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Chapter 2: Differentiation

Section 2.5: Implicit Differentiation

Example 2.5.3

The curve $y = y (x)$ defined implicitly by the equation $x^{3} + y^{3} = 3 a x y$ is called the Folium of Descartes.

a)	Obtain a graph of this curve. In particular, explore the effect of $a$ on the curve.

b)	Obtain $y' (x)$ by implicit differentiation.

c)	Find all points on this curve where its tangent line is horizontal.

d)	Find all points on this curve where its tangent line is vertical.

e)	Working numerically in the case $a = 3$ , find all points on the curve where its slope is 1. Obtain and graph the corresponding tangent lines.

f)	Solve for $y (x)$ explicitly in the case $a = 3$ . Graph the resulting branches.

Solution

Part (a)

Figure 2.5.3(a) contains an animation in which the slider in the animation toolbar controls the value of $a$ .

Figure 2.5.3(a) Folium of Descartes and its dependence on the parameter $a$

At $a = 0$ , the curve degenerates to the line $y = - x$ . For $a \neq 0$ , there are regions where, for a single value of $x$ , there are three points on the curve; and where there is a single point. This is to be expected because every cubic equation with real coefficients has three real roots, or one real root and a pair of complex conjugate roots.

Part (b)

Solution by Context Panel

•	Control-drag the equation of the Folium.

•	Context Panel: Differentiate≻Implicitly (Set $y$ as the dependent variable)

$x^{3} + y^{3} = 3 a x y$ $\overset{implicit differentiation}{\to}$ $\frac{a y - x^{2}}{- a x + y^{2}}$

Interactively implemented stepwise solution

•	Control-drag the equation of the Folium and press the Enter key.

•	Context Panel: Evaluate at a Point≻ $y = y (x)$

•	Context Panel: Differentiate≻Implicitly (Set $y$ as the dependent variable.)

•	Context Panel: Solve≻Isolate Expression for≻ $diff (y (x), x)$

$x^{3} + y^{3} = 3 a x y$

$\overset{evaluate at point}{\to}$

$x^{3} + {y (x)}^{3} = 3 a x y (x)$

$\overset{differentiate w.r.t. x}{\to}$

$3 x^{2} + 3 {y (x)}^{2} (\frac{&DifferentialD;}{&DifferentialD; x} y (x)) = 3 a y (x) + 3 a x (\frac{&DifferentialD;}{&DifferentialD; x} y (x))$

$\overset{isolate for diff(y(x),x)}{\to}$

$\frac{&DifferentialD;}{&DifferentialD; x} y (x) = \frac{- 3 x^{2} + 3 a y (x)}{3 {y (x)}^{2} - 3 a x}$

Solution via the implicitdiff command

$implicitdiff (x^{3} + y^{3} = 3 a x y, y, x)$ = $\frac{a y - x^{2}}{- a x + y^{2}}$

To launch the Differentiation Methods tutor with the Folium embedded, press .

Alternatively, to access the Implicit Differentiation task template, click here.

Part (c)

Horizontal Tangents

•

The derivative at a point where there is a horizontal tangent is zero, that is, $y' (c) = 0$ if $(c, y (c))$ is the point of contact with such a tangent. Hence, find $y' (x)$ by implicit differentiation, and simultaneously solve the equation of the Folium and $y' (x) = 0$ for the $(c, y (c))$ pair satisfying both equations. Setting the numerator of $y'$ to zero, instead of $y'$ itself, simplifies the algebra.

•	Control-drag (or type) the equation of the Folium, and press the Enter key.

•	Context Panel: Differentiate≻Implicitly (Set $y$ as the dependent variable.)

•	Context Panel: Numerator

$x^{3} + y^{3} = 3 a x y$

(1)

$\overset{implicit differentiation}{\to}$

$\frac{a y - x^{2}}{- a x + y^{2}}$

$\overset{numerator}{\to}$

$- a y + x^{2}$

(2)

•	Using equation labels, form the sequence of two equations: the Folium, and the numerator set to zero. Context Menu: Label≻Label Reference

•	Context Panel: Solve≻Solve for Variables≻ $\{x, y\}$

•	Context Panel: Conversions≻To Radical

$, = 0$

$x^{3} + y^{3} = 3 a x y, - a y + x^{2} = 0$

$\overset{solve (specified)}{\to}$

$\{x = 0, y = 0\}, \{x = RootOf ({_Z}^{3} - 2) a, y = {RootOf ({_Z}^{3} - 2)}^{2} a\}$

$\overset{radical form}{=}$

$\{\{x = 0, y = 0\}, \{x = 2^{1 / 3} a, y = 2^{2 / 3} a\}\}$

Figures 2.5.3(a-b) show the Folium and the horizontal tangents when $a = \pm 1$ , respectively.

