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Draw Contour Lines of a Quadratic Surface Model

Learn to plot level curves of a quadratic surface by evaluating the function on a grid and connecting points of equal value.

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Four exercises
Worked answers

Algebra and Function Practice

A quadratic surface model assigns a height to each point on a plane. Contour lines, also called level curves, connect points that share the same height. Drawing them by hand builds intuition for how the surface rises and falls. This lesson uses a simple quadratic function, a coordinate grid, and a fixed set of level values. You will evaluate the function at grid intersections, mark where each level occurs, and connect those marks smoothly. The method works for any quadratic surface you can evaluate, and it keeps arithmetic checkable. A convention for lettering and line style is defined so that valid alternatives can be judged fairly.

Practice materials: Printable downloads are not available for this lesson yet. Use the instructions and four exercises below.

Set Up the Grid and Choose Levels

Begin with the quadratic surface f(x, y) = x^2 + 2y^2. This function is always zero or positive, and it grows faster in the y direction than in the x direction. Choose a rectangular grid with x from -2 to 2 and y from -2 to 2, using steps of 0.5. That gives nine vertical grid lines and nine horizontal grid lines, so there are 81 intersection points to evaluate. Label axes clearly and write the function at the top of the page.

Select level values that produce visible, separated curves. For this function, the levels 0, 1, 2, 4, and 8 are convenient because they fall on or near grid intersections. Level 0 occurs only at the origin. Level 1 passes through points such as (1, 0) and (0, 0.707), but 0.707 is not on the half-unit grid, so you will estimate its position between grid lines. Level 2 passes through (0, 1) and (1.414, 0), again requiring estimation. Level 4 passes through (2, 0) and (0, 1.414). Level 8 passes through (0, 2) and would pass through (2.828, 0); the latter lies outside the grid, so only part of that contour appears.

  1. Draw axes and mark x and y from -2 to 2 in steps of 0.5.
  2. Write f(x, y) = x^2 + 2y^2 beside the grid.
  3. List the chosen levels: 0, 1, 2, 4, 8.
  4. Evaluate f at each grid intersection and record values in a table or directly on the grid.

Mark Points Where Each Level Occurs

For each level, scan the grid for intersections where the computed value equals the level. Mark those points with a small dot and label the dot with the level number. When a level falls between two grid values, estimate the crossing by linear interpolation along a grid line. For example, along the x-axis where y = 0, f(x, 0) = x^2. To find where f = 1, solve x^2 = 1, giving x = 1 and x = -1, which are exact grid points. To find where f = 2, solve x^2 = 2, giving x ≈ 1.414 and x ≈ -1.414, which lie between 1.0 and 1.5 on the grid. Mark these estimated points with a small cross and note that they are approximate.

Along the y-axis where x = 0, f(0, y) = 2y^2. For level 1, solve 2y^2 = 1, giving y ≈ 0.707 and y ≈ -0.707, between 0.5 and 1.0. For level 2, y = 1 and y = -1 exactly. For level 4, y ≈ 1.414 and y ≈ -1.414. For level 8, y = 2 and y = -2 exactly. Mark all these points before drawing any curves. A consistent marking convention helps: use dots for exact grid intersections and crosses for estimated positions, and write the level value next to each mark.

  1. For each level, solve the function along the axes to find exact or estimated intercepts.
  2. Scan interior grid intersections and mark any that equal the level exactly.
  3. Use linear interpolation along grid lines to estimate crossings between intersections.
  4. Label every mark with its level value and note whether it is exact or estimated.

Connect Marks into Smooth Contours

Connect marks of the same level with a smooth curve. Do not use straight line segments between marks unless the marks are very close. The contours of this quadratic surface are ellipses centered at the origin, stretched along the x direction because the coefficient of x^2 is smaller than the coefficient of y^2. Level 0 is a single point at the origin, not a curve. Level 1 is a small ellipse passing through (±1, 0) and (0, ±0.707). Level 2 is a larger ellipse through (±1.414, 0) and (0, ±1). Level 4 passes through (±2, 0) and (0, ±1.414). Level 8 passes through (0, ±2) and would pass through (±2.828, 0), but those x values lie outside the grid, so the level 8 contour appears only as two arcs near the top and bottom edges.

Keep contours from crossing. If two levels appear to cross, recheck the arithmetic at the nearby grid points. Label each completed contour near one of its marks, for example write “f = 2” beside the ellipse. Use a consistent line style: solid lines for all contours, with the level value written in the same orientation as the curve where possible. The convention for lettering is that labels are written horizontally, not rotated, and placed outside the curve with a short leader line if space is tight. Handwriting style may vary; the rubric judges legibility and correct placement, not a single letterform.

