Finding dx & dy: Solving Math Homework Problem

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In summary: We want to find δx and δy in terms of δu and δv. So we can write:\begin{align*}\delta x & = \frac{\partial x}{\partial u} \delta u + \frac{\partial x}{\partial v} \delta v \\\delta y & = \frac{\partial y}{\partial u}
  • #1
renlok
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Homework Statement


x = (1/2)(u^2 - v^2), y = uv
find [tex]\delta{x}[/tex] and [tex]\delta{y}[/tex]

Homework Equations


-

The Attempt at a Solution


I think its simple enough but I don't know what sort of method to use to answer this.

Any help would be really awesome :)
 
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  • #2
With respect to what are you finding the partial derivatives - u or v?

In either case, treat the other variable as a constant. If with respect to u, then treat v as a constant. If with respect to v, then treat u as a constant.
 
  • #3
not in respect to something i don't need to find a gradient, i need to find the infintesimal change in x or y in terms of v & u.

I've found for finding [tex]\delta{y}[/tex] if you can treat v & u as functions of x you can just use the product rule and divide though by [tex]\delta{x}[/tex] to leave you with [tex]\delta{y} = u\delta{v} + v\delta{u}[/tex]
but I've still no clue how to get [tex]\delta{x}[/tex]
 
  • #4
I'm not sure what you did to find δy. What you said doesn't really make sense. Think of x and y as functions of u and v:

x = x(u,v)
y = y(u,v)

Then you have

[tex]\begin{align*}
dx & = \frac{\partial x}{\partial u} du + \frac{\partial x}{\partial v} dv \\
dy & = \frac{\partial y}{\partial u} du + \frac{\partial y}{\partial v} dv
\end{align*}
[/tex]
 

Related to Finding dx & dy: Solving Math Homework Problem

1. What is the process for finding dx and dy in a math homework problem?

The process for finding dx and dy in a math homework problem involves using the chain rule and implicit differentiation. First, identify the function that contains both x and y. Then, take the derivative of that function with respect to x, treating y as a constant. This will give you dx. Next, take the derivative of the function with respect to y, treating x as a constant. This will give you dy. Finally, use algebra to solve for dx and dy in terms of x, y, and any other variables given in the problem.

2. What is the purpose of finding dx and dy in a math homework problem?

The purpose of finding dx and dy is to find the rates of change of a function with respect to its independent variables. This is useful for understanding how small changes in one variable affect the overall function, and for solving more complex problems in calculus and physics.

3. Can you explain the chain rule and how it is used to find dx and dy?

The chain rule is a rule in calculus that allows us to take the derivative of a function within a function. In the context of finding dx and dy, we use the chain rule to take the derivative of the function that contains both x and y. This involves treating y as a constant while taking the derivative with respect to x. The result is dx, or the change in x. Then, we take the derivative of the function with respect to y, treating x as a constant. This gives us dy, or the change in y. Together, dx and dy represent the rates of change of the function with respect to x and y, respectively.

4. Are there any common mistakes to avoid when finding dx and dy in a math homework problem?

One common mistake to avoid is forgetting to use the chain rule and treating both x and y as independent variables. Another mistake is not simplifying the final expressions for dx and dy, which can make the problem more difficult to solve. It's also important to pay attention to any given variables and constants in the problem and include them in the final expressions for dx and dy.

5. Can you provide an example of finding dx and dy in a math homework problem?

Sure! Let's say we have the equation x^2 + y^2 = 25, and we want to find dx and dy when x = 3. First, we take the derivative of the equation with respect to x, treating y as a constant: 2x + 2y(dy/dx) = 0. This gives us dx = -2x/(2y(dy/dx)). Then, we take the derivative of the equation with respect to y, treating x as a constant: 2x(dx/dy) + 2y = 0. This gives us dy = -2y/(2x(dx/dy)). Finally, we solve for dx and dy in terms of x, y, and any given constants: dx = -3/4, dy = -4/3. So, when x = 3, dx = -3/4 and dy = -4/3.

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