# Find g(x)/h(y) for a given F(x,y)

• songoku
In summary, the equation ##\frac{\partial F}{\partial x} . \frac{1}{4x} = \frac{\partial F}{\partial y} . \frac{1}{10y}## can be rearranged to ##g(x)/h(y)=-F_x/F_y## where ##g(x)## and ##h(y)## satisfy the equation ##h(y)F_x+g(x)F_y=0##. This proves that the ratio of ##g(x)## and ##h(y)## can be used to find the solution for any choice of ##F_x## and ##F_y## that satisfy the equation.
songoku
Homework Statement
Given that ##F(x,y)=f(2x^2 +5y^2)## and satisfies ##h(y) \frac{\partial F}{\partial x}+g(x) \frac{\partial F}{\partial y}=0##, find ##\frac{g(x)}{h(y)}##
Relevant Equations
Partial derivative
$$F(x,y)=f(2x^2+5y^2)$$

$$\frac{\partial F}{\partial x}=f'(2x^2+5y^2) . (4x)$$

$$\frac{\partial F}{\partial y}=f'(2x^2+5y^2).(10y)$$

##f'(2x^2+5y^2)=\frac{\partial F}{\partial x} . \frac{1}{4x} = \frac{\partial F}{\partial y} . \frac{1}{10y}##

So
$$\frac{\partial F}{\partial x} . \frac{1}{4x} = \frac{\partial F}{\partial y} . \frac{1}{10y}$$
$$\frac{\partial F}{\partial x} . 10y=\frac{\partial F}{\partial y} . 4x$$
$$10y . \frac{\partial F}{\partial x}-4x . \frac{\partial F}{\partial y}=0$$

Then
$$\frac{g(x)}{h(y)}=-\frac{4x}{10y}$$

But the answer is ##\frac{4x}{10y}##. Where is my mistake?

Thanks

I agree with your minus result.

songoku
I agree with the minus sign. I disagree with your approach. You demonstrated that ##h(y)=10y## and ##g(x)=4x## happen to satisfy the equation, and then computed the ratio of them. But that doesn't really prove that ratio works for any choice of ##g## and##h## that satisfy the equation (of which there are many)

I think what they wanted you to do is simply algebra

##h(y)F_x+g(x)F_y=0##
##g(x)F_y=-h(y)F_x##
##g(x)/h(y)=-F_x/F_y##

PeroK and songoku
Office_Shredder said:
I agree with the minus sign. I disagree with your approach. You demonstrated that ##h(y)=10y## and ##g(x)=4x## happen to satisfy the equation, and then computed the ratio of them. But that doesn't really prove that ratio works for any choice of ##g## and##h## that satisfy the equation (of which there are many)

I think what they wanted you to do is simply algebra

##h(y)F_x+g(x)F_y=0##
##g(x)F_y=-h(y)F_x##
##g(x)/h(y)=-F_x/F_y##
Ah I see it is possible to start from there. I thought I had to start from partial derivative and tried to rearrange the equation to fit the question.

Thank you very much anuttarasammyak and Office_Shredder

## 1. What is the purpose of finding g(x)/h(y) for a given F(x,y)?

The purpose of finding g(x)/h(y) for a given F(x,y) is to simplify the function and make it easier to analyze. By breaking down the function into smaller components, it becomes easier to understand the relationship between the variables and make predictions about their behavior.

## 2. How do you find g(x)/h(y) for a given F(x,y)?

To find g(x)/h(y) for a given F(x,y), you first need to identify the individual functions g(x) and h(y) within the larger function F(x,y). Then, you can use algebraic manipulation and techniques such as factoring, cancelling, and substitution to simplify the function and express it in the form of g(x)/h(y).

## 3. Can g(x)/h(y) be simplified further?

Yes, g(x)/h(y) can potentially be simplified further by using more advanced techniques such as differentiation, integration, or trigonometric identities. However, the level of simplification depends on the complexity of the original function and the techniques used.

## 4. Are there any restrictions when finding g(x)/h(y) for a given F(x,y)?

Yes, there may be restrictions when finding g(x)/h(y) for a given F(x,y). For example, the function may be undefined for certain values of x or y, which would result in a division by zero error. It is important to identify and address these restrictions when simplifying the function.

## 5. How does finding g(x)/h(y) for a given F(x,y) help in solving real-world problems?

By finding g(x)/h(y) for a given F(x,y), we can gain a better understanding of the underlying relationships between variables in a real-world problem. This can help in making predictions, optimizing solutions, and identifying patterns and trends in data. It also allows for more efficient and accurate calculations and simplifications of complex functions.

• Calculus and Beyond Homework Help
Replies
4
Views
690
• Calculus and Beyond Homework Help
Replies
6
Views
852
• Calculus and Beyond Homework Help
Replies
5
Views
763
• Calculus and Beyond Homework Help
Replies
5
Views
619
• Calculus and Beyond Homework Help
Replies
6
Views
548
• Calculus and Beyond Homework Help
Replies
2
Views
543
• Calculus and Beyond Homework Help
Replies
18
Views
1K
• Calculus and Beyond Homework Help
Replies
2
Views
462
• Calculus and Beyond Homework Help
Replies
8
Views
874
• Calculus and Beyond Homework Help
Replies
1
Views
440