# Angle between two surfaces

1. Sep 22, 2010

In Marion & Thorton problem 1.29 asks to find the angle between two surfaces $$(x^2 +y^2 + z^2)^2 = 9$$ and $$x + y + z^2 = 1$$ at a point.

The solution takes the gradient of $$(x^2 +y^2 + z^2)^2 - 9$$ and $$x + y + z^2 - 1$$, and using the dot product between the two vectors at that point gets the angle.

My question is, isn't $$(x^2 +y^2 + z^2)^2 - 9$$ and $$x + y + z^2 - 1$$ both zero and hence taking the gradient would give you 0. shouldn't you rather take one variable as the dependent variable so you have z(x,y) for example and then take the gradient of that? I'm confused as to why they took the gradient of the way they did.

2. Sep 22, 2010

### Eynstone

The intersection of two surfaces is a curve & rarely a straight line. Hence, the notion of the 'angle' between two surfaces is invalid. The author probably means what he calculated by 'the angle between the surfaces '.

3. Sep 22, 2010

### HallsofIvy

Staff Emeritus
The fact that a function is 0 for a specific value of x does NOT mean it is a constant nor that its derivative must be 0! Similarly, a function of three variables, that is equal to 0 on some specific subset of R3, is NOT necessarily a constant and its gradient is not necessarily constant on that subset.

In fact, what is true is that if f(x,y,z)= constant on some subset, then its gradient is perpendicular to that subset. Recall that the gradient vector always points in the direction of fastest increase and, further, that the "rate of change" of the function in a particular direction is the projection of the gradient vector on that direction. If the function is constant in a given direction, the projection of the gradient vector in that direction is 0 which does NOT mean the gradient is 0, only that it is perpendicular to that direction.

Here, defining $f(x,y,z)= (x^2+ y^2+ z^2)^2$ (or, equivalently, [/itex]f(x, y, z)= (x^2+ y^2+ z^2)^2- 9)[/itex] says that on the subset of R3 defined by $(x^2+ y^2+ z^2)^2= 9$ f(x,y,z)= 9 (or, equivalently f(x,y,z)= 0). Since f is constant on that surface, its gradient is perpendicular to that surface, not necessarily 0.

Similarly, $g(x,y,z)= x + y + z^2$ take on the constant value "1" on the surface $x+ y+ z^2$ and so its gradient is perpendicular to the surface at every point on the surface.

That is, at every point on the respective surfaces, $\nabla f= 2(x^2+ y^2+ x^2)(2x\vec{i}+ 2y\vec{j}+ 2z\vec{k})$ and $\nable g= \vec{i}+ \vec{j}+ \vec{k}$ are perpendicular to their respective surfaces. The intersection of the two surfaces, being a curve that lies in both surfaces, must be perpendicular to both vectors. That is, we can find a vector in the direction of the intersection curve by taking the cross product of the two gradients.