Derivative of Functions in Parametric Forms

Sometimes it is not possible (or not convenient) to write y directly as a function of x. In such cases, we introduce a third variable—called a parameter—usually denoted by t.
Then we express both x and y in terms of this parameter:

x = x(t), y = y(t)

This way of describing a curve using a third variable is called a parametric representation or writing the function in parametric form.

Normal Differentiation vs Implicit Differentiation vs Parametric Differentiation

(A) Explicit form

If a function is given as y = f(x)

Then differentiation is simple: dy\dx

(B) Implicit form

If a function is given as f(x, y) = 0. Then both x and y appear in one equation.

We either:

• convert to explicit form, or
• use implicit differentiation (differentiate both sides w.r.t x).

(C) Parametric form

Sometimes it is impossible or inconvenient to write y directly in terms of x. So we introduce a parameter t and write:

x = f(t), y = g(t)

This is called a parametric equation.

Parametric Equation

Parametric Equation is a equation in which the two variables are equated to a third variable usually denoted by 't'. This third variable is called parameter and hence the name is Parametric Equation. This type of equation is usually used in replacing the coordinates from the cartesian form of equation of a curve and convert it into a variable form called as parametric form.

In Parametric Function the two variables are given as x = f(t) and y = g(t) i.e. both are a function of another independent variable 't'.

How to Find Derivatives of Functions in Parametric Forms?

Let's say we have two variables x and y, usually, such variables are related to each other in an implicit or an explicit manner. But in some cases, these variables are related to each other through a third variable. This form is called the parametric form of the equation and the variable is called a parameter. We have x = f(t) and y = g(t) here, t is a parameter.

We will first differentiate x and y with respect to 't' separatly. On differentiating x with respect to 't' we get dx/dt and on differentiating y by 't' we get dy/dt.

Now to the derivative of y with respect to x is given by the formula dy/dx = (dy/dt).(dt/dx). The formula for parametric differentiation is also expressed in the image attached below:

The derivative of such functions is given by chain rule,

Using Chain Rule, dy/dt can be written as

\frac{dy}{dt} = \frac{dy}{dx}\cdot \frac{dx}{dt}

Now, rearranging the terms, we get

\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}, where\frac{dx}{dt} \ne 0

Thus,\frac{dy}{dx}= \frac{g'(t)}{f'(t)}  [as\frac{dy}{dx} = g'(t)and\frac{dx}{dt} = f'(t) ]

Proof of Parametric Differentiation Formula

Since y and x are dependent on 't', then any change in 't' would also cause a change in 'y' and 'x'. Hence, for small change in 't' given as Δt the corresponding changes in x and y are Δx and Δy.

We can write Δy/Δx = (Δy/Δt)/(Δx/Δt)

Taking limit on both sides

lim _Δx→0 Δy/Δx = lim _Δt→0 (Δy/Δt)/lim _Δt→0 (Δx/Δt)

Using the Concept of Limit and Derivatives, we have

dy/dx = (dy/dt)/(dx/dt)

Parametric Differentiation Examples

Example 1: Find dy/dx if x = 4t² + 10, y =t²at t = 1.

Solution:

x = 4t² + 10, y =t²

dx/dt = 8t

dy/dt = 2t

Now, let's find out

dy/dx = (dy/dt)/(dx/dt)

⇒ dy/dx = 2t/8t

At t = 1

dydx = 1/4

Example 2: Find dy/dx, if x = 4t, y = 1/t at t = 1.

Solution:

x = 4t, y = 1/t

dx/dt = 4

dy/dt = -1/t²

Now, let's find out

dy/dx = (dy/dt)/(dx/dt)

⇒\frac{dy}{dx} = \frac{-\frac{1}{t^2}}{4}

⇒ dy/dx = -1/4t²

At t = 1

dy/dx = -1/4

Example 3: Find dy/dx, if x = at² + 2t, y = t at t = 0.

