Introduction Introduction
We have primarily worked with explicit functions, where one variable is defined by another.
y = x^2 + 2x \quad\quad x = \sin(y)
We had two lessons on implicit relations, where the relationship between variables is shown via an equation.
x^2 + y^2 = 1 \quad\quad xy + y^2 = x
For parametric functions, multiple variables are defined by a parameter: an independent variable. Below is one parametric function:
\begin{align*}
x(t) &= 2\cos(t) \\
y(t) &= 2\sin(t)
\end{align*}
We can plot this function by graphing
x(t)
and
y(t)
as
t
changes.
First Derivatives
We can define multiple derivatives for a parametric function:
dy/dt
, the rate of change of y
with respect to t
dx/dt
, the rate of change of x
with respect to t
dy/dx
, the rate of change of y
with respect to x
dx/dy
, the rate of change of x
with respect to y
Generally, we are looking to find
dy/dx
, the slope of the tangent
line to the curve at
(x,y)
.
This is done via the chain rule:
\frac{dy}{dx} = \frac{dy/dt}{dx/dt} \quad \text{(when } \frac{dx}{dt} \neq 0\text{)}
When
dx/dt=0
, the tangent line is vertical. There is no change in
x
, and there cannot be a slope.
Example: For the following parametric equation, at
t = 1
, find
The location of the function
The slope of the tangent line.
\begin{align*}
x(t) &= 3t - 1 \\
y(t) &= t^2
\end{align*}
\frac{dy}{dt} = 2t \\ \text[] \\
\frac{dx}{dt} = 3 \\ \text{} \\
\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{2t}{3} \\ \text{} \\
x(1) = 3(1) - 1 = 2, \\ \text{} \\
y(1) = 1^2 = 1
(x(1),\,y(1))=(2, 1) \\ \text{} \\
\left. \frac{dy}{dx} \right|_{t=1} = \frac{2}{3}
Second Derivatives
To find
d^2y/dx^2
, we again use the chain rule:
\frac{d^2y}{dx^2} = \frac{d\left(\frac{dy}{dx}\right)}{dx} = \frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{\frac{dx}{dt}}
Example: Continuing the previous example:
\frac{d^2y}{dx^2} = \frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{\frac{dx}{dt}} =
\frac{\frac{d}{dt}\left(\frac{2t}{3}\right)}{3} =
\frac{2/3}{3} =
\frac{2}{9}
Note that, same as with explicit functions, the sign of
d^2y/dx^2
denotes concavity.
Since
d^2y/dx^2 > 0
, the curve is concave up at
t=1
.