From a Height Formula to an Equation
A surface given by \(z=f(x,y)\) assigns a height \(z\) to each permitted pair \((x,y)\). The same surface can also be described by a relation among all three coordinates: \(z-f(x,y)=0\). This second form is called an implicit equation because the height is specified through an equation rather than isolated on one side.
For example, the formula \(z=x^2+y^2\) describes all points whose height equals the sum of the squares of their first two coordinates. Subtracting the right-hand side gives \(z-x^2-y^2=0\). The subtraction changes the form of the description, not the points it describes.
The crucial detail is that the function’s domain must be preserved. If \(f\) is defined only for certain pairs \((x,y)\), its graph contains no points above pairs outside that domain. An equation involving \(z-f(x,y)\) must therefore be considered on the same domain.
Explicit and Implicit Descriptions
Let \(D\) be a subset of \(\mathbb{R}^2\), and let \(f:D\to\mathbb{R}\) be a function. Its graph is the set of points in \(\mathbb{R}^3\) formed by pairing each input \((x,y)\) with its output \(f(x,y)\).
In the explicit description, the height is isolated: \(z=f(x,y)\). In the implicit description, the coordinates satisfy a relation: \(F(x,y,z)=0\). In this specific construction, \(F\) is obtained by moving the formula for the height to the left side of the equation.
The equivalence is a statement about sets of points, not merely about two algebraically similar formulas. It says that a point belongs to one description exactly when it belongs to the other, with the same restriction \((x,y)\in D\).
Why the Descriptions Are Equivalent
The justification follows directly from equality. For any permitted input \((x,y)\in D\) and any real number \(z\), the equation \(F(x,y,z)=0\) says \(z-f(x,y)=0\). Adding \(f(x,y)\) to both sides gives \(z=f(x,y)\). Conversely, if \(z=f(x,y)\), subtracting \(f(x,y)\) from both sides gives \(z-f(x,y)=0\). Thus neither description includes an extra point or omits a point from the graph.
A point \((x,y,z)\) is on the graph precisely when \((x,y)\in D\) and \(z=f(x,y)\).
Set \(F(x,y,z)=z-f(x,y)\), using the same permitted inputs \((x,y)\in D\).
The equality \(z=f(x,y)\) holds exactly when \(z-f(x,y)=0\), which is exactly \(F(x,y,z)=0\).
Every point satisfying the graph condition satisfies the implicit equation, and every point satisfying the implicit equation on the stated domain satisfies the graph condition.
This proof uses no special properties such as continuity or differentiability. It depends only on the definition of a function and the fact that subtracting the same quantity from both sides preserves equality. The construction works for any real-valued function on any domain \(D\subseteq\mathbb{R}^2\).
Worked Examples
Worked Example: A Paraboloid
Convert the graph \(z=x^2+y^2\), defined for all \((x,y)\in\mathbb{R}^2\), into an implicit equation. Then check the point \((1,2,5)\).
Here \(f(x,y)=x^2+y^2\). By definition, form \(F=z-f(x,y)\):
The implicit equation is therefore \(z-x^2-y^2=0\). To check the proposed point, substitute \(x=1\), \(y=2\), and \(z=5\):
The point satisfies the implicit equation. The explicit formula gives the same check: \(x^2+y^2=1+4=5=z\). Thus \((1,2,5)\) lies on both descriptions. More generally, for every real \(x,y\), the equation \(z-x^2-y^2=0\) rearranges to \(z=x^2+y^2\), so it describes exactly the graph.
Worked Example: A Square-Root Graph and Its Domain
Convert \(z=\sqrt{9-x^2-y^2}\) to an implicit description, keeping the domain correct. Determine whether \((1,2,2)\) is on the graph.
The square root is real only when \(9-x^2-y^2\geq 0\), so the domain is \(D=\{(x,y):x^2+y^2\leq 9\}\). With \(f(x,y)=\sqrt{9-x^2-y^2}\), the direct implicit equation is
For the proposed point, \(1^2+2^2=5\leq 9\), so its first two coordinates are in the domain. Its height from the explicit formula is \(\sqrt{9-1-4}=\sqrt{4}=2\), as proposed. Substitution in the implicit equation confirms this:
One may also square the equation to obtain \(z^2+x^2+y^2=9\), but squaring alone loses information: it permits negative \(z\) as well. To preserve the original graph, the squared equation must be accompanied by \(z\geq 0\). With those restrictions, the equivalence can be checked in both directions. The original equation implies \(z\geq0\) and \(z^2=9-x^2-y^2\). Conversely, if \(z^2+x^2+y^2=9\) and \(z\geq0\), then \(z=\sqrt{9-x^2-y^2}\). The domain restriction is also satisfied because \(x^2+y^2=9-z^2\leq9\).
