Let f: {(x,y) ∈ R^2 : x > 0, y > 0} → R be given by f(x,y) = (1/3)x^2 y^4 tan^(-1)(y/x) + (1/sqrt(x^2 + y^2)). Then the value of g(x,y) = (x f_x(x,y) + y f_y(x,y)) / f(x,y). How do... Let f: {(x,y) ∈ R^2 : x > 0, y > 0} → R be given by f(x,y) = (1/3)x^2 y^4 tan^(-1)(y/x) + (1/sqrt(x^2 + y^2)). Then the value of g(x,y) = (x f_x(x,y) + y f_y(x,y)) / f(x,y). How does g(x,y) change with x and y?

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Understand the Problem

The question asks to evaluate a function g(x, y) based on an initial function f(x, y) and determine how g(x, y) changes with respect to the variables x and y. This involves understanding the relationships between the functions and performing some mathematical analysis.

Answer

Changes with $x$ and also with $y$.
Answer for screen readers

The value of ( g(x,y) ) changes with ( x ) and also with ( y ).

Steps to Solve

  1. Define the function and its components

The given function is

$$ f(x,y) = \frac{1}{3} x^2 y^4 \tan^{-1}\left(\frac{y}{x}\right) + \frac{1}{\sqrt{x^2+y^2}} $$

We need to compute the partial derivatives ( f_x ) and ( f_y ).

  1. Compute the partial derivative with respect to x

The partial derivative ( f_x(x,y) ) can be found by differentiating ( f(x,y) ) with respect to ( x ):

$$ f_x(x,y) = \frac{\partial}{\partial x} \left(\frac{1}{3} x^2 y^4 \tan^{-1}\left(\frac{y}{x}\right)\right) + \frac{\partial}{\partial x} \left(\frac{1}{\sqrt{x^2 + y^2}}\right) $$

  1. Compute the partial derivative with respect to y

Now, find the partial derivative ( f_y(x,y) ):

$$ f_y(x,y) = \frac{\partial}{\partial y} \left(\frac{1}{3} x^2 y^4 \tan^{-1}\left(\frac{y}{x}\right)\right) + \frac{\partial}{\partial y} \left(\frac{1}{\sqrt{x^2 + y^2}}\right) $$

  1. Substituting into g(x, y)

Now substitute ( f_x ) and ( f_y ) into the equation for ( g(x,y) ):

$$ g(x,y) = \frac{x f_x(x,y) + y f_y(x,y)}{f(x,y)} $$

  1. Analyze the result of g(x, y)

After substitution, analyze how ( g(x,y) ) behaves upon changes in ( x ) and ( y ). You may find that both ( f ) and the derivatives contain terms dependent on both variables.

The value of ( g(x,y) ) changes with ( x ) and also with ( y ).

More Information

Understanding changes in multivariable functions through partial derivatives is essential in calculus, especially in applications involving rates of change in several dimensions.

Tips

  • Not computing partial derivatives correctly: Make sure to use the product and chain rules properly.
  • Ignoring dependency on variables: Always check how each term in ( g(x,y) ) relates to both ( x ) and ( y ).

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