Worked examples · Sheet 2

Level curves

Theory used: Level curve Function of two or more variables

Exercise 1

Determine the level curves of

f (x, y) = 2 x - y + 1.

2 x - y + 1 = k

y = 2 x + 1 - k

The level curve $k$ is the solution set of the equation $f (x, y) = k$ , so one solves $2 x - y + 1 = k$ for $y$ . For every $k \in R$ one obtains the line $y = 2 x + 1 - k$ : the level curves are the lines with slope $2$ , one for each value of $k$ , all parallel to one another. As $k$ increases the line shifts downward, since the intercept $1 - k$ decreases. The level curves of a first-degree function are always parallel lines, because the graph of the function is a plane.

Exercise 2

Determine the level curves of

f (x, y) = x^{2} + y^{2} - 4 x .

x^{2} + y^{2} - 4 x = k

(x - 2)^{2} + y^{2} = k + 4

$k > - 4$ : circle with center $(2, 0)$ and radius $k + 4$
$k = - 4$ : the point $(2, 0)$
$k < - 4$ : empty set

Level curves of

f (x, y) = x^{2} + y^{2} - 4 x

: circles with center

(2, 0)

and radii

1, 2, 3

for

k = - 3, 0, 5

; for

k = - 4

the center alone

The equation $x^{2} + y^{2} - 4 x = k$ is rewritten by completing the square in the variable $x$ : since $x^{2} - 4 x = (x - 2)^{2} - 4$ , the equation becomes $(x - 2)^{2} + y^{2} = k + 4$ . The left-hand side is a sum of squares, so the equation has solutions only if $k + 4 \geq 0$ . For $k > - 4$ the level curve is the circle with center $(2, 0)$ and radius $k + 4$ ; for $k = - 4$ the sum of squares is zero and only the point $(2, 0)$ remains; for $k < - 4$ the level set is empty. The value $k = - 4$ is the minimum of the function, attained at the common center of the circles.

Exercise 3

Determine the level curves of

f (x, y) = e^{y - x^{2}} .

e^{y - x^{2}} = k, k > 0

y - x^{2} = ln k

y = x^{2} + ln k

$k > 0$ : parabola $y = x^{2} + ln k$
$k \leq 0$ : empty set

y = x^{2} - 1

y = x^{2}

y = x^{2} + 1

Level curves of

f (x, y) = e^{y - x^{2}}

: the parabolas

y = x^{2} + ln k

for

k = \frac{1}{e}, 1, e

, that is,

ln k = - 1, 0, 1

The exponential takes only strictly positive values, so for $k \leq 0$ the equation $f (x, y) = k$ has no solutions and the level set is empty. For $k > 0$ one applies the logarithm to both sides, an operation that is legitimate because both sides are positive, and obtains $y - x^{2} = ln k$ , that is, the parabola $y = x^{2} + ln k$ . The level curves are therefore the vertical translates of the parabola $y = x^{2}$ : for $k = 1$ one obtains the parabola $y = x^{2}$ , since $ln 1 = 0$ , for $k > 1$ the translates upward and for $0 < k < 1$ the translates downward, since in that case $ln k < 0$ .