Why Fix One Coordinate?
A surface \(z=f(x,y)\) assigns a height \(z\) to each allowed input pair \((x,y)\). To understand its shape, it can help to examine what happens along a plane that holds one coordinate constant. Fixing \(x=a\) restricts attention to the points of the surface directly above the line \(x=a\) in the \(xy\)-plane.
The intersection is a curve, called a vertical trace at fixed \(x\). Its height is found by substituting \(a\) for \(x\), while \(y\) remains free to vary. The trace can then be sketched in the vertical plane \(x=a\), using \(y\) and \(z\) as the coordinates within that plane.
Definition and Geometric Meaning
Let \(f:D\to\mathbb{R}\), where \(D\) is a subset of \(\mathbb{R}^2\). The graph of \(f\) is the set of points \((x,y,z)\) with \((x,y)\in D\) and \(z=f(x,y)\). For a real number \(a\), define the allowed \(y\)-values in the slice by
The vertical trace at fixed \(x=a\) consists of the graph points whose \(x\)-coordinate is \(a\). If \(D_a\) is empty, there is no trace at that value of \(a\). Otherwise, its points have the form \((a,y,f(a,y))\), with \(y\in D_a\). In the plane \(x=a\), the trace can be drawn using the pair \((y,z)\), where \(z=f(a,y)\).
The word “vertical” describes the slicing plane: the plane \(x=a\) contains the vertical \(z\)-direction. It does not mean that the trace itself must be a vertical line. The trace may curve, rise, fall, or have multiple pieces, depending on the formula and its domain.
Why Substitution Gives the Trace
The description follows directly from the definition of a graph and of the plane \(x=a\). A point lies on the graph precisely when it has the form \((x,y,f(x,y))\) for an allowed input. It lies in the plane \(x=a\) precisely when its first coordinate equals \(a\). Imposing both conditions gives the trace.
It has coordinates \((x,y,f(x,y))\), with \((x,y)\in D\).
The condition \(x=a\) changes the point to \((a,y,f(a,y))\), provided \((a,y)\in D\).
Since \(x\) is fixed, use \(y\) and \(z\) to plot \(z=f(a,y)\).
This also proves the formula in both directions. Every point of the surface-plane intersection must have \(x=a\) and satisfy \(z=f(a,y)\), so it appears in the stated parametrization. Conversely, every point \((a,y,f(a,y))\) for which \((a,y)\in D\) lies on the graph and in the plane \(x=a\). Thus the parametrization describes exactly the intersection, with no extra points and none omitted.
Worked Example: Traces of an Upward-Opening Surface
Worked Example: Traces of \(z=x^2+y^2\)
Find and sketch the traces at \(x=0\), \(x=1\), and \(x=2\). The domain is all of \(\mathbb{R}^2\), so every real value of \(y\) is allowed for each slice.
For \(x=0\), substitute \(0\) for \(x\):
This is a parabola in the plane \(x=0\), with vertex \((0,0,0)\). For instance, at \(y=1\), \(z=1\), giving the point \((0,1,1)\). Substitution into the original formula confirms \(0^2+1^2=1\).
For \(x=1\):
The trace is again a parabola, now in the plane \(x=1\). Its lowest point occurs at \(y=0\), where \(z=1\), so the vertex is \((1,0,1)\). At \(y=1\), the trace gives \(z=2\), and the original formula gives \(1^2+1^2=2\), as required.
For \(x=2\):
This trace is a parabola in the plane \(x=2\) with vertex \((2,0,4)\). At \(y=1\), it gives \(z=5\); checking in the original surface formula, \(2^2+1^2=5\).
To sketch the family, draw a parabola in each of the three vertical planes. Each opens upward and has the same shape in its own \(yz\)-coordinates. As \(x\) increases from \(0\) to \(1\) to \(2\), the lowest height increases from \(0\) to \(1\) to \(4\). Each trace’s vertex is therefore at \((a,0,a^2)\).
Worked Example: Traces That Shift Vertically
Worked Example: Traces of \(z=y^2+2x\)
Find traces at \(x=-1\), \(x=0\), and \(x=1\), and describe how they compare. Again, every real \(y\) is allowed.
At \(x=-1\), substitute carefully:
The trace is an upward-opening parabola with vertex \((-1,0,-2)\). At \(y=1\), \(z=1-2=-1\). The original formula gives \(1^2+2(-1)=1-2=-1\), confirming the point \((-1,1,-1)\).
