From Vertical Slices to a Surface
A formula \(z=f(x,y)\) assigns a height to each allowed point \((x,y)\) in the domain. Its graph is a surface in three-dimensional space, but a perspective drawing can make the shape difficult to infer from the formula alone. Vertical traces give a more manageable view: each one is a curve formed by holding one input coordinate fixed while allowing the other input coordinate to vary.
The aim is not to draw several unrelated curves. They must fit together as parts of one surface. Their locations, heights, shared points, intercepts, and the domain all constrain the sketch. These constraints help reveal a plausible shape and expose inconsistent drawings.
For an x-fixed trace, \(x\) stays constant and \(y\) varies. For a y-fixed trace, \(y\) stays constant and \(x\) varies. Each trace must be drawn in its actual vertical plane. A curve’s equation in terms of \(y\) and \(z\), for instance, describes its shape inside the plane \(x=a\); it does not remove the requirement that \(x=a\).
A Consistency Principle for Traces
Any x-fixed trace at \(x=a\) and y-fixed trace at \(y=b\) meet above the same input point \((a,b)\), provided \((a,b)\) belongs to the domain. Their shared point must have height \(f(a,b)\). This gives a direct way to check whether the traces can belong to one graph.
This follows directly from the definition of the graph. Its points have the form \((x,y,f(x,y))\), with \((x,y)\in D\). Imposing \(x=a\) and \(y=b\) at once leaves exactly the point \((a,b,f(a,b))\). The domain condition matters: if \((a,b)\notin D\), the function has no assigned height there, so there is no shared graph point.
The coordinate intercepts provide further checks. The graph meets the \(z\)-axis only if \((0,0)\) is in the domain, at \((0,0,f(0,0))\). Its intersections with the \(x\)-axis or \(y\)-axis must also have \(z=0\): they can be found by solving \(f(x,0)=0\) or \(f(0,y)=0\), respectively, while respecting the domain.
A Practical Construction Method
For a hand sketch, select a few useful fixed values of \(x\) and \(y\), often including zero and symmetric positive and negative values. Substitute each value into the function, simplify, and identify the resulting curve in the appropriate vertical plane. Then plot the curves in their correct positions and check the points where x-fixed and y-fixed traces cross in the domain.
Decide which input points are allowed. Show any boundary beyond which no graph is present.
Select several values of \(x\) and \(y\) that make the restricted formulas easy to interpret.
Substitute the fixed coordinate, preserve its plane, and mark useful points such as vertices and intercepts.
At every intersection of chosen fixed-coordinate planes in the domain, verify that both traces give the same height.
Use the traces and domain to guide the surface between them, without treating a few slices as a complete description.
Worked Example: An Upward-Bending Bowl
Worked Example: An Upward-Bending Bowl
Sketch the surface \(z=x^2+2y^2\), whose domain is all of \(\mathbb{R}^2\), using traces at \(x=0\), \(x=1\), \(y=0\), and \(y=1\).
First fix \(x=0\). Substitution gives \(z=0^2+2y^2=2y^2\), so this is an upward-opening parabola in the plane \(x=0\). Its vertex is \((0,0,0)\). At \(y=1\) and \(y=-1\), the heights are \(2\), giving points \((0,1,2)\) and \((0,-1,2)\).
At \(x=1\), the trace is \(z=1^2+2y^2=1+2y^2\), also an upward-opening parabola, now in the plane \(x=1\). Its minimum occurs at \(y=0\), with point \((1,0,1)\). At \(y=1\), \(z=1+2=3\), giving \((1,1,3)\).
For the y-fixed traces, \(y=0\) gives \(z=x^2+2(0)^2=x^2\) in the plane \(y=0\). It passes through \((0,0,0)\) and \((1,0,1)\). At \(y=1\), the trace is \(z=x^2+2(1)^2=x^2+2\) in the plane \(y=1\). It passes through \((0,1,2)\) and \((1,1,3)\).
Check the shared point above \((1,1)\): the x-fixed trace at \(x=1\) gives \(z=1+2(1)^2=3\), and the y-fixed trace at \(y=1\) gives \(z=(1)^2+2=3\). Both contain \((1,1,3)\). At the origin, the height is \(0\). Since squares are nonnegative, \(x^2+2y^2\geq 0\), so the origin is the lowest point. The traces indicate a bowl rising away from it, more sharply in the \(y\)-direction than in the \(x\)-direction.
As a check, substituting \((1,1)\) into the original function gives \(1^2+2(1)^2=3\), matching both traces. The domain has no boundary, so the graph does not stop at a finite edge.
Worked Example: A Saddle Shape
Worked Example: A Saddle Shape
Sketch \(z=x^2-y^2\) over \(\mathbb{R}^2\), using traces through \(x=0\), \(x=1\), \(y=0\), and \(y=1\).
At \(x=0\), substitution gives \(z=0^2-y^2=-y^2\). This is a downward-opening parabola in the plane \(x=0\), with vertex \((0,0,0)\). At \(y=1\) it has height \(-1\), so \((0,1,-1)\) lies on the trace.
