Behavior of Implicit Relations

Horizontal and vertical tangents

Setting

N = 0

gives an equation in

x

and

y

with infinitely many solutions, and only the ones on the curve count. So solve it together with the curve's equation, then check

D

at each point you find.

The curve

C

given by

y^2 - 3xy + x^3 = 0

carries Examples 1, 2, and 4. It closes a loop between

x = 0

and

x = \frac{9}{4}

and crosses itself at the origin.

Example 1

Find the points of

C

where the tangent line is horizontal or vertical. Then find the point where the formula for

\frac{dy}{dx}

can't decide the tangent, and describe the curve there.

Differentiate, with the product rule on

3xy

\begin{aligned} 2yy' - 3y - 3xy' + 3x^2 &= 0, \\ (2y - 3x)\,y' &= 3y - 3x^2, \\ \frac{dy}{dx} &= \frac{3y - 3x^2}{2y - 3x}. \end{aligned}

Horizontal: the numerator is

0

when

y = x^2

. Substituting into the curve's equation,

\begin{aligned} x^4 - 3x^3 + x^3 &= 0, \\ x^3(x - 2) &= 0. \end{aligned}

The root

x = 2

gives

(2, 4)

, where the denominator is

2(4) - 3(2) = 2 \ne 0

, so the tangent there is horizontal.

Vertical: the denominator is

0

when

y = \frac{3}{2}x

. Substituting,

\begin{aligned} \tfrac{9}{4}x^2 - \tfrac{9}{2}x^2 + x^3 &= 0, \\ x^2\left(x - \tfrac{9}{4}\right) &= 0. \end{aligned}

The root

x = \frac{9}{4}

gives

\left(\frac{9}{4}, \frac{27}{8}\right)

, and the numerator there is

3\left(\frac{27}{8}\right) - 3\left(\frac{81}{16}\right) = -\frac{81}{16} \ne 0

, so the tangent is vertical.

The origin: both conditions also give

x = 0

, the point

(0, 0)

, where the numerator and the denominator are both

0

. The formula says nothing there. Solving the equation for

y

as a quadratic gives the two branches

y = \frac{1}{2}x\left(3 \pm \sqrt{9 - 4x}\right)

, which cross at the origin with slopes

3

and

0

. So

C

has two tangent lines at the origin, and one of them is horizontal.

Away from the origin, the tangent to

C

is horizontal only at

(2, 4)

and vertical only at

\left(\frac{9}{4}, \frac{27}{8}\right)

The curve y² − 3xy + x³ = 0. Its tangent is horizontal at (2, 4) and vertical at (9/4, 27/8), and the curve crosses itself at the origin.

Concavity at a point

There are two ways to find

\frac{d^2y}{dx^2}

. One differentiates the formula for

\frac{dy}{dx}

with the quotient rule, then substitutes

\frac{dy}{dx}

back into the result. The other differentiates the equation a second time before solving for anything. At a single point the second is often shorter, since the point and the slope go in as numbers.

Example 2

Determine whether

C

is concave up or concave down at

(2, 2)

The point is on

C

, since

4 - 12 + 8 = 0

. From Example 1 the slope there is

\frac{3(2) - 3(4)}{2(2) - 3(2)} = \frac{-6}{-2} = 3

. Differentiate the equation

2yy' - 3y - 3xy' + 3x^2 = 0

again, with the product rule on

2yy'

and on

3xy'

\begin{aligned} &2(y')^2 + 2yy'' - 3y' \\ &\qquad - 3y' - 3xy'' + 6x = 0. \end{aligned}

Substitute

x = 2

y = 2

, and

y' = 3

\begin{aligned} 18 + 4y'' - 18 - 6y'' + 12 &= 0, \\ y'' &= 6. \end{aligned}

Since

\frac{d^2y}{dx^2} = 6 > 0

(2, 2)

, the curve

C

is concave up there.

The curve y² − 3xy + x³ = 0 and its tangent line at (2, 2), with slope 3. Near (2, 2) the curve lies above the line.

Example 3

The curve

xe^y + y = 2

passes through

(2, 0)

. Find

\frac{d^2y}{dx^2}

(2, 0)

, and determine whether the curve is concave up or concave down there.

