证明两曲面在点处正交相交及具体曲面正交性验证技术问询

Hey there! Let's work through this problem step by step. First, I'll explain the general condition for two surfaces to intersect orthogonally at a point, then we'll verify it for the specific surfaces you mentioned.

Two surfaces intersect orthogonally at a point if and only if their normal vectors at that point are perpendicular to each other.

Here's the reasoning: A surface's tangent plane at a point is always perpendicular to the surface's normal vector at that point. So if the two normal vectors are perpendicular, their corresponding tangent planes will also be perpendicular—meaning the surfaces intersect orthogonally.

Mathematically, for two surfaces defined implicitly as:

• ( F(x, y, z) = 0 )
• ( G(x, y, z) = 0 )

Their normal vectors at a point ( P(x_0, y_0, z_0) ) are given by the gradient vectors:

• ( \nabla F|_P = \left( \frac{\partial F}{\partial x}, \frac{\partial F}{\partial y}, \frac{\partial F}{\partial z} \right) \bigg|_P )
• ( \nabla G|_P = \left( \frac{\partial G}{\partial x}, \frac{\partial G}{\partial y}, \frac{\partial G}{\partial z} \right) \bigg|_P )

If the dot product of these two gradients equals zero (( \nabla F|_P \cdot \nabla G|_P = 0 )), the surfaces are orthogonal at ( P ).

Let's apply this to your surfaces at the point ( (2, 1, -1) ):

Step 1: Rewrite surfaces in implicit form #

1. The first surface ( z = 7x^2 - 12x - 5y^2 ) can be rewritten as:
   ( F(x, y, z) = 7x^2 - 12x - 5y^2 - z = 0 )
2. The second surface ( xyz^2 = 2 ) becomes:
   ( G(x, y, z) = xyz^2 - 2 = 0 )

Step 2: Compute gradients for both surfaces #

For ( F(x, y, z) ):

• ( \frac{\partial F}{\partial x} = 14x - 12 )
• ( \frac{\partial F}{\partial y} = -10y )
• ( \frac{\partial F}{\partial z} = -1 )

Evaluating at ( (2, 1, -1) ):
( \nabla F|_{(2,1,-1)} = (14(2)-12, -10(1), -1) = (16, -10, -1) )

For ( G(x, y, z) ):

• ( \frac{\partial G}{\partial x} = yz^2 )
• ( \frac{\partial G}{\partial y} = xz^2 )
• ( \frac{\partial G}{\partial z} = 2xyz )

Evaluating at ( (2, 1, -1) ):

• ( \frac{\partial G}{\partial x} = 1(-1)^2 = 1 )
• ( \frac{\partial G}{\partial y} = 2(-1)^2 = 2 )
• ( \frac{\partial G}{\partial z} = 2(2)(1)(-1) = -4 )

So ( \nabla G|_{(2,1,-1)} = (1, 2, -4) )

Step 3: Calculate the dot product of the gradients #

Compute ( \nabla F \cdot \nabla G ):
( (16)(1) + (-10)(2) + (-1)(-4) = 16 - 20 + 4 = 0 )

Since the dot product is zero, the normal vectors are perpendicular. This means the tangent planes of the two surfaces at ( (2,1,-1) ) are perpendicular, so the surfaces intersect orthogonally at that point.

内容的提问来源于stack exchange，提问作者confusedmathstudent