Level Curves

Cut the surface at one height

Take the bowl z = x² + y² and cut it with the flat plane z = 4. The two meet along every point of the bowl at height 4, which is where x² + y² = 4: a circle of radius 2, lying 4 units up.

That curve is a level curve of the function f(x, y) = x² + y², the curve along which f keeps one value. In general the level curve at height k is the set of points with f(x, y) = k.

Cut it again, at equal steps

Cut the same bowl at z = 8, 12 and 16 as well, every 4 units up to a rim at 16. Each cut is a circle: x² + y² = 8, 12 and 16, with radii √8 ≈ 2.828, √12 ≈ 3.464 and √16 = 4.

The cuts are equally spaced in height, but the circles are not equally spaced in radius. The gaps between them are 0.828, 0.636 and 0.536: the higher the ring, the closer it sits to the one below.

The bowl z = x² + y² up to height 16, cut at 4, 8, 12 and 16. The gold plane is the cut at height 4, and the gold ring where it meets the bowl is the circle x² + y² = 4.

Look straight down

Seen from directly above, each cut drops onto the xy-plane as a curve in x and y alone, labeled with its height. The level curve at height 4 is x² + y² = 4, radius 2; at height 25 it would be x² + y² = 25, radius 5, the circle through (3, 4).

The radius is the square root of the height, not the height: the level curve at z = 16 has radius 4, and its diameter is 8.

The contour map of z = x² + y²: rings at heights 4, 8, 12 and 16, each labeled. The gold ring is x² + y² = 4, with its radius 2 marked.

Close rings mean steep ground

Between two neighboring rings the ground rises by the same 4 units. Where the rings are close, that rise happens over a short distance, so the ground is steep. Dividing the rise by the gap gives the average steepness: 4 ÷ 0.828 ≈ 4.83 from the first ring to the second, 4 ÷ 0.636 ≈ 6.29 to the third, and 4 ÷ 0.536 ≈ 7.46 to the rim.

That matches the bowl itself. Along a line out from the center, z = r², so the slope is 2r: 4 at the first ring, where r = 2, and 8 at the rim, where r = 4. The bowl steepens as it rises, and its rings crowd together.

The steps must be equal

Spacing means steepness only when the heights are equally spaced. The bowl cut at heights 1, 4, 9 and 16 gives rings of radius 1, 2, 3 and 4, evenly spaced, though the bowl is just as curved as before. That is why a map states its contour interval.

A surface that does rise at a steady rate has evenly spaced rings at equal steps. The cone z = √(x² + y²) has level curve x² + y² = k² at height k, a circle of radius k, so at heights 1, 2, 3 and 4 its rings are 1 apart: its sides are straight, with slope 1 everywhere.

The level curves of the cone z = √(x² + y²) at heights 1, 2, 3 and 4: circles of radius 1, 2, 3 and 4, one unit apart, because the cone rises at the same rate everywhere.

Other surfaces, other curves

The level curves of z = x² − y² are x² − y² = k. At k = 0 that is x² = y², the two lines y = x and y = −x crossing at the saddle point. For k > 0 it is a hyperbola crossing the x-axis at ±√k, and for k < 0 a hyperbola crossing the y-axis at ±√(−k).

The level curves of the plane z = x + y are x + y = k, parallel straight lines, equally spaced at equal steps because a plane rises at the same rate everywhere.

The surface z = x² + 4y² has level curves x² + 4y² = k, ellipses half as wide along y as along x, so its rings are closer together along y and it is steeper in that direction. The hill in the first application below has contours of exactly this shape.

The contour map of z = x² − y² over the square with |x| ≤ 2 and |y| ≤ 2, at heights −3 to 3 in steps of 1. The two lines of height 0 cross at the saddle point; the positive heights lie to the left and right of them and the negative heights above and below.

The usual mistakes

Giving the height as the radius. The level curve of x² + y² at z = 25 has radius 5, not 25.

Giving the diameter. The circle x² + y² = 25 is 10 across, but its radius is 5.

Reading close rings as high ground. Close rings mean steep ground; the heights come from the labels.