  1. Start with the smallest nonzero level and connect its marks in a closed loop.
  2. Draw each contour as a smooth curve, adjusting the shape to pass through all marks of that level.
  3. Label each contour with its level value using horizontal text and a leader line if needed.
  4. Check that no two contours cross and that the spacing between contours reflects the function’s rate of change.

Check the Drawing Against the Function

After drawing, verify a few points on each contour by substituting their coordinates into f(x, y). For example, the point (1.2, 0.8) gives f ≈ 1.44 + 1.28 = 2.72, which is not 2, so that point is not on the level 2 contour. A correct point on level 2 is (1.414, 0), where f = 2.000. Use exact arithmetic where possible and label estimated coordinates with a squiggle or “approx.” to show they are not exact grid values.

Check the spacing between contours. Along the x-axis, the contour intercepts are x = sqrt(c/1) = sqrt(c). The level 1 and level 2 contours are separated by sqrt(2) - sqrt(1) ≈ 0.414 units, while level 4 and level 8 are separated by sqrt(8) - sqrt(4) ≈ 0.828 units, so the spacing increases with level along the x-axis; the level 8 contour is partly outside the grid. Do not assume contours are evenly spaced: the gap depends on the level values and on the direction you measure. If your drawing shows contours evenly spaced everywhere, recheck the evaluation table. A final check is to confirm that each contour is symmetric about both axes, since f(x, y) = f(-x, y) = f(x, -y).

  1. Substitute one marked point per contour into the function and confirm the result equals the level.
  2. Check symmetry of each contour about the x-axis and y-axis.
  3. Compare spacing between adjacent contours near the origin and near the edges.
  4. Mark any estimated points clearly and confirm that exact points lie on grid intersections.

Worked example

For f(x, y) = x^2 + 2y^2 on the grid x, y from -2 to 2 with step 0.5, draw the contour for level 4. List the exact and estimated points you would mark, and describe the shape.

Worked answer

Along the x-axis (y = 0), solve x^2 = 4, giving x = 2 and x = -2, both exact grid points. Along the y-axis (x = 0), solve 2y^2 = 4, giving y^2 = 2, so y ≈ 1.414 and y ≈ -1.414, estimated between 1.0 and 1.5 and between -1.5 and -1.0. Check interior grid points: (1.5, 1.0) gives 2.25 + 2 = 4.25, not 4; (1.0, 1.0) gives 1 + 2 = 3; (1.5, 0.5) gives 2.25 + 0.5 = 2.75. No interior grid intersection equals 4 exactly. Mark the four axis points and connect them with a smooth ellipse centered at the origin, wider along the x direction. Label the contour “f = 4” horizontally near the point (2, 0).

Why: The level 4 contour is the set of points where x^2 + 2y^2 = 4. Solving along the axes gives the intercepts, and checking interior grid points confirms that no other exact marks are available. The ellipse is wider along x because the x^2 coefficient is smaller, so the contour extends farther in the x direction than in the y direction.

Check: Substitute (2, 0) into f: 4 + 0 = 4, correct. Substitute (0, 1.414) approximately: 0 + 2(2) = 4, correct within estimation. Confirm symmetry about both axes and that the contour does not cross the level 2 or level 8 contours.

Four practice exercises

Exercise 1

For f(x, y) = x^2 + 2y^2 on the grid x, y from -2 to 2 with step 0.5, draw the contour for level 1. List the exact and estimated points you would mark, and describe the shape.

Show answer and check

Along the x-axis, x^2 = 1 gives x = 1 and x = -1, both exact. Along the y-axis, 2y^2 = 1 gives y ≈ 0.707 and y ≈ -0.707, estimated between 0.5 and 1.0 and between -1.0 and -0.5. No interior grid intersection equals 1 exactly: (0.5, 0.5) gives 0.25 + 0.5 = 0.75; (1.0, 0.5) gives 1 + 0.5 = 1.5. Mark the four axis points and connect them with a small ellipse centered at the origin, wider along x. Label “f = 1” horizontally near (1, 0).

Why: The level 1 contour satisfies x^2 + 2y^2 = 1. Axis intercepts give the marks, and interior checks confirm no exact grid intersections. The ellipse is small and stretched along x because the x^2 coefficient is smaller.

Check: Substitute (1, 0): 1 + 0 = 1, correct. Substitute (0, 0.707) approximately: 0 + 2(0.5) = 1, correct within estimation. Confirm symmetry and that the contour lies inside the level 2 contour.

