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# What is the equation for the gradient of a function?

Wiki User

2010-11-14 04:09:41

The equation for the gradient of a linear function mapped in a two dimensional, Cartesian coordinate space is as follows.

The easiest way is to either derive the function you use the gradient formula

(y2 - y1) / (x2 - x1)

were one co-ordinate is (x1, y1) and a second co-ordinate is (x2, y2)

This, however, is almost always referred to as the slope of the function and is a very specific example of a gradient. When one talks about the gradient of a scalar function, they are almost always referring to the vector field that results from taking the spacial partial derivatives of a scalar function, as shown below.

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The equation for the gradient of a function, symbolized ∇f, depends on the coordinate system being used.

For the Cartesian coordinate system:

f(x,y,z) = ∂f/∂x i + ∂f/∂y j + ∂f/∂z k where ∂f/(∂x, ∂y, ∂z) is the partial derivative of f with respect to (x, y, z) and i, j, and k are the unit vectors in the x, y, and z directions, respectively.

For the cylindrical coordinate system:

f(ρ,θ,z) = ∂f/∂ρ iρ + (1/ρ)∂f/∂θ jθ + ∂f/∂z kz where ∂f/(∂ρ, ∂θ, ∂z) is the partial derivative of f with respect to (ρ, θ, z) and iρ, jθ, and kz are the unit vectors in the ρ, θ, and z directions, respectively.

For the spherical coordinate system:

f(r,θ,φ) = ∂f/∂r ir + (1/r)∂f/∂θ jθ + [1/(r sin(θ))]∂f/∂φ kφ where ∂f/(∂r, ∂θ, ∂φ) is the partial derivative of f with respect to (r, θ, φ) and ir, jθ, and kφare the unit vectors in the r, θ, and φ directions, respectively.

Of course, the equation for ∇f can be generalized to any coordinate system in any n-dimensional space, but that is beyond the scope of this answer.

Wiki User

2010-11-14 04:09:41