Introduction
Implicit relations are not solved for
y
. Instead, they are solved for
x
and
y
together.
Here are some examples to refresh our memory:
x^2+y^2=1
x^3+y^3=6xy
What has been covered in this chapter regarding explicit functions is also true for implicit relations. The only difference is some calculation.
For example:
If dy/dx=0
, then there is a critical point.
If d^2y/dx^2=0
, then there is an inflection point.Examples
Example 1: For the implicit relation
x^2 + y^2 = 25
:
1. Find dy/dx
.
2. Determine the critical points.
3. Determine the intervals on which the curve is increasing and decreasing.
1. First, we need to find
dy/dx
. To do this, we need to differentiate both sides of the equation with respect to
x
.
\begin{align*}
\frac{d}{dx}[x^2 + y^2]&=\frac{d}{dx}25 \\\\
2x\frac{dx}{dx} + 2y\frac{dy}{dx}&=0 \\\\
2y\frac{dy}{dx}&=-2x \\\\
\frac{dy}{dx}&=\frac{-2x}{2y} \\\\
\frac{dy}{dx}&=-\frac{x}{y}
\end{align*}
2. To determine the critical points, we need to find where
dy/dx=0
or is undefined.
-x/y
is
0
when
x
is
0
and is undefined when
y
is
0
.
Therefore, the critical points are
(0,-5)
,
(0,5)
,
(-5,0)
, and
(5,0)
.
3. To determine the intervals on which the curve is increasing and decreasing, we need to know when the sign of
dy/dx
is positive or negative. Since
dy/dx=-x/y
,
dy/dx
is positive when
x
and
y
have different signs and negative when they have the same sign.
dy/dx
is positive in the second and fourth quadrants and negative in the first and third quadrants.
The function is increasing in the second and fourth quadrants and decreasing in the first and third quadrants. It is flat at the x-axis and vertical at the y-axis.
Example 2: x^3 + y^3 = 6xy
Find
d^2y/dx^2
.
\begin{align*}
\frac{d}{dx}[x^3 + y^3]&=\frac{d}{dx}6xy \\\\
3x^2\frac{dx}{dx} + 3y^2\frac{dy}{dx}&=6y\frac{dx}{dx} + 6x\frac{dy}{dx} \\\\
[3y^2 - 6x]\frac{dy}{dx}&=[6y - 3x^2] \\\\
\frac{dy}{dx}&=\frac{6y - 3x^2}{3y^2 - 6x} \\\\
\frac{d}{dx}\frac{dy}{dx}&=\frac{d}{dx}\frac{6y - 3x^2}{3y^2 - 6x} \\\\
\frac{d^2y}{dx^2}&=\frac{[3y^2 - 6x][6\frac{dy}{dx} - 6x] - [6y-3x^2][6y\frac{dy}{dx} - 6]}{[3y^2-6x]^2}
\end{align*}