Figure 2.5.3(a) Folium and horizontal tangents with $a = - 1$

Figure 2.5.3(b) Folium and horizontal tangents with $a = 1$

Part (d)

Vertical Tangents

•

The derivative at a point where there is a vertical tangent is not defined, that is, $y' (c) = \infty$ if $(c, y (c))$ is the point of contact with such a tangent. Alternatively, the denominator of $y'$ becomes zero at a point of contact with a vertical tangent. Hence, find $y' (x)$ by implicit differentiation, and simultaneously solve the equation of the Folium and the denominator of $y' (x)$ set to zero for the $(c, y (c))$ pair satisfying both equations.

•	Control-drag (or type) the equation of the Folium, and press the Enter key.

•	Context Panel: Differentiate≻Implicitly (Set $y$ as the dependent variable.)

•	Context Panel: Denominator

$x^{3} + y^{3} = 3 a x y$

(3)

$\overset{implicit differentiation}{\to}$

$\frac{a y - x^{2}}{- a x + y^{2}}$

$\overset{denominator}{\to}$

$a x - y^{2}$

(4)

•	Using equation labels, form the sequence of two equations: the Folium, and the denominator set to zero. Context Menu: Label≻Label Reference

•	Context Panel: Solve≻Solve for Variables≻ $\{x, y\}$

•	Context Panel: Conversions≻To Radical

$, = 0$

$x^{3} + y^{3} = 3 a x y, a x - y^{2} = 0$

$\overset{solve (specified)}{\to}$

$\{x = 0, y = 0\}, \{x = {RootOf ({_Z}^{3} - 2)}^{2} a, y = RootOf ({_Z}^{3} - 2) a\}$

$\overset{radical form}{=}$

$\{\{x = 0, y = 0\}, \{x = 2^{2 / 3} a, y = 2^{1 / 3} a\}\}$

Figures 2.5.3(c-d) show the Folium and the horizontal tangents when $a = \pm 1$ , respectively.

Figure 2.5.3(c) Folium and vertical tangents with $a = - 1$

Figure 2.5.3(d) Folium and vertical tangents with $a = 1$

Part (e)

Tangent Lines with Slope 1

•	Type the equation of the Folium, setting $a = 3$ . Press the Enter key.

•	Context Panel: Differentiate≻Implicitly (Set $y$ as the dependent variable)

$x^{3} + y^{3} = 9 x y$

(5)

$\overset{implicit differentiation}{\to}$

$\frac{- x^{2} + 3 y}{y^{2} - 3 x}$

(6)

•	Using equation labels, form the sequence of equations consisting of the equation of the Folium and the implicit derivative set equal to 1.

•	Context Panel: Evaluate at a Point≻ $x \to u, y \to v$

•	Context Panel: Solve≻Numerically Solve

$, = 1$

$x^{3} + y^{3} = 9 x y, \frac{- x^{2} + 3 y}{y^{2} - 3 x} = 1$

$\overset{evaluate at point}{\to}$

$u^{3} + v^{3} = 9 u v, \frac{- u^{2} + 3 v}{v^{2} - 3 u} = 1$

(7)

$\overset{solve}{\to}$

$\{u = 3.619951269, v = 1.576201153\}$

(8)

•	Using the equation label, reference the sequence of equations written in terms of $u$ and $v$ . Press the Enter key.

•	Context Panel: Solve≻ Numerically Solve from point≻ $u = 1, v = 3$ (See Figure 2.5.3(e).)

$u^{3} + v^{3} = 9 u v, \frac{- u^{2} + 3 v}{v^{2} - 3 u} = 1$

$\overset{solve}{\to}$

$\{u = 1.576201153, v = 3.619951269\}$

(9)

•	Write the point-slope form of the line through $(u, v)$ and with slope 1. Press the Enter key.

•	Expression palette: Evaluation template Evaluate the line at the first $(u, v)$ solution. Press the Enter key.

•	Repeat for the second $(u, v)$ solution.

