(1,-1)$ equals to 4.

So far, finding the $\nabla f$ at $(1,-1)$ gives the result of $(3,-4)$.

As $|\hat n| =1,$ $$|\nabla f| cos{\theta}=4$$ $$cos{\theta} =0.8$$ $$\hat n = (cos{\theta},sin{\theta})= (\frac45, \frac 35)$$ $$\nabla f \cdot \hat n = (3,-4)\cdot(\frac45, \frac 35)=0\not=4 $$

As far as I know, solving for the dot product $\nabla f \cdot \hat n =4 $ is also another way. However, solving for $(a,b)$ with $ (3,-4)\cdot(a,b)=4$ doesn't seem to give me my answer. Any help is appreciated.

Answer 1

HINT

We need to find the unknown $\hat n=(\cos \theta, \sin \theta)$ by

$$(\cos \theta, \sin \theta)\cdot (3,-4)=3\cos\theta-4\sin \theta=4$$

Answer 2

If $\hat n=(a,b)$ is a unit vector, you get a system with two equations:

$3a-4b=4$,

$a^2+b^2=1$.

Answer 3

Your first method finds the cosine of the angle correctly, but this angle is measured from $\nabla f$. You computed a vector that makes this angle with the positive $x$-axis instead. If you rotate $\nabla f$ by this angle and normalize to get a unit vector, you’ll find that the derivative in that direction has the correct value. Note that there are two solutions, since there are two values of $\theta$ for which $\cos\theta = 4/5$.

Alternatively, one solution is obviously $(0,-1)$. The other solution is its reflection in $\nabla f$.