Parametric Differentiation | Summary & Examples

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1 Summary

2 What is parameter?

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