Solution:

x = at² + 2t, y = t

dx/dt = 2at + 2

dy/dt = 1

Now, let's find out

dy/dx = (dy/dt)/(dx/dt)

⇒\frac{dy}{dx} = \frac{1}{2at + 2}

⇒ dy/dx = 1/2

Example 4: Find dy/dx if x = at², y =2at at t = 1.

Solution:

x = at², y =2at

dx/dt = 2at

dy/dt = 2a

Now, let's find out

dy/dx = (dy/dt)/(dx/dt)

dy/dx = 2a/2at

At t = 1

dy/dx = 1

Example 5: Find dy/dx, if x = at³ + 2t², y = t² at t = 1.

Solution:

x = at³ + 2t², y = t²

dx/dt = 3at² + 4t

dy/dt = 2t

Now, let's find out

dy/dx = (dy/dt)/(dx/dt)

⇒\frac{dy}{dx} = \frac{2t}{3at^2 + 4}

⇒\frac{dy}{dx} = \frac{2}{3a + 4}

Example 6: Find dy/dx, if x = acos(θ) , y = asin(θ).

Solution:

x = acos(θ) and y = asin(θ)

dx/dθ ​= −a sin(θ)

dy/dθ = a cosθ

Now, let's find out

dy/dx = (dy/dt)/(dx/dt)

⇒ dy/dx = (dy/dθ)/(dx/dθ)

⇒ dy/dx = acosθ/-asinθ

⇒ dy/dx = -cot θ

Example 7: Find dy/dx, if x = acos²(θ) , y = asin²(θ).

Solution:

x = a cos²(θ) , y = a sin²(θ)

dx/dθ = -2a cosθsinθ

dy/dθ = 2a cosθsinθ

Now, let's find out

dy/dx = (dy/dt)/(dx/dt)

⇒ dy/dx = (dy/dθ)/(dx/dθ)

⇒\frac{dy}{dx} = \frac{-2acos(\theta)sin(\theta)}{2acos(\theta)sin(\theta)}

⇒ dy/dx = -1

Example 8: Find dy/dx , if x = tsin(t), y = cos(t), at t = π/2.

Solution:

x = tsin(t), y = cos(t)

dx/dt = t.cos(t) + sin(t)

dy/dt = -sin(t)

Now, let's find out

dy/dx = (dy/dt)/(dx/dt)

⇒\frac{dy}{dx} = \frac{-sin(t)}{tcos(t) + sin(t)}

At t = π/2

⇒\frac{dy}{dx} = \frac{-sin(\frac{\pi}{2})}{\frac{\pi}{2}cos(\frac{\pi}{2}) + sin(\frac{\pi}{2})}

⇒ dy/dx = -1

Example 9: Find dy/dx, if x = e^t + sin(t), y = t², at = 0.

Solution:

x = e^t + sin(t), y =t²

dx/dt = e^t + cos(t)

dy/dt = 2t

Now, let's find out

dy/dx = (dy/dt)/(dx/dt)

⇒\frac{dy}{dx} = \frac{2t}{e^t + cos(t)}

At t = 0

⇒ dy/dx = 0

Example 10: Find dy/dx, if x = 4e^t, y = cos(t)at t = 1.

Solution:

x = 4e^t, y =cos(t)

dx/dt = 4et

dy/dt = -sin(t)

Now, let's find out\

dy/dx = (dy/dt)/(dx/dt)

⇒\frac{dy}{dx} = \frac{-sin(t)}{4e^t}

At t = 1

⇒\frac{dy}{dx} = -\frac{sin(1)}{4e}

Practice Questions on Parametric Differentiation

Q2: Find dy/dx if x = at³ and y = 3at

Q4: Find dy/dx if x = a sec θ and y = b tan θ.

Q3: Find dy/dx if x = 3 log t and y = e^2t

Q1: Find dy/dx if x = sin²t and y = cos³t

Related Articles:

• Derivative of Inverse Trigonometric Function
• Chain Rule
• Application of Derivatives