Implicit Functions

From the equation $3x + 2y + 1 = 0$, we can find $y$ in terms of $x$. In other words, we can find $${y = -\frac{1 - 3x}{2}}$$. In general, given a quadratic function $f(x,y)$, if $y = g(x)$ always satisfies $f(x,g(x)) = 0$, then we say the equation $y = g(x)$ is an implicit function determined by the equation $f(x,y) = 0$.

Understanding

Finding the implicit function $y = g(x)$ determined by the equation $f(x,y) = 0$ is the same as solving the equation $f(x,y) = 0$ for $y$. But for some type of $f$ and some $x$, there may not be a $y$ satisfying the equation $f(x,y) = 0$.

Theorem 4..8   Suppose that $f(x,y)$ is the class $C^{1}$ at $\boldsymbol{x}_{0} = (x_{0},y_{0})$ in the region $D$. If

$$f(x_{0},y_{0}) = 0, f_{y}(x_{0},y_{0}) \neq 0$$

, then there is an unique implicit function $y = g(x)$ determined by $f(x,y) = 0$ in the neighborhood of $x = x_{0}$ which satisfies

$$1. f(x,g(x)) = 0$$

$$2. g(x_{0}) = y_{0}$$

$$3. g(x) {\rm is} C^{1}, \frac{dy}{dx} = -\frac{f_{x}(x,y)}{f_{y}(x,y)}$$

$f_{x}(x,y)$ is a funciton of $x,y$ and $y$ is a function of $x$. Thus we have the following figure.

Figure 4.29: Check

Thus, $\frac{df_{x}(x,y)}{dx} = f_{xx}(x,y) + f_{xy}(x,y)\frac{dy}{dx}$

NOTE In the neighborhood of the point satisfying $f_{y}(x,y) \neq 0$, the implicit function $y$ exists and the implcit function is differentiable. Thus the total diffetential of $f$ is

$$f_{x}dx + f_{y}dy = 0$$

Now using the condition $f_{y} \neq 0$,

$$\frac{dy}{dx} = -\frac{f_{x}}{f_{y}}$$

This is 3. Furthermore, differentiate the above equation with respect to $x$, we have

$$\frac{d^{2}y}{dx^2} = -\frac{\frac{df_{x}}{dx}f_{y} - f_{x}\frac{df_{y}}{dx}}{f_{y}^{2}}$$

Now we use

$$\frac{df_{x}}{dx} = f_{xx} + f_{xy}\frac{dy}{dx} = f_{xx} - f_{xy}\frac{f_{x}}{f_{y}}$$

$$\frac{df_{y}}{dx} = f_{yx} + f_{yy}\frac{dy}{dx} = f_{xy} - f_{yy}\frac{f_{x}}{f_{y}}$$

to obtain

$$\frac{d^{2}y}{dx^2} = -\frac{(f_{xx} - f_{xy}\frac{f_{x}}{f_{y}})f_{y} - f_{x}(f_{xy} - f_{yy}\frac{f_{x}}{f_{y}})}{f_{y}^{2}}$$

$$= -\frac{f_{xx}f_{y}^{2} - 2f_{xy}f_{x}f_{y} + f_{yy}f_{x}^{2}}{f_{y}^{3}}$$

Implicit Functions

$\frac{d^{2}y}{dx^2} = -\frac{f_{xx}f_{y}^{2} - 2f_{xy}f_{x}f_{y} + f_{yy}f_{x}^{2}}{f_{y}^{3}}$ is called 2nd order derivative of implicit function

Example 4..19   Find the derivative of a implicit function $y$ such as $${\frac{dy}{dx}, \frac{d^{2}y}{dx^{2}}}$$ deteremined by

$$x^{2} + y^{2} = 3xy$$

SOLUTION Set $f(x,y) = x^{2} + y^{2} - 3xy$ and take total differential of $f(x,y)$,

$$df = f_{x}dx + f_{y}dy = (2x - 3y)dx + (2y - 3x)dy = 0$$

Thus,

$$\frac{dy}{dx} = -\frac{2x-3y}{2y-3x}.$$

Since $f_{xx} = 2, f_{xy} = -3, f_{yy} = 2$,

$$\frac{d^{2}y}{dx^{2}} = -\frac{2(2y-3x)^{2} + 6(2x-3y)(2y-3x) + 2(2x-3y)^{2}}{(2y-3x)^{3}}$$

$$= \frac{10(x^2 - 3xy + y^2)}{(2y - 3x)^{2}} = 0\ensuremath{ \blacksquare}$$

A problem of finding a differentiation of an implicit function is solved by taking total differentiatial

Check

$x^2 +y^2 =3xy$

Exercise 4..19   Find the derivative of a implicit function $y$ such as $${\frac{dy}{dx}, \frac{d^{2}y}{dx^{2}}}$$ deteremined by

$$2x^2 + 5xy - 3y^2 = 1$$

Exercise4-19

Since $4(5x-6y)^2 - 10(4x+5y)(5x-6y) - 6(4x+5y)^2 = -98(2x^2 + 5xy-3y^2)$ and $2x^2 + 5xy -3y^2 = 1$.

