A Reliable Route from Formula to Sketch
A formula \(z=f(x,y)\) gives the height of a surface above each point \((x,y)\) in the plane. A useful sketch does not come from one isolated feature: it combines intercepts, vertical traces, and level curves. Each provides a different kind of information, and each can be checked directly against the formula.
We will apply this method to three surfaces: \(z=x^2+y^2\), \(z=x^2-y^2\), and \(z=\sin x+\sin y\). The first is a bowl, the second rises in one coordinate direction and falls in the other, and the third repeats in both horizontal directions. The goal is to identify the features that determine each sketch and to label representative contours accurately.
These are direct consequences of the graph equation. In the plane \(y=a\), substitute \(a\) for \(y\) to get the trace \(z=f(x,a)\). In the horizontal plane \(z=c\), the surface equation becomes \(f(x,y)=c\), which gives the level curve. Thus traces describe vertical slices, while level curves record where the surface meets horizontal planes.
Worked Example: The Bowl \(z=x^2+y^2\)
Worked Example: Sketching \(z=x^2+y^2\)
Step 1: Find the intercepts. On the \(x\)-axis of three-dimensional space, \(y=0\) and \(z=0\). The formula then gives \(0=x^2\), so \(x=0\). The only intercept on that axis is \((0,0,0)\). On the \(y\)-axis, \(x=0\) and \(z=0\), so \(0=y^2\), giving the same point. On the \(z\)-axis, \(x=y=0\), and \(z=0\). The only intercept on all three coordinate axes is the origin.
Step 2: Find the coordinate-plane traces. In the plane \(y=0\), the trace is \(z=x^2\), an upward-opening parabola. In the plane \(x=0\), it is \(z=y^2\), also an upward-opening parabola. These two traces show that the surface rises away from the origin in both coordinate directions.
Step 3: Check parallel vertical traces. In the plane \(y=2\), substitution gives \(z=x^2+2^2=x^2+4\), a parabola with minimum height \(4\) at \(x=0\). In the plane \(x=1\), it gives \(z=1^2+y^2=y^2+1\), with minimum height \(1\) at \(y=0\). The offset of each trace agrees with the formula.
Step 4: Find labeled level curves. At height \(c\), the equation is \(x^2+y^2=c\). If \(c<0\), there are no points because a sum of squares cannot be negative. At \(c=0\), the level curve is just \((0,0)\). At \(c=1\), it is a circle of radius \(1\); at \(c=4\), it is a circle of radius \(2\). These are concentric circles centered at the origin, with larger radii at higher levels.
Check: The point \((1,0)\) on the \(c=1\) circle gives \(z=1^2+0^2=1\). The point \((0,2)\) on the \(c=4\) circle gives \(z=0^2+2^2=4\). Both lie at the heights assigned to their contours.
Sketch conclusion: Draw a smooth bowl with its lowest point at the origin. Its two coordinate-plane slices are upward parabolas, and its horizontal contours are nested circles. The formula is nonnegative everywhere, so the surface never goes below \(z=0\).
Worked Example: The Saddle \(z=x^2-y^2\)
Worked Example: Sketching \(z=x^2-y^2\)
Step 1: Find the intercepts. On the \(x\)-axis, \(y=0\) and \(z=0\), so \(0=x^2\) and \(x=0\). On the \(y\)-axis, \(x=0\) and \(z=0\), so \(0=-y^2\) and \(y=0\). On the \(z\)-axis, \(x=y=0\), giving \(z=0\). Again, the only coordinate-axis intercept is \((0,0,0)\).
Step 2: Compare the coordinate-plane traces. In the plane \(y=0\), the trace is \(z=x^2\), which opens upward. In the plane \(x=0\), the trace is \(z=-y^2\), which opens downward. These opposite directions are the key feature: the surface rises along the \(x\)-direction and falls along the \(y\)-direction.
Step 3: Check parallel vertical traces. In the plane \(y=1\), the trace is \(z=x^2-1\), with minimum \(-1\) at \(x=0\). In the plane \(x=2\), it is \(z=4-y^2\), with maximum \(4\) at \(y=0\). Each trace is a parabola, but its orientation depends on which variable remains free.
Step 4: Find and classify the level curves. At height \(c\), the equation is \(x^2-y^2=c\). For \(c=1\), the point \((1,0)\) lies on the curve because \(1^2-0^2=1\); the curve is a hyperbola opening in the positive and negative \(x\)-directions. For \(c=-1\), the point \((0,1)\) satisfies \(0^2-1^2=-1\); this hyperbola opens in the positive and negative \(y\)-directions. For \(c=0\), factor the equation:
The zero-level curve consists of two intersecting diagonal lines. The sign of \(c\) determines which coordinate direction the hyperbola opens toward: positive levels open along \(x\), and negative levels along \(y\).
Check: On the positive-level curve, \((\sqrt{2},1)\) gives \(x^2-y^2=2-1=1\). On the negative-level curve, \((1,\sqrt{2})\) gives \(x^2-y^2=1-2=-1\). Both points satisfy the levels assigned to them.
