Parametric Differentiation

Summary

  • When x and y are expressed in terms of a third variable it is called a parameter.
  • To differentiate parametric equations to find derivative of y with respect to x, we use the chain rule method.
  • When the parameter in the equations is “t”, the chain rule is defined as:     \frac { dy }{ dx } =\quad \frac { dy }{ dt } \quad \times \quad \frac { dt }{ dx }

We now completely understand that differentiation is the process that we use to find the gradient of a curve and it is denoted by  \frac { dy }{ dx } ,  we call it a derivative or differential coefficient of y with respect to x.

Moving forward, we know that a relation between the coordinates x and y is known as ”cartesian form”. However, there are times when it is convenient to express x and y in terms of a third variable which is called a parameter.

Suppose, the equation of a curve is given by two parametric equations. For example, two parametric equations of a circle with centre zero and radius a are given by:
x\quad =\quad acos(t)  and  y\quad =\quad asin(t)  here t is the parameter.

Now to differentiate these parametric equations, we must understand that in each case we can’t differentiate with respect to x, hence we cannot directly find  \frac { dy }{ dx } ,  we will use the chain rule here:

  • For each of the above equations we will only be able to find  \frac { dx }{ dt } ,  and  \frac { dy }{ dt }   and then using the chain rule we find  \frac { dy }{ dx } .
  • The chain rule will be applied like this:      \frac { dy }{ dx } =\quad \frac { dy }{ dt } \quad \times \quad \frac { dt }{ dx }

Let’s do an example now to see how we carry out parametric differentiation.

Example #1

Q. Find  \frac { dy }{ dx }   when  x\quad =\quad cos(t)  and  y\quad =\quad sin(t).

Solution:

\frac { dx }{ dt } \quad =\quad -sin(t)  and  \frac { dy }{ dt } \quad =\quad cos(t)

Use the chain rule:   \frac { dy }{ dx } =\quad \frac { dy }{ dt } \quad \times \quad \frac { dt }{ dx }

We know:   \frac { dy }{ dt } \quad =\quad cos(t)

But since:   \frac { dx }{ dt } \quad =\quad -sin(t),\quad \frac { dt }{ dx } \quad =\quad \frac { -1 }{ sin(t) }

Therefore:

\frac { dy }{ dx } \quad =\quad cos(t)\quad \times \quad \frac { -1 }{ sin(t) }

\frac { dy }{ dx } \quad =\quad \frac { -cos(t) }{ sin(t) } \quad =\quad -cot(t)         Ans

Example #2

Q. Let  x\quad =\quad 4sin(t)  and  y\quad =\quad 3cos(t),  find  \frac { dy }{ dx }   at  t\quad =\quad \frac { \pi }{ 4 } .

Solution:

\frac { dx }{ dt } \quad =\quad 4cos(t)  and  \frac { dy }{ dt } \quad =\quad -3sin(t)

\frac { dy }{ dx } \quad =\quad \frac { -3sin(t) }{ 4cos(t) }

\frac { dy }{ dx } \quad =\quad \frac { -3 }{ 4 } tan(t)

Put  t\quad =\quad \frac { \pi }{ 4 }   in  \frac { dy }{ dx } .

\frac { dy }{ dx } \quad =\quad \frac { -3 }{ 4 } tan(\frac { \pi }{ 4 } )

Hence:

\frac { dy }{ dx } \quad =\quad \frac { -3 }{ 4 }              Ans

Example #3

Q. Find  \frac { dy }{ dx } when  x\quad =\quad { r }^{ 3 }\quad -\quad r  and  y\quad =\quad { 4\quad -\quad r }^{ 2 }

Solution:

Here the parameter given is r:

\frac { dx }{ dr } \quad =\quad { 3r }^{ 2 }\quad -\quad 1  and  \frac { dy }{ dr } \quad =\quad { -2r }

Now use the chain rule:  \frac { dy }{ dx } =\quad \frac { dy }{ dr } \quad \times \quad \frac { dr }{ dx }

We know:  \frac { dy }{ dr } =\quad -2r

But:  \frac { dr }{ dx } \quad =\quad \frac { 1 }{ 3{ r }^{ 2 }\quad -\quad 1 }

\frac { dy }{ dx } \quad =\quad -2r\quad \times \quad \frac { 1 }{ 3{ r }^{ 2 }\quad -\quad 1 }

\frac { dy }{ dx } \quad =\quad \frac { -2r }{ 3{ r }^{ 2 }\quad -\quad 1 }         Ans

 

Reference
  1. https://studylib.net/doc/13474624/parametric-differentiation
  2. https://slideplayer.com/slide/10093157/