Exercise 2

For f(x, y) = x^2 + 2y^2 on the grid x, y from -2 to 2 with step 0.5, draw the contour for level 2. List the exact and estimated points you would mark, and describe the shape.

Show answer and check

Along the x-axis, x^2 = 2 gives x ≈ 1.414 and x ≈ -1.414, estimated between 1.0 and 1.5 and between -1.5 and -1.0. Along the y-axis, 2y^2 = 2 gives y = 1 and y = -1, both exact. Check interior grid points: (1.0, 0.5) gives 1 + 0.5 = 1.5; (1.5, 0.5) gives 2.25 + 0.5 = 2.75; (1.0, 1.0) gives 1 + 2 = 3. No interior grid intersection equals 2 exactly. Mark the four axis points and connect them with an ellipse centered at the origin, wider along x. Label “f = 2” horizontally near (0, 1).

Why: The level 2 contour satisfies x^2 + 2y^2 = 2. Axis intercepts give the marks, and interior checks confirm no exact grid intersections. The ellipse is larger than level 1 and still stretched along x.

Check: Substitute (0, 1): 0 + 2 = 2, correct. Substitute (1.414, 0) approximately: 2 + 0 = 2, correct within estimation. Confirm symmetry and that the contour lies between level 1 and level 4.

Exercise 3

For f(x, y) = x^2 + 2y^2 on the grid x, y from -2 to 2 with step 0.5, draw the contour for level 8. Describe which parts of the contour are visible on the grid and list the exact and estimated points you would mark.

Show answer and check

Along the y-axis, 2y^2 = 8 gives y = 2 and y = -2, both exact grid points at the top and bottom edges. Along the x-axis, x^2 = 8 gives x ≈ 2.828 and x ≈ -2.828, which lie outside the grid, so those intercepts are not visible. Check interior grid points: (2, 1) gives 4 + 2 = 6; (2, 1.5) gives 4 + 4.5 = 8.5; (1.5, 1.5) gives 2.25 + 4.5 = 6.75. No interior grid intersection equals 8 exactly. Mark (0, 2) and (0, -2) and draw two arcs near the top and bottom edges, curving outward toward the left and right edges but not reaching them. Label “f = 8” horizontally near (0, 2).

Why: The level 8 contour satisfies x^2 + 2y^2 = 8. The y-intercepts are exact grid points, but the x-intercepts lie outside the grid, so only two arcs are visible. Interior checks confirm no other exact marks.

Check: Substitute (0, 2): 0 + 8 = 8, correct. Substitute (2, 1.5): 4 + 4.5 = 8.5, not 8, so that point is not on the contour. Confirm the arcs are symmetric about both axes and lie outside the level 4 contour.

Exercise 4

For f(x, y) = x^2 + 2y^2 on the grid x, y from -2 to 2 with step 0.5, explain why the level 0 contour is a single point and describe how you would mark it. Then state one check that confirms your marking is correct.

Show answer and check

The level 0 contour satisfies x^2 + 2y^2 = 0. Since both terms are nonnegative, the only solution is x = 0 and y = 0. Mark the origin with a dot and label it “f = 0” horizontally nearby. Do not draw a curve. A check is to substitute (0, 0) into the function: 0 + 0 = 0, correct. Also confirm that no other grid intersection gives 0, since any nonzero x or y makes at least one term positive.

Why: The sum of two nonnegative terms is zero only when both terms are zero. The origin is the only point on the level 0 contour, so it is a degenerate contour, not a curve.

Check: Substitute (0, 0) into f and confirm the result is 0. Check a nearby grid point such as (0.5, 0) gives 0.25, not 0, confirming the contour is a single point.

Common mistakes

Self-check rubric

Questions and answers

Why are the contours ellipses rather than circles?

The coefficients of x^2 and y^2 are different (1 and 2), so the surface rises faster in the y direction. Level curves of x^2 + 2y^2 = c are ellipses stretched along the x direction.

What if a level value does not appear on the grid?

Estimate its position by solving the function along grid lines, for example x^2 = c when y = 0, and mark the estimated point with a cross. If the level lies entirely outside the grid, note that only part of the contour is visible.

How many points do I need to mark before drawing a contour?

At least four points per closed contour, one in each quadrant if possible, plus any exact grid intersections. More marks improve accuracy, but four well-placed marks are enough to sketch the ellipse.

Can I use a different set of level values?

Yes, as long as the levels are positive and produce visible contours on your grid. The method is the same; just evaluate and mark the new levels consistently.

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