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

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.

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.

• Ah so we want, all the sum of the components squared to be equal to 1. Thanks! – Bobby Aug 20 '13 at 16:59
• Yes. The square root of the sum of the components squared is the vector's magnitude: think of a two-dimensional vector: this corresponds to the length hypotenuse of the right-angled triangle formed by the vector, the x and the y axes. (Actually, you can define different norms ('magnitudes') on a vector; this one is called the Euclidean norm; but you shouldn't worry about this for now - at this stage you're unlikely to come across any other norm). – MGA Aug 20 '13 at 17:04

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.

• Thank you for helping out! :-) I understand now. – Bobby Aug 20 '13 at 17:00

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.)

• Sorry I am not well verse with the notion of a unit vector, care to explain? I can see 5^2+3^2+(-1)^2 = 35 so I suppose thats where that comes from? – Bobby Aug 20 '13 at 16:57
• That's right; see MGA's answer below. – user84413 Aug 20 '13 at 16:59
• Thank you for helping out! :) – Bobby Aug 20 '13 at 17:00

Your answer is correct I think but I want to highlight what's actually going on in the solution to the problem. It seems like you've accounted for it by setting $z=f(x,y)$ and then just putting -1 as the $z-$component of the normal vector but I think there's a more intuitive approach. I'm a student by the way so take this with a grain of salt. The key to this approach is the observation that gradients are normal to level sets.

This observation becomes intuitively clear if you consider zooming in on two level sets that are very close to eachother. As you zoom in, they'll start to look like straight, parallel lines (assuming the function has continuous partials) and so the direction of steepest ascent is the line perpendicular to both. A way of making this formal is to consider the parameterization of the level curve $c(t)$. $df(c(t))/dt=0$, because it's a level set. But $d(f(c(t))/dt=\nabla f(c(t)) dot c'(t)$. Thus, the gradient is normal to the tangent vector.

Now to the calculation. The way we leverage this idea is to write a new function which has a level set corresponding to the original function (the one we care about). In this example, the level set $f^{-1}(x,y,z)=0$ of $f=x^{4}y+xy^{2}-z$ suffices. The gradient of this level set is normal to it (and thus also the surface we care about). $\nabla f(1,1,2)/||\nabla f(1,1,2)||$ is the vector we care about.