$y = 1 \cdot (x - u) + v$

$y = x - u + v$

(10)

$\binom{}{} | \binom{}{}$

$y = x - 2.043750116$

$\binom{}{} | \binom{}{}$

$y = x + 2.043750116$

•	Code for Figure 2.5.3(e) is hidden in the cell containing the graph.

•	Construct Figure 2.5.3(e) interactively via the Plot Builder, launched from the Context Panel applied to a sequence of (5), (the equation of the Folium with $a = 3$ ), and the equations of the two tangent lines.

•	The relevant options are then the setting of the ranges for both x and y to $[- 5, 5]$ .

Figure 2.5.3(e) Folium and tangent lines of slope 1

Part (f)

Table 2.5.3(a) lists the three branches of the Folium when $a = 3$ . Figure 2.5.3(f) color-codes the graphs of the three branches.

$A = {(- 4 x^{3} + 4 \sqrt{- 108 x^{3} + x^{6}})}^{1 / 3}$

Branch	Color
$y_{1} = \frac{1}{2} A + \frac{6 x}{A}$	black
$y_{2} = - \frac{1}{4} A - \frac{3 x}{A} + \frac{1}{2} I \sqrt{3} (\frac{1}{2} A - \frac{6 x}{A})$	red
$y_{3} = - \frac{1}{4} A - \frac{3 x}{A} - \frac{1}{2} I \sqrt{3} (\frac{1}{2} A - \frac{6 x}{A})$	green
Table 2.5.3(a) Three explicit branches of the Folium

>	plot([solve(x^3+y^3=9xy,y)],x=-5..5,y=-5..5,color=[black,red,green],scaling=constrained,thickness=2);

Figure 2.5.3(f) Folium branches

Branch $y_{1}$ is real for $- \infty < x \leq 3 \cdot 2^{2 / 3}$ , the right endpoint being the point where the tangent line is vertical. Branch $y_{2}$ is real for $0 \leq x < \infty$ , while branch $y_{3}$ is real for $0 \leq x \leq 3 \cdot 2^{2 / 3}$ .

Table 2.5.3(b) lists the interactive steps by which a figure comparable to Figure 2.5.3(f) could be obtained. Since the conversion to a list repeats the display of the solutions, just the steps leading to the solutions are implemented.

•	Type the equation of the Folium when $a = 3$ , the press the Enter key. Context Panel: Solve≻Obtain Solutions for≻ $y$

•	Context Panel: Plots≻2-D Implicit Plot≻ $x, y$

•	Control-drag onto graph of the folium the expressions for the tangent lines

$x^{3} + y^{3} = 9 x y$

$\overset{solutions for y}{\to}$

$\frac{1}{2} {(- 4 x^{3} + 4 \sqrt{x^{6} - 108 x^{3}})}^{1 / 3} + \frac{6 x}{{(- 4 x^{3} + 4 \sqrt{x^{6} - 108 x^{3}})}^{1 / 3}}, - \frac{1}{4} {(- 4 x^{3} + 4 \sqrt{x^{6} - 108 x^{3}})}^{1 / 3} - \frac{3 x}{{(- 4 x^{3} + 4 \sqrt{x^{6} - 108 x^{3}})}^{1 / 3}} + \frac{1}{2} I \sqrt{3} (\frac{1}{2} {(- 4 x^{3} + 4 \sqrt{x^{6} - 108 x^{3}})}^{1 / 3} - \frac{6 x}{{(- 4 x^{3} + 4 \sqrt{x^{6} - 108 x^{3}})}^{1 / 3}}), - \frac{1}{4} {(- 4 x^{3} + 4 \sqrt{x^{6} - 108 x^{3}})}^{1 / 3} - \frac{3 x}{{(- 4 x^{3} + 4 \sqrt{x^{6} - 108 x^{3}})}^{1 / 3}} - \frac{1}{2} I \sqrt{3} (\frac{1}{2} {(- 4 x^{3} + 4 \sqrt{x^{6} - 108 x^{3}})}^{1 / 3} - \frac{6 x}{{(- 4 x^{3} + 4 \sqrt{x^{6} - 108 x^{3}})}^{1 / 3}})$

Table 2.5.3(b) Interactive steps for obtaining Figure 2.5.3(f)

The complexity of the explicit expressions for $y (x)$ suggests that differentiating after solving is far more tedious than implicit differentiation.

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