Thus the equivalent restricted polynomial description is \(z^2+x^2+y^2=9\) with \(z\geq0\). The sign condition is essential; the direct equation \(z-\sqrt{9-x^2-y^2}=0\) already retains it.
Worked Example: A Function with Mixed Terms
Write the graph \(z=xy+2x-3y\) implicitly and test the point \((1,2,-2)\).
Here \(f(x,y)=xy+2x-3y\), defined for all real \(x,y\). Subtract \(f\) from \(z\):
Hence the implicit equation is \(z-xy-2x+3y=0\). At \((1,2,-2)\), evaluate each term:
The explicit formula gives \(xy+2x-3y=(1)(2)+2(1)-3(2)=2+2-6=-2\), agreeing with \(z=-2\). The point therefore satisfies both descriptions. Conversely, rearranging \(z-xy-2x+3y=0\) gives \(z=xy+2x-3y\), so every solution is on the stated graph.
Keeping Restrictions and Equivalence Clear
When converting a graph to an implicit equation, begin by identifying the function and its domain. Then form \(z-f(x,y)\), set it equal to zero, and retain any domain restriction. If you perform additional algebraic operations, check whether they are reversible under the conditions you have stated.
| Explicit description | Implicit description | Restriction to retain |
|---|---|---|
| \(z=x^2+y^2\) | \(z-x^2-y^2=0\) | All real \(x,y\) |
| \(z=\sqrt{9-x^2-y^2}\) | \(z-\sqrt{9-x^2-y^2}=0\) | \(x^2+y^2\leq9\) |
| \(z=xy+2x-3y\) | \(z-xy-2x+3y=0\) | All real \(x,y\) |
A general equation \(G(x,y,z)=0\) is an implicit description of a set of points, but it need not have the particular form \(z-f(x,y)=0\). The special form studied here guarantees that each permitted input \((x,y)\) has exactly one corresponding value of \(z\), namely \(f(x,y)\). Other implicit equations may fail to determine a single height for each input. Therefore, when converting a known graph, use the defining function itself to construct the equation rather than assuming that any equation involving \(x,y,z\) will have the same meaning.
A Common Sign Error
A frequent mistake is to change the sign incorrectly while moving terms. Starting from \(z=f(x,y)\), one might incorrectly write the implicit equation as \(z+f(x,y)=0\). That equation instead means \(z=-f(x,y)\), which generally describes a different graph.
For a concrete check, take \(f(x,y)=x^2+y^2\) and the graph point \((1,1,2)\). The incorrect equation gives
So the point on the original graph fails the supposed implicit equation. The correct conversion subtracts \(f\) from both sides: \(z=f(x,y)\) becomes \(z-f(x,y)=0\). At the same point, \(2-(1^2+1^2)=2-2=0\), as required. A reliable check is to substitute a point known to satisfy the explicit formula into the proposed implicit equation.
Why This Form Is Useful
The explicit form emphasizes a height rule: choose \((x,y)\), then calculate \(z\). The implicit form emphasizes a condition that points in space must satisfy. Both are useful descriptions of one surface, and switching between them lets us choose the form suited to the question at hand. For this conversion, the key is not a complicated technique: it is careful translation of the equality, together with preservation of the domain.
Check Your Understanding
For each question, pay attention to the domain and to whether the proposed equation has exactly the intended meaning.
- Write an implicit equation for \(z=3x-y^2\), and state its domain.
- If \(f(x,y)=\sqrt{x-y}\), what restriction on \(x,y\) is needed when writing \(z-f(x,y)=0\)?
- Explain why \(z=f(x,y)\) and \(z-f(x,y)=0\) describe the same points.
- For \(z=\sqrt{4-x^2-y^2}\), what additional restriction is needed if you replace the square-root equation by \(z^2+x^2+y^2=4\)?
- Check whether \((2,1,5)\) satisfies the implicit equation obtained from \(z=xy+2x-y\).