Why Tangent Planes to Implicit Surfaces Matter
A surface in three-dimensional space is sometimes given as a graph, such as \(z=g(x,y)\). In other problems, it is defined by a single equation involving all three coordinates: \(F(x,y,z)=c\). The second description is called implicit because it does not necessarily isolate one variable as a function of the other two.
At a suitable point on an implicit surface, the gradient of \(F\) determines a plane that approximates the surface nearby. That plane is useful for understanding the surface’s local orientation and for making first-order approximations. To use the formula correctly, however, we must know that the level set really is a differentiable surface near the point.
Level Surfaces and Regular Points
Let \(F:\mathbb{R}^3\to\mathbb{R}\) be a scalar-valued function, and let \(c\) be a real number. The level surface at level \(c\) is the set of points satisfying \(F(x,y,z)=c\). A point \(\mathbf{a}=(a_1,a_2,a_3)\) lies on this level surface when \(F(\mathbf{a})=c\).
The gradient at \(\mathbf{a}\) is the vector of first partial derivatives evaluated there:
A point on the level surface is called a regular point if its gradient is nonzero. The condition says that at least one of the three first partial derivatives is nonzero.
Both parts of the condition matter. The nonzero gradient identifies a direction in which the function changes to first order, while continuous first partial derivatives provide the regularity needed to guarantee a local surface. Differentiability at the single point alone is not enough to justify the tangent-plane conclusion.
The Tangent-Plane Formula
The formula uses the gradient at the point of tangency, not at a general point \((x,y,z)\). It describes a plane through \(\mathbf{a}\), since substituting \(x=a_1\), \(y=a_2\), and \(z=a_3\) makes every coordinate difference zero.
Why the Formula Works
Here is a derivation using the implicit function theorem. Since the gradient is nonzero, at least one partial derivative is nonzero. Suppose, for the derivation, that \(F_z(\mathbf{a})\ne 0\). If instead \(F_x(\mathbf{a})\) or \(F_y(\mathbf{a})\) is nonzero, the same argument applies after choosing that coordinate as the dependent variable.
The implicit function theorem gives a neighborhood of \((a_1,a_2)\) and a continuously differentiable function \(z=g(x,y)\) whose graph describes the level surface near \(\mathbf{a}\). Thus \(g(a_1,a_2)=a_3\), and throughout this local graph, \(F(x,y,g(x,y))=c\). Differentiating this identity with respect to each independent variable gives the following steps.
By the chain rule, \(F_x+F_z g_x=0\) along the graph. At \(\mathbf{a}\), this gives \(g_x(a_1,a_2)=-F_x(\mathbf{a})/F_z(\mathbf{a})\).
Similarly, \(F_y+F_z g_y=0\), so \(g_y(a_1,a_2)=-F_y(\mathbf{a})/F_z(\mathbf{a})\).
The first-order equation for the graph at \((a_1,a_2,a_3)\) is \(z-a_3=g_x(a_1,a_2)(x-a_1)+g_y(a_1,a_2)(y-a_2)\).
Insert the two derivative formulas and multiply by the nonzero number \(F_z(\mathbf{a})\). Rearranging gives \(F_x(\mathbf{a})(x-a_1)+F_y(\mathbf{a})(y-a_2)+F_z(\mathbf{a})(z-a_3)=0\), the stated plane.
This establishes the formula where \(F_z(\mathbf{a})\ne 0\). If a different partial derivative is nonzero, solve locally for its coordinate instead; the same calculation yields the same symmetric equation. The gradient is therefore a normal vector to this tangent plane, consistent with the gradient’s role on a regular level surface.
Worked Examples
Worked Example: Tangent Plane to a Sphere
Find the tangent plane to \(x^2+y^2+z^2=25\) at \(\mathbf{a}=(3,4,0)\).
Define \(F(x,y,z)=x^2+y^2+z^2\) and \(c=25\). First check that the point is on the level surface:
The function has continuous first partial derivatives everywhere. Its gradient is \(\nabla F=\langle 2x,2y,2z\rangle\), so at the point, \(\nabla F(3,4,0)=\langle 6,8,0\rangle\ne\mathbf{0}\). The tangent-plane formula gives
Expanding and simplifying, \(6x-18+8y-32=0\), hence \(6x+8y=50\), or \(3x+4y=25\). Check the point in the plane: \(3(3)+4(4)=9+16=25\). It lies on both the sphere and the calculated plane.
