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Calculating the normal vector to a surface

  1. Aug 10, 2011 #1
    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. jcsd
  3. Aug 10, 2011 #2


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    Is the vector (-3, -5, 1) a unit vector? How would you make it a unit vector?
  4. Aug 10, 2011 #3
    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) = [itex]\sqrt{35}[/itex]

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

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

    Which is a lot closer than my last attempt. Did I miss something somewhere?
  5. Aug 10, 2011 #4
    Nvm, I see it's the same answer, just a different form.
  6. Aug 10, 2011 #5


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    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 [itex]f(x, y, z)= z- x^2y + xy^4= 0[/itex], then find [itex]\nabla f[/itex]. Finally, calculate the length of that vector and divide by its length.
  7. Aug 10, 2011 #6


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

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