Sketch conclusion: Draw a saddle through the origin. Its \(x\)-trace rises and its \(y\)-trace falls; the diagonal zero-level lines separate the positive-height regions from the negative-height regions.
Worked Example: The Periodic Surface \(z=\sin x+\sin y\)
Worked Example: Sketching \(z=\sin x+\sin y\)
Step 1: Find the intercepts and coordinate-plane traces. On the \(x\)-axis, \(y=0\) and \(z=0\), so the formula requires \(\sin x=0\), or \(x=n\pi\) for an integer \(n\). Thus the graph intersects that axis at \((n\pi,0,0)\). On the \(y\)-axis the intersections are \((0,n\pi,0)\), since \(\sin y=0\). On the \(z\)-axis, \(x=y=0\) gives \(z=0\), so the intercept there is the origin. In the vertical plane \(y=0\), the trace is \(z=\sin x\); in \(x=0\), it is \(z=\sin y\).
Step 2: Check parallel vertical traces. In the plane \(x=\pi/2\), the trace is \(z=1+\sin y\), which ranges from \(0\) to \(2\). In the plane \(x=-\pi/2\), the trace is \(z=-1+\sin y\), which ranges from \(-2\) to \(0\). Likewise, each plane \(y=a\) gives the trace \(z=\sin x+\sin a\), a sine wave shifted vertically by \(\sin a\).
Step 3: Determine the range and representative levels. Since \(-1\leq\sin x\leq1\) and \(-1\leq\sin y\leq1\), their sum lies between \(-2\) and \(2\). The maximum \(2\) occurs when both sines are \(1\), for example at \((x,y)=(\pi/2,\pi/2)\). The minimum \(-2\) occurs at \((-\pi/2,-\pi/2)\). Therefore levels outside \([-2,2]\) have no points.
At level \(c=0\), use the sum-to-product identity to factor the equation:
This product is zero when either factor is zero. The resulting zero-level lines can be written \(y=-x+2k\pi\) or \(y=x+(2k+1)\pi\), where \(k\) is any integer. For levels \(c=1\) and \(c=-1\), useful points are \((\pi/2,0)\) and \((-\pi/2,0)\), respectively: substituting gives \(1+0=1\) and \(-1+0=-1\). The contours repeat periodically in both coordinate directions, with period \(2\pi\) in each variable.
Check: At \((\pi/2,\pi/2)\), the height is \(1+1=2\); at \((-\pi/2,-\pi/2)\), it is \(-1-1=-2\). At \((\pi,0)\), it is \(0+0=0\), agreeing with both the intercept calculation and the zero-level equation.
Sketch conclusion: Draw a repeating field of hills and depressions. The coordinate traces are sine waves, and the pattern repeats every \(2\pi\) horizontally in either direction. Mark the maximum and minimum heights \(2\) and \(-2\), and draw representative zero, positive, and negative contours.
How the Pieces Constrain the Sketch
The sketches above use a general principle. Every point on a vertical trace must satisfy the original surface equation with one coordinate fixed. Every point on a level curve must satisfy it with the height fixed. These are not merely visual hints: they are exact slices of the graph. A proposed surface that disagrees with even one such slice cannot represent the formula.
Check where the graph meets the coordinate axes and whether changing signs or swapping variables preserves its height.
Set \(x=0\) and \(y=0\) to see the basic cross-sections through the origin.
Fix nonzero values of \(x\) or \(y\) to see how the cross-sections shift or change.
Solve \(f(x,y)=c\) for several heights, including zero when useful, and check points by substitution.
For \(x^2+y^2\), the contours are circles and the surface rises in every direction from its minimum. For \(x^2-y^2\), the contours change orientation across height zero, matching the opposing vertical traces. For \(\sin x+\sin y\), the traces and contours repeat, so a sketch of one period must be extended periodically rather than treated as the whole surface.
A Common Mistake: Treating Both Squares as Positive
A tempting but incorrect sketch of \(z=x^2-y^2\) is a bowl, based only on seeing two squared variables. That reasoning effectively treats the formula as though both terms raise the height. A direct check exposes the error:
The surface is above the \(xy\)-plane at \((1,0)\) and below it at \((0,1)\). Its trace in the \(yz\)-plane is \(z=-y^2\), not \(z=y^2\). The correction is to retain the minus sign in every substitution and check at least one point in each coordinate direction before deciding the overall shape.
A second useful check is to substitute a point from each proposed contour into the original formula. A level curve labeled \(c\) must produce height exactly \(c\), not merely look as if it belongs at that height.
Check Your Understanding
Use the formula as well as the shape of each trace when answering.
- For \(z=x^2+y^2\), what are the level curves at heights \(0\), \(1\), and \(4\), and what is their center?
- For \(z=x^2-y^2\), what are the traces in the planes \(y=0\) and \(x=0\)?
- Why does the zero-level curve of \(z=x^2-y^2\) consist of two lines?
- What is the range of \(z=\sin x+\sin y\), and at which points does it attain its maximum and minimum?
- What vertical trace results from fixing \(x=\pi/2\) for \(z=\sin x+\sin y\)?