Worked Example: A Product and a Square
Find the tangent plane to \(xy+z^2=3\) at \(\mathbf{a}=(1,2,1)\).
Set \(F(x,y,z)=xy+z^2\) and \(c=3\). The point satisfies the level equation because \(F(1,2,1)=(1)(2)+1^2=2+1=3\). The partial derivatives are \(F_x=y\), \(F_y=x\), and \(F_z=2z\). Therefore, \(\nabla F(1,2,1)=\langle 2,1,2\rangle\), which is nonzero. The tangent plane is
Combining terms gives \(2x-2+y-2+2z-2=0\), so \(2x+y+2z=6\). Substitute the point into the simplified equation: \(2(1)+2+2(1)=2+2+2=6\). The original level equation also gives \(1\cdot2+1^2=3\), confirming that the point used in the calculation is on the required surface.
Worked Example: Solving the Tangent Plane for \(z\)
Find the tangent plane to \(x^2+y^2-z=0\) at \(\mathbf{a}=(1,2,5)\), and write it with \(z\) isolated.
Let \(F(x,y,z)=x^2+y^2-z\) and \(c=0\). Check the level equation: \(F(1,2,5)=1^2+2^2-5=1+4-5=0\). The partial derivatives are \(F_x=2x\), \(F_y=2y\), and \(F_z=-1\). Thus \(\nabla F(1,2,5)=\langle 2,4,-1\rangle\ne\mathbf{0}\), and the formula gives
Expanding, \(2x-2+4y-8-z+5=0\), so \(2x+4y-z-5=0\). Isolating \(z\) gives \(z=2x+4y-5\). Check the point in this plane: \(2(1)+4(2)-5=2+8-5=5\), as required. As an additional check, the original surface can be written \(z=x^2+y^2\). Its first-order graph at \((1,2)\) is \(z-5=2(x-1)+4(y-2)\), which simplifies to the same plane.
A Common Mistake: Omitting the Point of Tangency
The equation of a plane with normal vector \(\langle A,B,C\rangle\) through a point \((a_1,a_2,a_3)\) uses the coordinate differences \(x-a_1\), \(y-a_2\), and \(z-a_3\). A frequent error is to use the gradient components as coefficients but omit these differences.
For the sphere example, that error would give \(6x+8y=0\). But the point \((3,4,0)\) does not satisfy it: \(6(3)+8(4)=18+32=50\ne0\). The proposed plane therefore does not even pass through the point of tangency. The correct calculation keeps the offsets: \(6(x-3)+8(y-4)=0\), which simplifies to \(6x+8y=50\). Substitution now gives \(18+32=50\), so the point lies on the plane.
There is also an important assumption check. A nonzero gradient at one point, without continuous first partial derivatives in a neighborhood, does not by itself guarantee the local surface regularity needed for this tangent-plane theorem. Verify the hypotheses before applying the formula; do not treat the equation as justified by a pointwise gradient calculation alone.
Interpreting the Result
The tangent plane is a first-order approximation to the level surface near the point. In the graph case \(z=g(x,y)\), the plane records the value and the two first partial derivatives of \(g\) at the point. For an implicit equation, the gradient formula packages that information without requiring the final plane to be written as a graph.
This is especially useful when solving explicitly for a coordinate would be difficult or impossible in a convenient form. The level equation may still be easy to differentiate, and its gradient at the point gives the needed coefficients directly. The regularity conditions ensure that this local linear approximation represents an actual differentiable surface rather than merely a formal equation.
Check Your Understanding
Use the level-surface conditions and tangent-plane formula to answer these questions.
- What conditions on \(F\) and \(\mathbf{a}\) ensure that the tangent-plane theorem applies?
- For \(F(x,y,z)=x^2+y^2+z^2\), find the gradient at \((0,0,2)\) and write the tangent plane to the level surface through that point.
- Why must the tangent-plane equation include the coordinates of the point of tangency?
- If \(F_z(\mathbf{a})=0\) but \(F_y(\mathbf{a})\ne0\), does the gradient vanish? Which coordinate could the implicit function theorem locally express in terms of the others?
- Explain briefly how the implicit function theorem and the chain rule lead to the gradient form of the plane.