Question

Find the angle between the xy-plane and the tangent plane to the ellipsoid
(x^2)/12 + (y^2)/12 + (z^2)/3 = 1
at the point (2, 2, 1).

Answer 1

Looks like all we need to do is find the gradient at the given point (which is the normal vector of the tangent plane at that point), and use the formula for the dot-product to find the angle between that vector and the normal vector of the xy-plane, which is \(\hat k\)

Answer 2

I found the gradient which was (1/6)x + (1/6)y + (2/3)z ... what I'm confused about is what parametrization I should use for the xy-plane ... i+j+0k? If so, my answer comes out to arccos(1/3) when it should be arccos\[\sqrt{2/3}\]

Answer 3

the normal to the xy-plane is in the z-line, which is \(\hat k\) as I said before

Answer 4

sorry >w< i've never seen that notation before :[

Answer 5

the \(\hat k\) notation you mean?

Answer 6

yep ... at least not with that weird caret ... sorry, i'm a little math retarded lol ... is that the k part of the i+j+k?

Answer 7

yeah
actually your gradient is wrong because you did not write i, j, and k, which is what makes \(\nabla f\) a vector and not a scalar

Answer 8

\[\nabla f=\langle f_x,f_y,f_z\rangle=f_x\hat i+f_y\hat j+f_z\hat k\neq f_x+f_y+f_z\]if you write it with '+' between each partial and don't write the direction vectors i, j, and k you make the gradient a scalar instead of a vector, which is no good.

Answer 9

since in the xy-plane, the normal vector is perpendicular to that plane, which means it's in the z-direction. the vector that points in that direction is \(\hat k\)

Answer 10

the dot product of the gradient of the function (which is a \(vector\)) dotted with the normal vector to the xy-plane (which is \(\hat k\) ) will be \(\cos\theta\) according to the well-known formula for the dot product\[\vec u\cdot\vec v=\|\vec u\|\|\vec v\|\cos\theta_{uv}\]

Answer 11

here is a rough sketch of what I imagine this looks like with the xy-plane drawn in, and the tangent plane at point \(P\) which has normal vector \(\nabla f\)