Comparing spacings on a map cut at unequal steps. The spacing says nothing about steepness unless the heights go up in equal steps.

A hill, a weather map and a workshop

In the applications below, the contours of a hill are ellipses, and the path that meets the next contour sooner is the steeper one. The isobars on a weather map are the level curves of the pressure, here parallel lines, and the distance between two of them is measured at right angles. And the mixes of workers and machines that make the same output lie on one level curve, along which the cheapest mix is found.

Worked example: A Hill on a Walking Map: The Height of a Hut, and the Contour Through It

Question Above the point x kilometers east and y kilometers north of its summit, a hill has a height of h(x, y) = 450 − 200(x² + 4y²) meters. The model holds down to the foot of the hill, where h = 0. (a) A mountain hut stands at (0.6, 0.4). Find its height. (b) Find the equation of the contour through the hut and where it crosses the paths that run due east and due north from the summit. Compare the average gradients of those two paths from the summit down to that contour.

1.(a) At the hut, x² + 4y² = 0.36 + 4 × 0.16 = 1, so its height is h = 450 − 200 × 1 = 250 meters.

(a) The summit is at the center of the map; the contours drawn are at 400, 350 and 300 meters, and the hut stands at 250 meters.
2.(b) The contour through the hut joins every point at a height of 250 meters: 450 − 200(x² + 4y²) = 250, which gives x² + 4y² = 1. This is an ellipse centered on the summit, and every contour of this hill is an ellipse of the same shape.

(b) The contour through the hut is the level curve h = 250: the ellipse x² + 4y² = 1.
3.On the path due east, y = 0, so x² = 1 and the path meets the contour 1 kilometer from the summit. On the path due north, x = 0, so 4y² = 1 and the path meets the contour 0.5 kilometers from the summit.

It crosses the path due east 1 kilometer from the summit and the path due north 0.5 kilometers from it.
4.Both paths drop 450 − 250 = 200 meters to the contour. The average gradient is 2001000 = 0.2 due east and 200500 = 0.4 due north, so the path due north is on average twice as steep.

Both paths drop 200 meters, so the path due north, half as long, is on average twice as steep.
5.Check: the point (0, 0.5) gives h = 450 − 200 × 4 × 0.25 = 250 meters, as it should. On the map the contours lie closer together to the north of the summit, which is what steeper ground looks like.

The contours lie closer together to the north of the summit, where the ground is steeper.

Answer: (a) 250 meters; (b) x² + 4y² = 1, an ellipse that crosses the east path 1 kilometer and the north path 0.5 kilometers from the summit; the average gradients are 0.2 due east and 0.4 due north, so the north path is twice as steep

Common mistakes

Mixing kilometers and meters in the gradient, as in 2001 = 200. The drop is in meters, so the distance must be too: 1 kilometer is 1000 meters.
Reading the crossing of the north path as 1 kilometer because the right-hand side of x² + 4y² = 1 is 1. With x = 0 the equation is 4y² = 1, so y = 0.5.

More functions of several variables problems, worked step by step →

Worked example: Isobars on a Weather Map: The Pressure at a Town, and the Distance Between Two Isobars

Question Over a region around a weather station, the air pressure at sea level x kilometers east and y kilometers north of the station is modeled by p(x, y) = 1012 + 0.02x − 0.015y hectopascals. (a) Find the pressure at a town 100 kilometers east and 200 kilometers north of the station, and the equation of the isobar through the town. (b) Find the distance between the 1008 hectopascal isobar and the 1012 hectopascal isobar, measured at right angles to them.

1.(a) At the town, p(100, 200) = 1012 + 0.02 × 100 − 0.015 × 200 = 1012 + 2 − 3 = 1011 hectopascals.

(a) The town, 100 kilometers east and 200 north of the station, has a pressure of 1011 hectopascals.
2.The isobar through the town joins the points where p = 1011: 0.02x − 0.015y = −1. Multiplying by 200 gives 4x − 3y = −200. Check: 4 × 100 − 3 × 200 = −200.

