How Do You Find a Normal Vector to a Graph?

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SUMMARY

To find a normal vector to the surface defined by the function f(x,y) = ax^y + yx^y^x + b at a point (X_o, Y_o), one must utilize the gradient of the function. The normal vector can be derived from the coefficients of the tangent plane equation n1x + n2y + n3z + c = 0, where the normal vector is represented as n = (n1, n2, n3). The gradient ∇(f(x,y,z)) = fx*i + fy*j - k provides a direct method to obtain the normal vector at the specified point on the surface.

PREREQUISITES
  • Understanding of gradient vectors in multivariable calculus
  • Familiarity with tangent planes and their equations
  • Knowledge of vector operations, specifically cross-products
  • Basic concepts of level surfaces in three-dimensional space
NEXT STEPS
  • Study the properties of gradient vectors in multivariable functions
  • Learn how to derive tangent planes from functions of two variables
  • Explore vector calculus techniques, including cross-products
  • Investigate level surfaces and their implications in three-dimensional geometry
USEFUL FOR

Students and professionals in mathematics, particularly those focusing on multivariable calculus, as well as engineers and physicists working with surface normals in three-dimensional modeling.

RaulTheUCSCSlug
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How do you find a normal vector of a function at a point, such as f(x,y)= ax^y+yx^y^x+b at (X_o,Y_o)

where a and b are just arbitrary constants, and the function is an arbitrary function. So I guess, what is the general steps you take to find the normal? I thought it had to do with the gradient, but I'm still confused.
 
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It's a normal to the surface defined by the function, not a normal to the function.

Off the top of my head, one way would be to find the tangent vectors to the surface in the x direction and in the y direction, and take the cross-product..
 
You can find the tangent plane of the graph at the point ##\left(x_{0},y_{0}\right)##. The coefficients of the tangent plane ## n_{1}x+n_{2}y+n_{3}z+c=0## are the component of the normal vector ## \vec{n}=\left(n_{1},n_{2},n_{3}\right)## of the plane that is a normal vector of ##f(x,y)## at the given point ...
 
Given a surface described by z= f(x, y),the surface can be thought of as a "level surface for the function \phi(x, y, z)= f(x, y)- z= 0 so \nabla (f(x,y- z)= f_x\vec{i}+ f_y\vec{j}- \vec{k} is immediately a vector normal to the surface.
 
Relativistic Momentum, Mass, and Energy Momentum and mass (...), the classic equations for conserving momentum and energy are not adequate for the analysis of high-speed collisions. (...) The momentum of a particle moving with velocity ##v## is given by $$p=\cfrac{mv}{\sqrt{1-(v^2/c^2)}}\qquad{R-10}$$ ENERGY In relativistic mechanics, as in classic mechanics, the net force on a particle is equal to the time rate of change of the momentum of the particle. Considering one-dimensional...

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