Chapter 9: Problem 18

$(x+y)^{1 / 2}$ Hint : Make the change of variables (45^\circ rotation) $$ X=\frac{1}{\sqrt{2}}(x+y), \quad Y=\frac{1}{\sqrt{2}}(x-y) ; \quad \text { what is } \quad d X^{2}+d Y^{2} ? $$

Short Answer

Expert verified

$dX^2 + dY^2 = dx^2 + dy^2$

Step by step solution

- Understand the Variable Change

We need to make the change of variables: $X=\frac{1}{\sqrt{2}}(x+y)$ and $Y=\frac{1}{\sqrt{2}}(x-y)$. These represent a 45° rotation in the coordinate system.

- Compute Differentiate X and Y

Next, we find the differentials of $X$ and $Y$: $dX = \frac{1}{\sqrt{2}}(dx + dy)$ and $dY = \frac{1}{\sqrt{2}}(dx - dy)$.

- Square the Differentials

Square the differentials of $X$ and $Y$: $(dX)^2 = \left(\frac{1}{\sqrt{2}}(dx + dy)\right)^2$ and $(dY)^2 = \left(\frac{1}{\sqrt{2}}(dx - dy)\right)^2$.

- Expand the Squared Differentials

Expand the squared differentials using the distributive property: $(dX)^2 = \frac{1}{2}(dx^2 + 2dx \, dy + dy^2)$ and $(dY)^2 = \frac{1}{2}(dx^2 - 2dx \, dy + dy^2)$.

- Add the Squared Differentials

Add $dX^2$ and $dY^2$ together: $dX^2 + dY^2 = \frac{1}{2}(dx^2 + 2dx \, dy + dy^2) + \frac{1}{2}(dx^2 - 2dx \, dy + dy^2)$.

- Simplify the Expression

Combine the terms to simplify: $dX^2 + dY^2 = \frac{1}{2}(2dx^2 + 2dy^2) = dx^2 + dy^2$.

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Key Concepts

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

Variable Change

A variable change is crucial in mathematical problems to simplify the equations or transform them into a more tractable form. In this exercise, we use the variables:

$X = \frac{1}{\sqrt{2}}(x + y)$
$Y = \frac{1}{\sqrt{2}}(x - y)$

This change effectively rotates the coordinate system by 45 degrees. A variable change can often simplify differential equations, integrals, or other expressions in multivariable calculus.

Differential Forms

Differential forms involve the differentials of functions, which are represented as $dX$ and $dY$ in this problem. Here, they help us determine how small changes in the input ($x$ and $y$) affect the output ($X$ and $Y$). By differentiating the new variables, we get:

$dX = \frac{1}{\sqrt{2}}(dx + dy)$
$dY = \frac{1}{\sqrt{2}}(dx - dy)$

These differential forms are fundamental in calculus to find slopes, areas, and in this case, aid in coordinate transformation.

45-Degree Rotation

The change of variables mentioned represents a 45-degree rotation of the coordinate system. This is visually represented as spinning the axes by 45 degrees, which can simplify the problem at hand. Mathematically, it's shown using:

$X = \frac{1}{\sqrt{2}}(x + y)$
$Y = \frac{1}{\sqrt{2}}(x - y)$

This geometric interpretation simplifies expressions involving both variables because it aligns the problem along new axes, often orthogonal, and thereby easier to manipulate.

Squared Differentials

Squaring the differentials of $X$ and $Y$ gives us insight into distances and areas under differentials in a transformed coordinate system. Squaring results in:

$(dX)^2 = \left(\frac{1}{\sqrt{2}}(dx + dy)\right)^2$
$(dY)^2 = \left(\frac{1}{\sqrt{2}}(dx - dy)\right)^2$

Expanding these and adding them results in the problem's simplified form by ensuring that combining the differentials takes into account cross terms appropriately.

Simplification

The final step involves combining and simplifying the terms from the squared differentials expanded previously. By adding and combining like terms, we see the terms with cross products $dx \, dy$ cancel out, leaving us with:

$dX^2 + dY^2 = \frac{1}{2}(dx^2 + 2dx \, dy + dy^2) + \frac{1}{2}(dx^2 - 2dx \, dy + dy^2)$
After simplification:
$dX^2 + dY^2 = dx^2 + dy^2$

Thus, the problem is simplified into a more manageable form, demonstrating the power of variable change and other transformations in calculus.

\((x+y)^{1 / 2}\) Hint : Make the change of variables (45^\circ rotation) $$ X=\frac{1}{\sqrt{2}}(x+y), \quad Y=\frac{1}{\sqrt{2}}(x-y) ; \quad \text { what is } \quad d X^{2}+d Y^{2} ? $$