Deriviative of a multivariable function with respect to a constant?

In summary: This result somehow follow from the limit as h approaches zero of? Can anybody offer some insight? Thanks
  • #1
Fractal20
74
1

Homework Statement


So this is arising in my applied math course in solving the wave equation in n dimensions. So we have a function u([itex]\vec{x}[/itex]+r[itex]\vec{z}[/itex],t) and where x and z are n dimensional vectors and r is a scalar (also, u is a scalar function). Then when we take the partial derivative with respect to r we get:

∇u([itex]\vec{x}[/itex]+r[itex]\vec{z}[/itex],t)[itex]\bullet[/itex][itex]\vec{z}[/itex]
(sorry that huge dot is suppose to be a dot product)

I am just use to take derivatives of multivariable functions with respect to variables and then there is the old rule that the result is the gradient of the function dotted with the direction vector. But this is not case. I am okay with excepting this as a rule. It is the same as the single variable analog if ∇ was replaced with the partial with respect to r. But I would still like some sense of why it should be this way. Doe this result somehow follow from the limit as h approaches zero of

(1/h)*(u([itex]\vec{x}[/itex]+(r+h)[itex]\vec{z}[/itex],t)-u([itex]\vec{x}[/itex]+r[itex]\vec{z}[/itex],t)[itex])

? Can anybody offer some insight? Thanks

Homework Equations





The Attempt at a Solution


 
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  • #2
I think you must expand u into its vector components and differentiate term by term.
It might be possible also to use the chain rule.

Without knowing the nature of u or the constituent vectors, that is all the advice I can offer.
 
  • #3
Fractal20 said:

Homework Statement


So this is arising in my applied math course in solving the wave equation in n dimensions. So we have a function u([itex]\vec{x}[/itex]+r[itex]\vec{z}[/itex],t) and where x and z are n dimensional vectors and r is a scalar (also, u is a scalar function). Then when we take the partial derivative with respect to r we get:

∇u([itex]\vec{x}[/itex]+r[itex]\vec{z}[/itex],t)[itex]\bullet[/itex][itex]\vec{z}[/itex]
(sorry that huge dot is suppose to be a dot product)

Don't be sorry, use Tex: ##\nabla u(\vec x+r \vec z,t)\cdot \vec z##
Here's what I typed to get that, it's actually easier that what you did:
Code:
##\nabla u(\vec x+r \vec z,t)\cdot \vec z##
 

Related to Deriviative of a multivariable function with respect to a constant?

What is the definition of derivative of a multivariable function with respect to a constant?

The derivative of a multivariable function with respect to a constant is a measure of how the output of the function changes when the constant is varied. It represents the rate of change of the function with respect to the constant.

How do you find the derivative of a multivariable function with respect to a constant?

To find the derivative of a multivariable function with respect to a constant, one must hold all variables constant except for the one with respect to which the derivative is being taken. Then, the derivative can be calculated using the same rules as a single-variable function.

Why is the derivative of a multivariable function with respect to a constant important?

The derivative of a multivariable function with respect to a constant is important because it helps us understand how changes in the constant affect the overall function. It is also useful in optimization problems, where we want to find the optimal value of a variable while holding other variables constant.

What are some real-world applications of the derivative of a multivariable function with respect to a constant?

The derivative of a multivariable function with respect to a constant has many real-world applications, such as in economics, where it is used to study the relationship between multiple variables in a system. It is also used in physics, engineering, and other fields to analyze the behavior of complex systems.

Can the derivative of a multivariable function with respect to a constant be negative?

Yes, the derivative of a multivariable function with respect to a constant can be negative. This indicates that as the constant increases, the output of the function decreases. It is important to consider the sign of the derivative when interpreting its meaning in a real-world context.

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