Worked examples · Sheet 2

Level curves

Theory used: Level curves Graph of a function of two variables

Exercise 1

Find the level curves of $f(x, y) = 2x - y$ and the level curve that passes through the point $(1, 3)$ .

2x - y = k \iff y = 2x - k, \qquad k \in \mathbb{R}

f(1, 3) = 2 - 3 = -1 \qquad \Longrightarrow \qquad L_{-1} : \ y = 2x + 1

The level curve $k$ consists of the points where $f$ equals $k$ , so one solves the equation $2x - y = k$ for $y$ . For each $k$ one obtains the line $y = 2x - k$ , with slope $2$ and intercept $-k$ . The level curves are therefore parallel lines, one for each real value of $k$ , and none is empty. As $k$ increases the intercept decreases, so the lines of higher levels lie lower.

The point $(1, 3)$ lies on the level curve equal to the value of $f$ at that point, that is, $k = -1$ , which is the line $y = 2x + 1$ . As a check, $2 \cdot 1 + 1 = 3$ . The graph of $f$ is the plane $z = 2x - y$ , and its horizontal sections are lines, as found.

Exercise 2

Find the level curves of $f(x, y) = y - x^2$ and draw those that pass through the points $(0, 0)$ and $(1, 3)$ .

y - x^2 = k \iff y = x^2 + k, \qquad k \in \mathbb{R}

f(0, 0) = 0 \qquad \Longrightarrow \qquad L_0 : \ y = x^2

f(1, 3) = 3 - 1 = 2 \qquad \Longrightarrow \qquad L_2 : \ y = x^2 + 2

y = x^2

y = x^2 + 2

The level curves

0

and

2

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

are the parabola

y = x^{2}

and the same parabola shifted upward by

2

Solving the equation $y - x^2 = k$ for $y$ gives the parabola $y = x^2 + k$ , with vertex at $(0, k)$ and opening upward. The level curves are therefore identical parabolas, shifted vertically with respect to one another; for every $k$ the curve exists, because the parabola $y = x^2 + k$ is defined for every $x$ . The level increases going up, because, for fixed $x$ , $f$ increases as $y$ increases.

The value of $f$ at the point $(0, 0)$ is $0$ , so the curve through the origin is the parabola $y = x^2$ . At the point $(1, 3)$ the function equals $2$ , so the curve through $(1, 3)$ is $y = x^2 + 2$ ; indeed $1^2 + 2 = 3$ . The two curves have no points in common, as happens for every pair of different levels.

Exercise 3

Find the level curves of $f(x, y) = x^2 + y^2 - 4x$ , the level curve that passes through $(3, 2)$ , and the minimum value of $f$ .

x^2 - 4x + y^2 = k \iff (x - 2)^2 + y^2 = 4 + k

$k < -4$ : no points, $L_k = \emptyset$ ;
$k = -4$ : the single point $(2, 0)$ ;
$k > -4$ : the circle with center $(2, 0)$ and radius $\sqrt{4 + k}$ .

f(3, 2) = 9 + 4 - 12 = 1 \qquad \Longrightarrow \qquad L_1 : \ (x - 2)^2 + y^2 = 5

f(x, y) = (x - 2)^2 + y^2 - 4 \geq -4, \qquad \min f = f(2, 0) = -4

(x - 2)^2 + y^2 = 1

(x - 2)^2 + y^2 = 5

The level curves

k = - 3

and

k = 1

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

are circles with center

(2, 0)

; the second passes through

(3, 2)

To recognize the curve, one completes the square in the terms in $x$ , that is, one adds and subtracts $4$ , and obtains $x^2 - 4x = (x - 2)^2 - 4$ . The equation of the level curve becomes $(x - 2)^2 + y^2 = 4 + k$ . The left-hand side is the square of the distance of the point $(x, y)$ from $(2, 0)$ , which is never negative. Hence for $4 + k < 0$ the curve is empty, for $4 + k = 0$ it consists of the center alone, and for $4 + k > 0$ it is the circle with center $(2, 0)$ and radius $\sqrt{4 + k}$ .

The point $(3, 2)$ belongs to the level curve $k = f(3, 2) = 1$ , which is the circle of radius $\sqrt{5}$ . The level curves are empty for $k < -4$ , so $f$ takes no values less than $-4$ ; the value $-4$ is taken at the point $(2, 0)$ . Therefore the minimum value of $f$ is $-4$ , as confirmed by writing $f(x, y) = (x - 2)^2 + y^2 - 4$ , the sum of two squares and $-4$ .