Implicit Functions

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no formula

Let $y(x)$ be a function whose values are given by the equation $x^2 + y^2 = 1$.

What is the value of $y(0)$?

The value of $y(0)$ is the value of $y$ that makes the equation $0^2 + y^2 = 1$ true.

There are two of them: $y(0) = -1$ and $y(0) = 1$.

Whoops! $y(x)$ is not a function, which we can easily see from the graph. This violates our one and only rule for funcitons.

Most values for $x$ have two corresponding values for $y$.

Restrictions

Suppose we are really interested in negative values of $y(x)$ for domain values around $x = 0.6$.

We are interested in this part of the graph.

Then we could create a restricted version of $y$ by selecting an appropriate domain and range.

We have selected a domain and range the isolates the part of the curve around our point in such a way that each $x$-value has a single (unique) corresponding $y$-value.

$\blacktriangleright $ New domain is $(0.342, 0.766)$

$\blacktriangleright $ New range is $(-0.939, -0.642)$

With this new domain and range and our equation, $y(x)$ is a function whose pairs are described by $x^2 + y^2 = 1$.

Solving

We cannot solve $x^2 + y^2 = 1$, but we can solve it for two versions of $y$.

$\blacktriangleright $ Version 1: $y = \sqrt {1-x^2}$

$\blacktriangleright $ Version 2: $y = -\sqrt {1-x^2}$

For our situation above, we could have used $y = -\sqrt {1-x^2}$ with domain $[-1, 1]$. Our graph would look like

However, this is not always possible.

The graph of $x^3 + y^3 - 3 x y = 0$ looks like

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There is no way to get $y$ by itself on one side of the equation. We are stuck with the equation $x^3 + y^3 - 3xy = 0$.

The curve described by $(x^2 + y^2)^2 = 20 x y^2$ looks like

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We cannot isolate $y$. However, the equation involves squares and squares of squares. It might be possible to apply the quadratic formula.

\[ (x^2 + y^2)^2 = 20 x y^2 \]

\[ x^4 + 2 x^2 y^2 + y^4 = 20 x y^2 \]

\[ y^4 + (2 x^2 - 20 x) y^2 + x^4 = 0 \]

This is a quadratic in $y^2$. We could split it into two pieces.

\[ y^2 = \frac {-(2 x^2 - 20 x) + \sqrt {(2x^2-20x)^2 - 4 \cdot 1 \cdot x^4}}{2} \]

\[ y^2 = \frac {-(2 x^2 - 20 x) - \sqrt {(2x^2-20x)^2 - 4 \cdot 1 \cdot x^4}}{2} \]

And then, each of these can be broken down.

\[ y = \sqrt {\frac {-(2 x^2 - 20 x) + \sqrt {(2x^2-20x)^2 - 4 \cdot 1 \cdot x^4}}{2}} \]

\[ y = -\sqrt {\frac {-(2 x^2 - 20 x) + \sqrt {(2x^2-20x)^2 - 4 \cdot 1 \cdot x^4}}{2}} \]

\[ y^2 = \sqrt {\frac {-(2 x^2 - 20 x) - \sqrt {(2x^2-20x)^2 - 4 \cdot 1 \cdot x^4}}{2}} \]

\[ y^2 = -\sqrt {\frac {-(2 x^2 - 20 x) - \sqrt {(2x^2-20x)^2 - 4 \cdot 1 \cdot x^4}}{2}} \]

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If our equation only involves squares, then we have a chance to break it into pieces using the quadratic formula. However, we don’t have any other “formulas” for other types of expressions. Generally, other types of equations stay as they are.

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more examples can be found by following this link
More Examples of Implicit Functions

2025-05-17 23:31:47