For a scalar field $\displaystyle s(x,y,z) = x^{2} - y^{2} - z$
find the normal unit vector to a surface s = 0 at the point (1,1,0)
I know i coudl work out grad for s(x,y,z), but not entirely sure how to go beyond this to get a normal unit vector
For a scalar field $\displaystyle s(x,y,z) = x^{2} - y^{2} - z$
find the normal unit vector to a surface s = 0 at the point (1,1,0)
I know i coudl work out grad for s(x,y,z), but not entirely sure how to go beyond this to get a normal unit vector
Well i know how to do it for parametrics. My attempt for this would follow something i did at a level:
$\displaystyle z-z_{1} = m_{1}(x-x_{1}) + m_{2}(y-y_{1})$
$\displaystyle m_{1} = \frac{\partial dz}{\partial dx} = 2x$
$\displaystyle m_{2} = \frac{\partial dz}{\partial dy} = -2y$
So i assume the normal is one over these. And can sub in the coordinate points, thus obtaning:
$\displaystyle z - 0 = \frac{1}{2}(x-1) - \frac{1}{2}(y-1)$
$\displaystyle z = \frac{x}{2} - \frac{y}{2}$
Is this correct?
Yes,
$\displaystyle \nabla s = 2x \hat{i} - 2y \hat{j} - 1\hat{k}$
sub in values, do unit vector:
$\displaystyle \frac{2 \hat{i} - 2 \hat{j} - 1\hat{k}}{\sqrt{9}}$
hence:
$\displaystyle \frac{2}{3} \hat{i} - \frac{2}{3} \hat{j} - \frac{1}{3}\hat{k}$
(or could use -ve)
Im guessing this is the tangent unit vector?
Looking at a past question i think i can use the dot product here to ge the normal i.e.
$\displaystyle (x^{2}-1, y^{2}-1, z-0) . (\frac{2}{3},-\frac{2}{3},-\frac{1}{3}) = 0$
Which after expanding i get a similar answer as above :\ (though 2 instead of a half). The reason i chose this method is because this is what we were taught to use
$\displaystyle \frac{{\nabla s(1,1,0)}}{{\left\| {\nabla s(1,1,0)} \right\|}}=\frac{{\nabla s(1,1,0)}}{\tex{max}{Ds(1,1,0)}}=\frac{\frac{\part ial s}{\partial x}(1,1,0)i+\frac{\partial s}{\partial y}(1,1,0)j+\frac{\partial s}{\partial z}(1,1,0)k}{\sqrt{[\frac{\partial s}{\partial x}(1,1,0)]^2+[\frac{\partial s}{\partial y}(1,1,0)]^2+[\frac{\partial s}{\partial z}(1,1,0)]^2}} $
I assume your partial derivatives should go x, y, z not x,x ,x as that wouldn't make sense.
Isn't that what i did?
which is where the root 9 came from.
But that itself cannot be the normal unit vector surely. Is that not the tangent unit vector and hence this has to be used to obtain the unit normal vector that is at right angles to this tangent (gradient)
I see this is a special case in teh notes of a level surface:
A surface defined explicitly, e.g. z = f (x, y), can be considered as a level surface of the scalar field
U(x, y, z) = z − f (x, y) + c, c = const.
The reason i thoguht not was i was comparing it to an example that wasn't a level surface, though i did not see this.
Thanks
Yes, it was a level surface. You were asked to find the normal unit vector to a surface s = 0 which is clearly a level surface (the "level" being 0). To find a unit normal vector to a surface f(x,y,z)= constant, at a given point, evaluate $\displaystyle \nabla f$ at the given point and divide by its length.