At \(x=0\):
The vertex is \((0,0,0)\). At \(y=1\), \(z=1\), and the original formula gives \(1^2+2(0)=1\).
At \(x=1\):
The vertex is \((1,0,2)\). At \(y=1\), \(z=1+2=3\), matching \(1^2+2(1)=3\) in the original formula.
All three curves are the same parabola shape when plotted in their respective planes. Their vertex heights differ: \(-2\), \(0\), and \(2\). Comparing these fixed-\(x\) traces shows that increasing \(x\) shifts the curve upward in \(z\) for this surface.
Worked Example: A Trace with a Restricted Domain
Worked Example: Traces of a Hemisphere
Consider \(z=\sqrt{9-x^2-y^2}\), whose domain consists of input pairs satisfying \(x^2+y^2\leq 9\). Find the trace at \(x=2\), and determine its endpoints and highest point.
Substitute \(x=2\) into the equation:
The square root is real only when \(5-y^2\geq 0\). Hence \(y^2\leq 5\), or \(-\sqrt{5}\leq y\leq\sqrt{5}\). The trace is a curve in the plane \(x=2\), and its points are \((2,y,\sqrt{5-y^2})\) over this interval.
At \(y=-\sqrt{5}\), the height is \(z=\sqrt{5-5}=0\), giving \((2,-\sqrt{5},0)\). At \(y=\sqrt{5}\), it is also \(z=0\), giving \((2,\sqrt{5},0)\). At \(y=0\), \(z=\sqrt{5}\), so the highest point is \((2,0,\sqrt{5})\). These points satisfy the original equation: for either endpoint, \(9-2^2-(\sqrt{5})^2=9-4-5=0\); at the highest point, \(9-4-0=5\), whose square root is \(\sqrt{5}\).
The equation \(y^2+z^2=5\), together with \(z\geq 0\), shows that the trace is the upper half of a circle in the \(yz\)-coordinates of the plane \(x=2\). The domain restriction matters: values of \(y\) outside \([-\sqrt{5},\sqrt{5}]\) do not produce points on the surface at this \(x\).
How to Sketch a Fixed-\(x\) Trace
A practical sketch begins with the equation \(z=f(a,y)\) and the allowed \(y\)-values. Treat \(a\) as a constant and \(y\) as the variable. Identify the curve’s familiar shape when possible, then calculate a few points that locate it accurately. Mark the plane \(x=a\), since the curve is not floating in an unspecified vertical plane.
- Substitute \(x=a\) everywhere in the formula.
- Determine which values of \(y\) keep the input \((a,y)\) inside the function’s domain.
- Find useful points, such as intercepts, vertices, endpoints, or values at \(y=0\), when they are in the domain.
- Plot the resulting curve using \(y\) and \(z\) as coordinates within the plane \(x=a\).
- Repeat at other \(x\)-values and compare the curves in their corresponding planes.
A family of traces is useful because each curve records how height changes as \(y\) varies while \(x\) stays fixed. Comparing several values of \(a\) can reveal whether the curves translate, change size, change shape, or have different domains. It gives structured information about a surface, although a few selected traces alone do not specify every point on it.
Common Mistake: Substituting for the Wrong Variable
A frequent error is to substitute \(a\) for both \(x\) and \(y\). For example, for \(z=x^2+3y\), a mistaken calculation at \(x=2\) might be
This has also set \(y=2\), so it gives only the height at the single input \((2,2)\), not the trace. The correct substitution fixes \(x\) and leaves \(y\) variable:
For example, at \(y=0\), this gives \(z=4\), and the original formula gives \(2^2+3(0)=4\). At \(y=1\), it gives \(z=7\), and the original formula gives \(2^2+3(1)=7\). These are two points on the trace in the plane \(x=2\), rather than one height incorrectly treated as the entire curve.
Check Your Understanding
Use the definition of a fixed-\(x\) trace to answer these questions.
- For \(z=x^2+2y^2\), what equation describes the trace at \(x=3\), and what is its lowest point?
- For \(z=4-x^2-y^2\), which \(y\)-values are allowed in the trace at \(x=1\)?
- For \(z=xy+1\), write the trace equation at \(x=-2\). Which variable remains free?
- What plane contains the trace at \(x=a\), and which coordinates can be used to plot the curve within that plane?
- Why does substituting both \(x=a\) and \(y=a\) fail to describe the entire fixed-\(x\) trace?