SOLUTION Set $f(x,y) = 2x^2 + 5xy - 3y^2 - 1 = 0$ and find total differential of $f(x,y)$. Then

$$df = f_{x}dx + f_{y}dy = (4x + 5y)dx + (5x - 6y)dy = 0.$$

Thus,

$$\frac{dy}{dx} = -\frac{4x + 5y}{5x - 6y}.$$

$$\frac{d^2 y}{dx^2} = -\frac{(4 + 5\frac{dy}{dx})(5x - 6y) - (4x + 5y)(5 - 6\frac{dy}{dx})}{(5x - 6y)^2}$$

$$= -\frac{4(5x-6y)^2 - 10(4x+5y)(5x-6y) - 6(4x+5y)^2}{(5x - 6y)^3} = \frac{98}{(5x-6y)^3}\ensuremath{ \blacksquare}$$

Example 4..20   Find $${z_{x}, z_{y}}$$ for the implicit function $z = g(x,y)$ determined by the equation $xy + yz + zx = 1$.

By writing $z = \frac{1 - xy}{y+x}$, we can find $z_x = \frac{-y(y+x)-(1-xy)}{(y+x)^2} = -\frac{y^2 + 1}{(y+x)^2} = -\frac{y+z}{y+x}$.

SOLUTION For $f(x,y,z) = xy + yz + zx - 1 = 0$, set $x = t, y = s$. Then,

$$\frac{\partial f}{\partial t} = \frac{\partial f}{\partial x}\cdo... ...l f}{\partial z} \frac{\partial z}{\partial x}\frac{\partial x}{\partial t} = 0$$

Thus,

$$\frac{\partial z}{\partial x} = -\frac{\frac{\partial f}{\partial... ...\frac{\partial f}{\partial z}\frac{\partial x}{\partial t}} = - \frac{y+z}{y+x}$$

Similarly,

$$\frac{\partial z}{\partial y} = -\frac{\frac{\partial f}{\partial... ...}\frac{\partial y}{\partial t}} = - \frac{x+z}{y+x} \ensuremath{ \blacksquare}$$

Exercise 4..20   Find $${z_{x}, z_{y}}$$ for the implicit function $z = g(x,y)$ determined by the equation $x^2 + y^2 + z^2 = 4$.

SOLUTION For $f(x,y,z) = x^2 + y^2 + z^2 - 4$, set $x = t, y = s$.

$$\frac{\partial f}{\partial t} = \frac{\partial f}{\partial x}\fra... ...l f}{\partial z}\frac{\partial z}{\partial x} \frac{\partial x}{\partial t} = 0$$

Thus,

$$\frac{\partial z}{\partial x} = -\frac{\frac{\partial f}{\partial... ...l f}{\partial z}\frac{\partial x}{\partial t}} = - \frac{2x}{2z} = -\frac{x}{z}$$

Similarly,

$$\frac{\partial z}{\partial y} = -\frac{\frac{\partial f}{\partial... ...al y}{\partial t}} = - \frac{2y}{2z} = -\frac{y}{z} \ensuremath{ \blacksquare}$$

Theorem 4..9   If an implicit function $y,z$ of $x$ is determined by the equations $f(x,y,z) = 0, g(x,y,z) = 0$, then,

$$\frac{dy}{dx} = - \frac{f_x g_z - f_z g_x}{f_y g_z - f_zg_y}.$$

Theorem4-9

To find $\frac{dy}{dx}$, we first find $df,dg$ and then eliminate $dz$.

Example 4..21   Find $${\frac{dy}{dx}, \frac{dz}{dx}}$$ for the implicit functions $y,z$ of $x$ determined by the equations $${x^2 + y^2 + z^2 = 1, 2x + 3y - z = 1}$$.

SOLUTION Let $f(x,y,z) = x^2 + y^2 + z^2 - 1 = 0 , g(x,y,z) = 2x + 3y - z - 1 = 0$. Then totally differentiate $f$ and $g$ to get

$$df = 2xdx + 2ydy + 2zdz = 0, dg = 2dx + 3dy - dz = 0$$

Now delete $dz$ to get

$$(2x+4z)dx + (2y+6z)dy = 0$$

Check

From the equations $2xdx + 2ydy + 2zdz = 0, 2dx + 3dy - dz = 0$, we delete $dz$ by adding $4zdx + 6zdy -2zdz =0$ to the first equation,
$2xdx + 4zdx + 2ydy + 6zdy = (2x+4z)dx + (2y+6z)dy = 0$.

