Find unit normal vector to the surface $z=x^4y+xy^2$ at the point $(1,1,2)$

Question

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.

Answer 1

You just need to take the negative of this vector, to get a positive z-component, and then divide by its length to get a unit vector. (You'll also need to rationalize the denominator to get the answer in the form given.)

Answer 2

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.

Answer 3

In a word, yes. You found a vector that is normal to the surface. The question asked specifically for a unit vector (yours isn't) and that that vector have a positive $z$-component (yours doesn't).

They impose all these restrictions because there are many normal vectors to a surface at a given point, all of them scalar multiples of the others, and as such, the normal is not unique. Specifying these conditions makes for a unique answer that is convenient to put in an answer key.