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Determine whether the given function is homogeneous of degree zero. Rewrite those that are as functions of the single variable \(V=y / x\). $$f(x, y)=\frac{x+7}{2 y}$$

Short Answer

Expert verified
The given function is homogeneous of degree zero, as shown by evaluating \(f(tx, ty)\) and confirming that it equals \(f(x, y)\). Rewritten as a function of the single variable \(V\), where \(V = \frac{y}{x}\), it is given by: $$f(V) = \frac{1 + \frac{7}{x}}{2V}$$

Step by step solution

01

Check if the function is homogeneous of degree zero

To check if the function is homogeneous of degree zero, we need to evaluate \(f(tx,ty)\) and see if this equals \(f(x,y)\). Let's consider the given function \(f(x, y)=\frac{x+7}{2 y}\) and replace \(x\) with \(tx\) and \(y\) with \(ty\): $$f(tx, ty) = \frac{tx + 7}{2(ty)}$$ Now, simplify the function by dividing both the numerator and denominator by \(t\): $$\frac{t(x + 7)}{2t(y)} = \frac{x + 7}{2(y)}$$ Since this expression is equal to \(f(x, y)\), the given function is homogeneous of degree zero.
02

Rewrite f(x, y) as a function of V

Now we will rewrite the function as a function of variable \(V\), where \(V = \frac{y}{x}\). To do this, let's first solve for \(y\) in terms of \(x\) and \(V\): $$V = \frac{y}{x}$$ $$y = Vx$$ Now, substitute \(Vx\) for \(y\) in the original function: $$f(x, Vx) = \frac{x + 7}{2(Vx)}$$ Now, it's possible to cancel \(x\) from the numerator and denominator to rewrite the function as a function of only \(V\): $$f(V) = \frac{1 + \frac{7}{x}}{2V}$$ So the function, rewritten as a function of the single variable \(V\), is given by: $$f(V) = \frac{1 + \frac{7}{x}}{2V}$$

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Degree of Homogeneity
When we talk about homogeneity in functions, we are essentially talking about how the function reacts to scaling of variables. A function is said to be homogeneous of a certain degree if, when you multiply each input by a factor, all outputs scale by that factor raised to the degree. This can be quite useful in determining the nature and behavior of functions in economics and physics.

To assess the degree of homogeneity, we start with replacing each variable in the function with a scaled version, let's say multiplied by a constant factor \(t\). For a function \(f(x, y)\), this becomes \(f(tx, ty)\). For the function to be homogeneous of degree \(k\), \(f(tx, ty)\) should be equal to \(t^k f(x, y)\). In our exercise, when applying this process to \(f(x, y) = \frac{x+7}{2y}\), scaling \(x\) and \(y\) by \(t\) results in an expression identical to the original, showing homogeneity of degree zero because no additional factors of \(t\) appear in the result. This means that the function's value remains unchanged regardless of how much we scale \(x\) and \(y\).

Understanding this helps in many fields, such as economics, where homogeneous functions are used to model production with constant returns to scale.
Function of a Single Variable
Functions of a single variable can often simplify complex multivariable functions, making them easier to analyze and understand. When a function can be rewritten in terms of one variable, it helps highlight its core behavior without the clutter of extra variables. This is particularly useful in calculus, where derivative and integral calculations become much simpler.

In the context of the given problem, the function \(f(x, y) = \frac{x+7}{2y}\) is transformed to depend on a single variable \(V = \frac{y}{x}\). By recognising \(y\) as \(Vx\) and substituting this into the function, it is possible to express \(f\) entirely in terms of \(V\). This results in the single-variable function \(f(V) = \frac{1 + \frac{7}{x}}{2V}\). Despite this seeming complexity, transforming into single-variable form helps in analyzing specific relationships or constraints between \(x\) and \(y\).

This approach is widely used in optimization and estimation problems, and can equally aid in graphical analysis when a problem involves constraints that inherently bind two variables together, as seen here.
Variable Substitution
Variable substitution is a method used to simplify expressions and equations, which can make analysis more straightforward. It is particularly handy in algebra, calculus, and differential equations. Substitution involves expressing variables in terms of one another or a new variable altogether, hence simplifying the given function or equation.

In our exercise, the substitution \(V = \frac{y}{x}\) was used to reduce the function \(f(x, y)\) to a simpler form \(f(V)\). By substituting\( y\) with \(Vx\), we could express the entire function in terms of \(V\) without directly involving both \(x\) and \(y\).

The intention here is to streamline computations and analysis, especially useful when dealing with otherwise complicated equations. In practical applications, substitution can help in tasks like integration or finding extremum points under constraints. It's a key skill in problem-solving, making complex expressions more accessible and revealing hidden simple relationships between variables.

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Most popular questions from this chapter

A pyrotechnic rocket is to be launched vertically upwards from the ground. For optimal viewing, the rocket should reach a maximum height of 90 meters above the ground. Ignore frictional forces. (a) How fast must the rocket be launched in order to achieve optimal viewing? (b) Assuming the rocket is launched with the speed determined in part (a), how long after the rocket is launched will it reach its maximum height?

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Consider the family of curves $$ x^{2}+3 y^{2}=2 c y $$ (a) Show that the differential equation of this family is \(\frac{d y}{d x}=\frac{2 x y}{x^{2}-3 y^{2}}\) (b) Determine the orthogonal trajectories to the family (1.12.5).

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The pressure \(p,\) and density, \(\rho,\) of the atmosphere at a height \(y\) above the earth's surface are related by $$ d p=-g \rho d y $$ Assuming that \(p\) and \(\rho\) satisfy the adiabatic equation of state \(p=p_{0}\left(\frac{\rho}{\rho_{0}}\right)^{\gamma},\) where \(\gamma \neq 1\) is a constant and \(p_{0}\) and \(\rho_{0}\) denote the pressure and density at the earth's surface, respectively, show that $$ p=p_{0}\left[1-\frac{(\gamma-1)}{\gamma} \cdot \frac{\rho_{0} g y}{p_{0}}\right]^{\gamma /(\gamma-1)} $$.

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