Angle between two surfaces and gradient

In summary: The angle between the two surfaces at a particular point is just the angle between their respective normal vectors which is the angle between the two gradients at that point. In summary, in solving for the angle between two surfaces, we take the gradient of each surface and use the dot product of the two gradients to find the angle at a specific point. Just because the functions are 0 at that point does not mean their gradients are 0, but rather that they are perpendicular to the surface at that point. The angle between two surfaces is found by finding the angle between their respective gradients at the point of intersection.
  • #1
xanadu
3
0
In Marion & Thorton problem 1.29 asks to find the angle between two surfaces [tex](x^2 +y^2 + z^2)^2 = 9[/tex] and [tex]x + y + z^2 = 1[/tex] at a point.

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

My question is, isn't [tex](x^2 +y^2 + z^2)^2 - 9[/tex] and [tex]x + y + z^2 - 1[/tex] 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.

Thanks for your help.
 
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  • #2
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
xanadu said:
In Marion & Thorton problem 1.29 asks to find the angle between two surfaces [tex](https://www.physicsforums.com/showthread.php?p=2894895#post2894895[/tex] and [tex]x + y + z^2 = 1[/tex] at a point.

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

My question is, isn't [tex](x^2 +y^2 + z^2)^2 - 9[/tex] and [tex]x + y + z^2 - 1[/tex] 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.

Thanks for your help.
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 [itex]f(x,y,z)= (x^2+ y^2+ z^2)^2[/itex] (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 \(\displaystyle (x^2+ y^2+ z^2)^2= 9\) 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, [itex]g(x,y,z)= x + y + z^2[/itex] take on the constant value "1" on the surface [itex]x+ y+ z^2[/itex] and so its gradient is perpendicular to the surface at every point on the surface.

That is, at every point on the respective surfaces, [itex]\nabla f= 2(x^2+ y^2+ x^2)(2x\vec{i}+ 2y\vec{j}+ 2z\vec{k})[/itex] and [itex]\nable g= \vec{i}+ \vec{j}+ \vec{k}[/itex] 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.
 

Related to Angle between two surfaces and gradient

What is the angle between two surfaces?

The angle between two surfaces is the measurement of the deviation in direction of one surface from another. It is the angle formed at the point where the two surfaces meet.

How is the angle between two surfaces calculated?

The angle between two surfaces can be calculated using the dot product of their normal vectors. The formula is given by cosθ = (a•b) / (|a| * |b|), where a and b are the normal vectors of the two surfaces and θ is the angle between them.

What is the gradient of a surface?

The gradient of a surface is a vector that represents the direction and magnitude of the steepest slope of the surface at a given point. It is perpendicular to the tangent plane at that point.

How is the gradient of a surface calculated?

The gradient of a surface can be calculated using partial derivatives. The formula is given by ∇f(x,y) = (∂f/∂x, ∂f/∂y), where f(x,y) is the function representing the surface and ∂f/∂x and ∂f/∂y are the partial derivatives with respect to x and y, respectively.

What is the relationship between the angle between two surfaces and their gradients?

The angle between two surfaces is equal to the angle between their respective gradients at the point where they meet. This means that the angle between two surfaces can be found by calculating the angle between their gradient vectors.

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