# 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: $\frac { dy }{ dx } =\quad \frac { dy }{ dt } \quad \times \quad \frac { dt }{ dx }$

### What is parameter?

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.

### Chain Rule Method

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