Thus,

$$\frac{dy}{dx} = -\frac{2x+4z}{2y+6z}$$

Similarly, delete $dy$ to get

$$(6x-4y)dx + (6z + 2y)dz = 0$$

Thus,

$$\frac{dz}{dx} = -\frac{6x-4y}{6z+2y}\ensuremath{ \blacksquare}$$

Exercise 4..21   Find $${\frac{dy}{dx}, \frac{dz}{dx}}$$ for the implicit functions $y,z$ of $x$ determined by the equations $${xyz = 1, xy + yz + zx = 1}$$.

Check

$yzdx + xzdy + xydz = 0, (y + z)dx + (x + z) dy + (y + x)dz = 0$. Eliminate $dz$ by multiplying $(y+z)$ to the former equation and multiplyimg $xy$ to the latter equation.
$yz(y+x)dx + xz(y+x)dy - (xy(y+z)dx - xy(x+z)dy) = 0$. Simplifying, $(yz(y+x)-xy(y+z))dx + (xz(y+x)-xy(x+z))dy = 0$

SOLUTION Let $f(x,y,z) = xyz -1 = 0, g(x,y,z) = xy + yz + zx - 1 = 0$. Then find total derivatives.

$$df = f_{x}dx + f_{y}dy + f_{z}dz = yzdx + xzdy + xydz = 0$$ (4.3)

$$dg = g_{x}dx + g_{y}dy + g_{z}dz = (y + z)dx + (x + z) dy + (y + x)dz = 0$$ (4.4)

Now using the equation 4.3 and the equation 4.4 to eliminate $dz$.

$$((y+x)yz - xy(y+z))dx + ((y+x)xz - xy(x+z))dy = 0$$

Then

$$\frac{dy}{dx} = -\frac{y^2 z - xy^2}{x^2 z - x^2 y} = \frac{y^2 (x-z)}{x^2 (z-y)}.$$

This time, using the equation 4.3 and the equation 4.4 to eliminate $dy$.

$$((x+z)yz - xz(y+z))dx + ((x+z)xy - xz(y+x))dz = 0$$

$$z^2(y-x)dx + x^2(y-z)dz = 0$$

Thus

$$\frac{dz}{dx} = \frac{z^2 (x-y)}{x^2 (y-z)}\ensuremath{ \blacksquare}$$

Exercise A

1.

Find the $${\frac{dy}{dx}, \frac{d^{2}y}{dx^{2}}}$$ for the implicit functions $y$ of $x$ determined by the following functions．

(a) $${x - y^{2} = 1}$$ (b) $${x^{2} + xy + 2y^{2} = 1}$$ (c) $${x - e^{y} = 0}$$

(d) $${x^{3} - 3xy + y^{3} = 1}$$

2.

Find the $${\frac{dy}{dx}, \frac{dz}{dx}}$$ for the implicit functions $y,z$ of $x$ determined by the following functions．

(a) $${x^2 +y^2 + z^2 = 4, x + y + z = 1}$$ (b) $${xyz = 1, x + y + z = 1}$$

Exercise B

1.

Find $${\frac{dy}{dx}, \frac{d^{2}y}{dx^{2}}}$$ for the implicit function $y$ of $x$ determined by the following equations.

(a) $${2x^2+ 5xy - 3y^2 = 1}$$ (b) $${y = e^{x+y}}$$ (c) $${x^2 - y^2 = xy}$$

(d) $${\log{\sqrt{x^2 + y^2}} = \tan^{-1}{\frac{y}{x}}}$$

2.

Find $${\frac{dy}{dx}, \frac{dz}{dx}}$$ for the implicit function $y,z$ of $x$ determined by the following equations.

(a) $${x^2 +y^2 + z^2 = 4, x^2 + y^2 = 4x}$$ (b) $${xyz = 1, xy + yz + zx = 1}$$

3.

Find the equation of the tangent line and the normal line to the curve $2x^2 + 5y^2 = 12$ at the point $(1,\sqrt{2})$.

4.

Find the equation of the tangent plane to the surface $${z = \tan^{-1}\frac{y}{x}}$$ at the point $(1,1,\frac{\pi}{2})$．Find the equation of the normal line through $(1,1,\frac{\pi}{2})$.

5.

Find the local extrema of $y = g(x)$ implicitly defined by the following equations．

(a) $${8x^2 +4xy + 5y^2 = 36}$$ (b) $${x^{2}y + x + y = 0}$$

(c) $${x^3 + y^3 - 6xy = 0}$$