At \(x=1\), \(z=1^2-y^2=1-y^2\). This is another downward-opening parabola in the plane \(x=1\), with vertex \((1,0,1)\). It meets \(z=0\) when \(1-y^2=0\), or \(y=\pm1\). In particular, it contains \((1,1,0)\).
At \(y=0\), the trace is \(z=x^2-0^2=x^2\), an upward-opening parabola in the plane \(y=0\), with vertex \((0,0,0)\). At \(y=1\), it is \(z=x^2-1\), an upward-opening parabola in the plane \(y=1\). It passes through \((0,1,-1)\), and at \(x=1\) gives \(z=1-1=0\).
At the crossing above \((1,1)\), the x-fixed trace gives \(z=1-1^2=0\); the y-fixed trace gives \(z=1^2-1=0\). The original formula also gives \(1^2-1^2=0\). The traces bend upward in one coordinate direction and downward in the other. Together they describe a saddle: the origin is not a peak or a basin, since the graph rises along \(y=0\) and falls along \(x=0\).
The intercepts agree with this picture. On the \(x\)-axis, set \(y=0\) and \(z=0\): \(x^2=0\), so \(x=0\). On the \(y\)-axis, set \(x=0\) and \(z=0\): \(-y^2=0\), so \(y=0\). Thus the graph meets both input axes at the origin, which is also its \(z\)-axis intercept.
Worked Example: A Graph with a Circular Boundary
Worked Example: A Graph with a Circular Boundary
Sketch \(z=\sqrt{9-x^2-y^2}\), with domain \(D=\{(x,y):x^2+y^2\leq9\}\), using several vertical traces.
The domain is the disk of radius \(3\). At \(x=0\), the trace is \(z=\sqrt{9-y^2}\), for \(-3\leq y\leq3\). Squaring gives \(y^2+z^2=9\), and the original square root requires \(z\geq0\). Thus this trace is the upper semicircle of radius \(3\) in the plane \(x=0\), from \((0,-3,0)\) through \((0,0,3)\) to \((0,3,0)\).
At \(x=1\), the domain condition becomes \(1+y^2\leq9\), so \(-\sqrt{8}\leq y\leq\sqrt{8}\). The trace is \(z=\sqrt{8-y^2}\), or \(y^2+z^2=8\) with \(z\geq0\). Its highest point is \((1,0,\sqrt{8})\), and its endpoints are \((1,-\sqrt{8},0)\) and \((1,\sqrt{8},0)\). Since \(\sqrt{8}=2\sqrt{2}\), these endpoints are within the disk’s boundary.
At \(y=0\), the trace is \(z=\sqrt{9-x^2}\), for \(-3\leq x\leq3\). It is an upper semicircle in the plane \(y=0\), with center on the \(x\)-axis and radius \(3\). At \(y=2\), the domain condition gives \(x^2+4\leq9\), so \(-\sqrt{5}\leq x\leq\sqrt{5}\). The trace satisfies \(z=\sqrt{5-x^2}\), or \(x^2+z^2=5\) with \(z\geq0\).
Check the shared point above \((1,2)\). The x-fixed trace gives \(z=\sqrt{9-1^2-2^2}=\sqrt{4}=2\). The y-fixed trace gives \(z=\sqrt{5-1^2}=\sqrt{4}=2\). Substitution in the original function also gives \(z=\sqrt{9-1-4}=2\). The graph’s boundary occurs when \(x^2+y^2=9\), where \(z=0\). Its overall shape is the upper half of a sphere of radius \(3\); the vertical traces show semicircles that shrink as the fixed coordinate moves away from zero.
Why the Domain and Trace Placement Matter
A trace formula alone does not show where the curve sits in three-dimensional space. For instance, \(z=1-y^2\) is the curve obtained from \(z=x^2-y^2\) at \(x=1\), but the trace belongs in the plane \(x=1\). Drawing it in the plane \(y=1\) would place it incorrectly. The y-fixed trace at \(y=1\) is instead \(z=x^2-1\), in the plane \(y=1\).
The domain also determines where a trace exists and where the surface ends. In the hemisphere example, the expression under the square root must be nonnegative, so \(x^2+y^2\leq9\). A sketch that continues the graph beyond that disk would include points where the function is not defined. At the edge, the height is exactly zero, not an unspecified gap.
For example, if the only specified traces lie on a few lines, changing a function by adding a term that vanishes on those lines leaves all of those traces unchanged, even though the surface may change elsewhere. This is why a good sketch should distinguish what the chosen traces establish from what is being inferred between them.
Check Your Understanding
Use the trace and domain principles to answer the following questions.
- For \(z=x^2+3y^2\), what are the traces at \(x=0\) and \(y=0\), and in which planes do they lie?
- For \(z=x^2-y^2\), calculate the height where the traces \(x=2\) and \(y=1\) meet. How can you check that both traces agree?
- What information about the graph is provided by the condition \(x^2+y^2\leq9\) for \(z=\sqrt{9-x^2-y^2}\)?
- Why is it incorrect to draw the trace \(z=1-y^2\), obtained by fixing \(x=1\), in the plane \(y=1\)?
- Why can selected vertical traces guide a surface sketch without uniquely determining the whole surface?