Calculus Question: Gradients & Normals

[Lomion]

Geometry Question: Gradients & Normal lines & planes

This is a question from a past midterm that I'd appreciate some help with. It deals with the gradient as a normal. I'm not having trouble actually obtaining the gradient, but I am having trouble with some of the geometry involved, so any help would be appreciated!

Consider the function:

y = \sqrt{x^2 + z^2}

Give the equation for 2 planes whose intersection is the normal line to this surface at (1,4,\sqrt{15}).

I found the value: \nabla f(1,4,\sqrt{15}) = (1/4, -1, \sqrt{15}/4).

And the equation for the normal line is:

r(t) = (1,4,\sqrt{15}) + t(1/4, -1, \sqrt{15}/4)

My question is: How do I find two planes that intersect in this line?

I think I should parametrize the variables, so

x = 1 + 1/4t
y = 4 - t
z = \sqrt{15} + \sqrt{15}/4 * t

But that's where I get lost. Can someone just point me in the right direction in terms of what equations to set up?

Thanks!

 

[wais]

In calculus, the gradient is a vector that points in the direction of greatest increase of a function. In geometry, the normal line is a line that is perpendicular to a surface at a given point. The normal plane is a plane that is perpendicular to a surface at a given point.

To find the equation for two planes that intersect in the normal line to the given surface, we can use the fact that the normal line is perpendicular to the gradient vector at the given point. This means that the normal line is also perpendicular to any vector that lies in the plane tangent to the surface at that point. In other words, the normal line is perpendicular to the tangent plane at (1,4,\sqrt{15}).

To find the equation for the tangent plane, we can use the formula:

ax + by + cz = d

where (a,b,c) is the normal vector to the plane, and d is a constant.

Since the normal vector is perpendicular to the gradient vector, we can use the gradient vector we found earlier, (1/4, -1, \sqrt{15}/4), as the normal vector for the tangent plane. This gives us the equation:

(1/4)x - y + (\sqrt{15}/4)z = d

To find the value of d, we can plug in the point (1,4,\sqrt{15}) into the equation:

(1/4)(1) - (4) + (\sqrt{15}/4)(\sqrt{15}) = d

d = -3/4

So the equation for the tangent plane is:

(1/4)x - y + (\sqrt{15}/4)z = -3/4

To find the equations for the two planes that intersect in the normal line, we can use the fact that these planes are also perpendicular to the normal line. Since we already have one plane (the tangent plane) with a known normal vector, we can use the cross product to find another vector that is perpendicular to both the normal vector and the tangent plane. This will give us the normal vector for the second plane.

The cross product of (1/4, -1, \sqrt{15}/4) and (1/4, 0, \sqrt{15}/4) is:

(1/4)(\sqrt{15}/4) - (\sqrt{15}/4)(\sqrt{15}/4) = -3