Differentiating, with the product rule on

xe^y

, gives

e^y + xe^y y' + y' = 0

. So

\frac{dy}{dx} = -\frac{e^y}{xe^y + 1},

and at

(2, 0)

the slope is

-\frac{1}{3}

. This time use the quotient rule. Write

N = e^y

and

D = xe^y + 1

, remembering that

y

depends on

x

N' = e^y y'

and

D' = e^y + xe^y y'

. Then

\frac{d^2y}{dx^2} = -\frac{N'D - ND'}{D^2}.

(2, 0)

with

y' = -\frac{1}{3}

N = 1

D = 3

N' = -\frac{1}{3}

, and

D' = 1 + 2\left(-\frac{1}{3}\right) = \frac{1}{3}

. So

N'D = -1

ND' = \frac{1}{3}

, and

\frac{d^2y}{dx^2} = -\frac{-1 - \frac{1}{3}}{9} = \frac{4}{27}.

Since

\frac{d^2y}{dx^2} = \frac{4}{27} > 0

(2, 0)

, the curve is concave up there.

The curve x·eʸ + y = 2 and its tangent line at (2, 0), with slope −1/3. Near (2, 0) the curve lies above the line.

Relative extrema of y on a curve

This is the Second Derivative Test, applied to the function whose graph the curve is near

(a, b)

. For an equation built from polynomials, exponentials, and logarithms, that function exists near any point of the curve where the denominator of

\frac{dy}{dx}

isn't

0

. At a point where

\frac{dy}{dx} = 0

, every term that carries

y'

drops out of the second differentiation.

Example 5

Consider the curve

y^3 + 2xy - x^2 = 12

\begin{aligned} &\text{(a) Show that } \frac{dy}{dx} = \frac{2x - 2y}{3y^2 + 2x}. \\[4pt] &\text{(b) Find every point where the} \\ &\qquad \text{tangent line is horizontal.} \\[4pt] &\text{(c) Show that } (2, 2) \text{ is the lowest} \\ &\qquad \text{point of the curve.} \end{aligned}

(a) Differentiate, with the product rule on

2xy

\begin{aligned} 3y^2y' + 2y + 2xy' - 2x &= 0, \\ (3y^2 + 2x)\,y' &= 2x - 2y, \\ y' &= \frac{2x - 2y}{3y^2 + 2x}. \end{aligned}

(b) The numerator is

0

when

y = x

. Substituting into the curve's equation,

\begin{aligned} x^3 + 2x^2 - x^2 &= 12, \\ x^3 + x^2 - 12 &= 0, \\ (x - 2)\left(x^2 + 3x + 6\right) &= 0. \end{aligned}

The quadratic factor has discriminant

9 - 24 = -15 < 0

, so

x = 2

is the only solution, which gives

(2, 2)

. The denominator there is

3(4) + 2(2) = 16 \ne 0

. The tangent is horizontal at

(2, 2)

and nowhere else.

(c) Differentiate the first line of (a) again:

\begin{aligned} &6y(y')^2 + 3y^2y'' + 2y' \\ &\qquad + 2y' + 2xy'' - 2 = 0. \end{aligned}

(2, 2)

, where

y' = 0

, this is

12y'' + 4y'' - 2 = 0

, so

y'' = \frac{1}{8}

. Since

\frac{dy}{dx} = 0

and

\frac{d^2y}{dx^2} = \frac{1}{8} > 0

(2, 2)

, the Second Derivative Test gives

y

a relative minimum at

(2, 2)

A relative minimum is only the lowest point nearby. For the whole curve, treat the equation as a quadratic in

x

x^2 - 2yx + \left(12 - y^3\right) = 0

. Its solutions are

x = y \pm \sqrt{y^3 + y^2 - 12},

which are real only when

y^3 + y^2 - 12 \ge 0

. That expression factors:

\begin{aligned} y^3 + y^2 - 12 &= (y - 2) \\ &\quad \cdot \left(y^2 + 3y + 6\right). \end{aligned}

The quadratic factor has the negative discriminant from (b), so it's always positive, and the condition means

y \ge 2

. Every point of the curve has

y \ge 2

, and

y = 2

gives only

x = 2

. So

(2, 2)

is the lowest point of the curve.

The curve y³ + 2xy − x² = 12 and its horizontal tangent at (2, 2). No point of the curve lies below y = 2.