# Calculating the normal vector to a surface

1. Aug 10, 2011

### NewtonianAlch

1. The problem statement, all variables and given/known data
Find the unit vector with positive z component which is normal to the surface z = x^2*y + x*y^4 at the point (1,1,2) on the surface

3. The attempt at a solution

So I find:

dz/dx = 2xy +y^4
dz/dy = x^2 + 4xy^3

Substitute (1,1) in those equations and get dz/dx = 3, and dz/dy = 5

So now we have (3, 5, -1)

Multiply by -1 and now we have (-3, -5, 1), but it's apparently wrong.

However the answer was (-3/35*sqrt(35), -1/7*sqrt(35), 1/35*sqrt(35))

I haven't the faintest idea as to how that came about, well...it seems to have been multiplied by sqrt(35)/35, except for the second component which appears to have something else done to it.

I looked around for how the normal vector is calculated in my book, and online too; it's pretty much what I've done.

2. Aug 10, 2011

### SteamKing

Staff Emeritus
Is the vector (-3, -5, 1) a unit vector? How would you make it a unit vector?

3. Aug 10, 2011

### NewtonianAlch

Hmm, had to check this up, was not familiar with it. So it seems I have to divide the vector by the square root of the squares of the x,y,z components?

Which would mean square root(9 + 25 + 1) = $\sqrt{35}$

That leaves us with 1/$\sqrt{35}$
multiplied by the vector components multiplied by -1.

Which gives a vector of <-3/$\sqrt{35}$, -5/$\sqrt{35}$, 1/$\sqrt{35}$>

Which is a lot closer than my last attempt. Did I miss something somewhere?

4. Aug 10, 2011

### NewtonianAlch

Nvm, I see it's the same answer, just a different form.

5. Aug 10, 2011

### HallsofIvy

The gradient of a function, f, is always perpendicular to the surface f(x,y,z)= constant.
So the first thing I would do is write your equation as $f(x, y, z)= z- x^2y + xy^4= 0$, then find $\nabla f$. Finally, calculate the length of that vector and divide by its length.

6. Aug 10, 2011

### ehild

Both methods are used, and gives the same result as that of the OP.

ehild