How to Calculate the Gradient
Understand what the gradient is., Take the partial derivatives of the function with respect to each of the variables., Define function: f(x,y,z)=5xy2+2x3z2+6y{\displaystyle f(x,y,z)=5xy^{2}+2x^{3}z^{2}+6y} , Calculate partial derivatives., The...
Step-by-Step Guide
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Step 1: Understand what the gradient is.
The gradient generalizes the derivative to functions of multiple variables.
When you take a gradient, you must input a scalar function
- calculating the gradient then outputs a vector function, where the vectors point in the direction of greatest increase.
In the diagram above, the gradient is represented by the blue vector field.
The gradient is recognized by grad{\displaystyle \operatorname {grad} }, or the nabla symbol ∇{\displaystyle \nabla } called the del operator, which can be written out as ∇=(∂∂x,∂∂y,∂∂z).{\displaystyle \nabla =\left({\frac {\partial }{\partial x}},{\frac {\partial }{\partial y}},{\frac {\partial }{\partial z}}\right).} Gradients are most commonly taken in three dimensions.
The gradient is not defined for a vector field.
It only takes in scalar functions. -
Step 2: Take the partial derivatives of the function with respect to each of the variables.
In Cartesian coordinates, the gradient of a function f(x,y,z){\displaystyle f(x,y,z)} is ∇f=∂f∂xi+∂f∂yj+∂f∂zk,{\displaystyle \nabla f={\frac {\partial f}{\partial x}}{\mathbf {i} }+{\frac {\partial f}{\partial y}}{\mathbf {j} }+{\frac {\partial f}{\partial z}}{\mathbf {k} },} where i,j,k{\displaystyle {\mathbf {i} },{\mathbf {j} },{\mathbf {k} }} are unit vectors pointing in the x, y, and z axes, respectively. ,, ∂f∂x=5y2+6x2z2{\displaystyle {\frac {\partial f}{\partial x}}=5y^{2}+6x^{2}z^{2}} ∂f∂y=10xy+6{\displaystyle {\frac {\partial f}{\partial y}}=10xy+6} ∂f∂z=4x3z{\displaystyle {\frac {\partial f}{\partial z}}=4x^{3}z} , Therefore, ∇f=(5y2+6x2z2)i+(10xy+6)j+(4x3z)k{\displaystyle \nabla f=(5y^{2}+6x^{2}z^{2}){\mathbf {i} }+(10xy+6){\mathbf {j} }+(4x^{3}z){\mathbf {k} }} Note that we have mapped from a scalar field to a vector field. -
Step 3: Define function: f(x
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Step 4: z)=5xy2+2x3z2+6y{\displaystyle f(x
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Step 5: z)=5xy^{2}+2x^{3}z^{2}+6y}
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Step 6: Calculate partial derivatives.
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Step 7: The partial derivatives are the components of the gradient.
Detailed Guide
The gradient generalizes the derivative to functions of multiple variables.
When you take a gradient, you must input a scalar function
- calculating the gradient then outputs a vector function, where the vectors point in the direction of greatest increase.
In the diagram above, the gradient is represented by the blue vector field.
The gradient is recognized by grad{\displaystyle \operatorname {grad} }, or the nabla symbol ∇{\displaystyle \nabla } called the del operator, which can be written out as ∇=(∂∂x,∂∂y,∂∂z).{\displaystyle \nabla =\left({\frac {\partial }{\partial x}},{\frac {\partial }{\partial y}},{\frac {\partial }{\partial z}}\right).} Gradients are most commonly taken in three dimensions.
The gradient is not defined for a vector field.
It only takes in scalar functions.
In Cartesian coordinates, the gradient of a function f(x,y,z){\displaystyle f(x,y,z)} is ∇f=∂f∂xi+∂f∂yj+∂f∂zk,{\displaystyle \nabla f={\frac {\partial f}{\partial x}}{\mathbf {i} }+{\frac {\partial f}{\partial y}}{\mathbf {j} }+{\frac {\partial f}{\partial z}}{\mathbf {k} },} where i,j,k{\displaystyle {\mathbf {i} },{\mathbf {j} },{\mathbf {k} }} are unit vectors pointing in the x, y, and z axes, respectively. ,, ∂f∂x=5y2+6x2z2{\displaystyle {\frac {\partial f}{\partial x}}=5y^{2}+6x^{2}z^{2}} ∂f∂y=10xy+6{\displaystyle {\frac {\partial f}{\partial y}}=10xy+6} ∂f∂z=4x3z{\displaystyle {\frac {\partial f}{\partial z}}=4x^{3}z} , Therefore, ∇f=(5y2+6x2z2)i+(10xy+6)j+(4x3z)k{\displaystyle \nabla f=(5y^{2}+6x^{2}z^{2}){\mathbf {i} }+(10xy+6){\mathbf {j} }+(4x^{3}z){\mathbf {k} }} Note that we have mapped from a scalar field to a vector field.
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Abigail Davis
A seasoned expert in non profit, Abigail Davis combines 24 years of experience with a passion for teaching. Abigail's guides are known for their clarity and practical value.
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