Parametric Equations

For Parametric Equations, we will start to work with our usual variables (or parameters) like x and y. However, now we will use these variables/parameters where their value is dependent on an extra third variable/parameter. Often we use t, for time, as this third parameter.

Parametric Equations can be used for situations where for example, the x coordinate and the y coordinate are both based on time, t. Instead of writing the coordinates like (x, y), they can be written as x and y both as functions of t, like (x(t), y(t)) each of these parameters still plot on a graph the same way (x, y) would, it’s just the values of the x and y both depend on the value of t.

Definition - Parametric Equations

If we have a parametric equation, where we have the parameters that are differentiable to t, of x = x(t) and y = y(t). And dx/dt does not equal 0, then,

dy over dx derivative operator = 
dy/dt
dx/dt

Beginner Derivative of Parametric Equation

We have the parametric equation (x, y) = (t, t2 + 1)

Calculate dy over dx derivative operator using dy over dx derivative operator   =   
dy/dt
dx/dt

First, for Parametric equations, recognize that we have our x & y terms as functions of t; x(t) & y(t).

(x, y) = (t, t2 + 1)

x(t) = t

y(t) = t2 + 1

Now, to calculate dy over dx derivative operator we use
dy/dt
dx/dt

So differentiate x(t) to get dx/dt

And y(t) to get dy/dt

dx
dt
 =  1
dy
dt
 =  2t
dy over dx derivative operator   =   
dy/dt
dx/dt
  =   
2t
1
  = 2t

Differentiate the parametric equation (x, y) = (3t2 + 2t, sin(t))

using dy over dx derivative operator   =   
dy/dt
dx/dt

We have (x, y) = (3t2 + 2t, sin(t))

To calculate dy over dx derivative operator =
dy/dt
dx/dt
, first find
dy
dt
&
dy
dt

We have x(t) = 3t2 + 2t

dx/dt = (2)3t2-1 + 2t0

and

y(t) = sin(t)

dy/dt = cos(t)

dy over dx derivative operator   =   
dy/dt
dx/dt
  =   
cos(t)
6t + 2

Slope of Tangent / Derivatives with Parametric Equations

The example below show how to find the slope of a tangent by differentiating.

The first part of the solution to finding the slope of the tangent when t=1 for the graph of (x,y) = (t^3, sin(t))
The second part of the solution to finding the slope of the tangent when t=1 for the graph of (x,y) = (t^3, sin(t))

Slope of Tangent / Derivatives of Parametric Equations

Below is another worked examples showing how to find the slope of a tangent by differentiating.

The first part of the solution to finding the slope of the tangent when t=pi/4 for the graph of (x,y) = (4sin(t), 5cos(t))
The second part of the solution to finding the slope of the tangent when t=pi/4 for the graph of (x,y) = (4sin(t), 5cos(t))

Speed

For a graph that gives us the position of something based off time, we can use a method of differentiation of that graph to calculate its speed (or velocity)

Definition - Speed

If we have the variables x = x(t) and y = y(t) that give the position of something with regards to the time, t, and these variable functions of t are differentiable of t, then,

Speed = square rooot symbol
(dx/dt)2 + (dy/dt)2

Speed and Derivatives

Below a worked example showing how to find the speed of an object at given location and time.

The first part of the solution to finding the speed of an object when t=1 and t=2 for a location of (x,y) = (2t^3, cos(t))
The second part of the solution to finding the speed of an object when t=1 and t=2 for a location of (x,y) = (2t^3, cos(t))

Speed

Below is a worked examples showing how to determine at which location an object has greater speed.

The first part of the solution to finding at which location - t=pi/2 or t=0 is the greater speed of an object on (x,y) = (4sin(t), 5cos(t))
The second part of the solution to finding at which location - t=pi/2 or t=0 is the greater speed of an object on (x,y) = (4sin(t), 5cos(t))
Chain Rule - Part 2
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