The isobar through the town is the level curve p = 1011, the line 4x − 3y = −200.
3.(b) In the same way, the isobar p = c is the line 4x − 3y = 200(c − 1012). Every one of these lines has the gradient 43, so the isobars are parallel straight lines. The 1012 isobar is 4x − 3y = 0 and the 1008 isobar is 4x − 3y = −800.

(b) Every isobar has the gradient 43, so the isobars are parallel lines; here are the 1012 and 1008 isobars.
4.The distance between the parallel lines 4x − 3y = 0 and 4x − 3y = −800 is the difference of the constants divided by √4² + 3² = 5, which is 8005 = 160 kilometers.

The distance between them, at right angles, is 800√4² + 3² = 160 kilometers.
5.Check: the point of the 1008 isobar nearest the station is (−128, 96), since 4 × (−128) − 3 × 96 = −512 − 288 = −800, and its distance from the station is √16384 + 9216 = √25600 = 160 kilometers. The pressure falls by 4 hectopascals over those 160 kilometers, which is 2.5 hectopascals per 100 kilometers.

The nearest point of the 1008 isobar to the station is (−128, 96), 160 kilometers away: a fall of 2.5 hectopascals per 100 kilometers.

Answer: (a) 1011 hectopascals, on the isobar 4x − 3y = −200; (b) 160 kilometers

Common mistakes

Measuring the gap along the x-axis, from x = −200 to x = 0, and answering 200 kilometers. The axis crosses the isobars at a slant; the distance between two parallel lines is measured at right angles to them.
Dividing 800 by √0.02² + 0.015². The constant and the coefficients must come from the same form of the equation: either 8005 = 160 or 40.025 = 160.

More functions of several variables problems, worked step by step →

Worked example: A Workshop's Output From Its Workers and Machines: The Isoquant Through the Present Mix, and the Cheapest Mix on It

Question A workshop's weekly output is Q(L, K) = 20√LK units when it employs L workers and runs K machines. At present it has 16 workers and 25 machines. (a) Find the present weekly output, and the equation of the isoquant through the present mix: the level curve of every mix that gives the same output. (b) A worker costs $1200 a week and a machine costs $300 a week. Find the mix on that isoquant with the lowest weekly cost, and compare its cost with the present weekly cost.

1.(a) At present, Q(16, 25) = 20√16 × 25 = 20√400 = 20 × 20 = 400 units a week.

(a) The present mix, 16 workers and 25 machines, makes 20√400 = 400 units a week.
2.The isoquant is the level curve Q = 400: 20√LK = 400, so √LK = 20 and LK = 400. Every mix on it, such as 20 workers with 20 machines, makes the same 400 units.

The isoquant through it is the level curve LK = 400: every mix on it makes 400 units.
3.(b) The weekly cost is C = 1200L + 300K dollars. On the isoquant K = 400L, so along it C(L) = 1200L + 120000L.

(b) Along the isoquant the cost is a function of L alone. The dashed line joins every mix that costs the same as the present one, $26700.
4.Then dCdL = 1200 − 120000L², which is 0 when L² = 100, so L = 10, since a number of workers is positive. The second derivative, 240000L³, is positive there, so this is a minimum, and K = 40010 = 40 machines.

The cost is least at 10 workers and 40 machines, where a line of equal cost, here $24000, just touches the isoquant.
5.The cheapest mix costs 1200 × 10 + 300 × 40 = 12000 + 12000 = $24000 a week. The present mix costs 1200 × 16 + 300 × 25 = 19200 + 7500 = $26700 a week, so the cheapest mix saves $2700 a week for the same output. Check: Q(10, 40) = 20√400 = 400 units.

The cheapest mix saves $2700 a week and still makes 400 units.

Answer: (a) 400 units a week, on the isoquant LK = 400; (b) 10 workers and 40 machines, at $24000 a week against $26700 at present, a saving of $2700 a week

Common mistakes

Minimizing the cost 1200L + 300K with nothing linking L and K. The cost is then least with no workers and no machines, which makes nothing; the output must be held at 400 by putting K = 400L.
Writing the isoquant as L + K = 41 from the present 16 + 25. The output depends on the product LK, not the sum: 20 workers and 21 machines make 20√420, about 410 units, not 400.

More functions of several variables problems, worked step by step →