I've been trying to solve this question:
Find a unit vector with positive $z$ component which is normal to the surface $z=x^4y+xy^2$ at the point $(1,1,2)$ on the surface.
My working:
Let $z=f(x,y) = x^4 y + xy^2$,
$$\begin{split}\frac{\partial F}{\partial x} &= 4x^3y+y^2\\
\frac{\partial F}{\partial x}(1,1) &= 4(1)^3(1)+(1)^2 = 5\\
\frac{\partial F}{\partial y} &= x^4+2xy\\
\frac{\partial F}{\partial y} &= (1)^4+2(1)(1)=3
\end{split}
$$
Therefore the normal vector is $(5, 3, -1)^T$
But the answers say, $(-\frac{1}{7}\sqrt{35}, -\frac{3}{35}\sqrt{35}, \frac{1}{35}\sqrt{35})^T$
I assume the words unit, positive in the question has something to do with this? I also noticed that when I multiply the components of my answer by $-\frac{\sqrt{35}}{5}$ the answer pops out.
Best Answer
To normalize a vector means to make its magnitude equal to one. This is done by dividing every element in the vector by the vector's magnitude.
In your case, the magnitude is $\sqrt{5^2 + 3^2 + (-1)^2} = \sqrt{35}$.
The question asks you to give the vector with a positive z-component, so just multiply the vector you got by $-1$ to get $(-5, -3, 1)$ (this does not change the orientation of the vector, it only makes it point in the opposite direction).
Divide this vector by $\sqrt{35}$ to get a normalized (unit) vector. Finally, don't forget to rationalize - i.e. get rid of square roots in the denominators -, and there you go. This last step is only 'cosmetic', it doesn't change any of the values, but it's good practice.