Higher-Order Derivative of Functions In Parametric Form

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Parametric Form

The differentiation of a dependent variable with respect to another dependent variable when a third independent variable is given is termed as parametric differentiation. The third variable is termed as a parameter. Parametric differentiation is used for curves like an ellipse, parabola, hyperbola etc.

If x = ф(t) and y = Ψ(t), then the differentiation of y with respect to x is given by,

\[\frac{dy}{dx}\] = \[\frac{dy}{dt}\] х \[\frac{dt}{dx}\]

The differentiation of y with respect to t is multiplied by the reciprocal of differentiation of x with respect to t. Here, x and y are dependent variables and t is an independent variable. The above equation is derived using the chain rule for derivatives.


Higher-order Derivative:

If the first derivative dy/dx of a function y = f (x) is also a differentiable function, then it can be further differentiated with respect to x and this derivative is denoted by \[\frac{d^{2}y}{dx^{2}}\] which is called the second-order derivative of y with respect to x. Further, if \[\frac{d^{2}y}{dx^{2}}\] is also differentiable, then its derivative is called the third derivative of y which is denoted by \[\frac{d^{3}y}{dx^{3}}\]. Similarly, the nth derivative of y is denoted by \[\frac{d^{n}y}{dx^{n}}\]. All these derivatives are called the successive derivatives of y and this process is known as successive differentiation.


The Parametric Form of Higher-order Derivative:

The derivative of the first order in the parametric equation is given by,

\[\frac{dy}{dx}\] = \[\frac{dy}{dt}\] х \[\frac{dt}{dx}\] = \[\frac{y’(t)}{x’(t)}\]

y'(t) is the differentiation of y with respect to t and x'(t) is the differentiation of x with respect to t.

On differentiating this with respect to x, the second-order derivative will be obtained.

\[\frac{d^{2}y}{dx^{2}}\] = \[\frac{d}{dx}\](\[\frac{dy}{dx}\])

Now, substituting \[\frac{dy}{dt}\] х \[\frac{dt}{dx}\] for dy/dx,

\[\frac{d^{2}y}{dx^{2}}\] = \[\frac{d}{dx}\] (\[\frac{dy}{dt}\] х \[\frac{dt}{dx}\] )

Now, writing \[\frac{d}{dx}\] = \[\frac{\frac{d}{dt}}{\frac{dx}{dt}}\],

\[\frac{d^{2}y}{dx^{2}}\] = \[\frac{\frac{d}{dt}(\frac{dy}{dt} \times \frac{dt}{dx})}{\frac{dx}{dt}}\]

Multiplying numerator by reciprocal of the denominator,

\[\frac{d^{2}y}{dx^{2}}\] = \[\frac{d}{dt}\](\[\frac{dy}{dt}\]  х \[\frac{dt}{dx}\]) х \[\frac{dt}{dx}\] 

Therefore,

\[\frac{d^{2}y}{dx^{2}}\] = \[\frac{d}{dt}\](\[\frac{y’(t)}{x’(t)}\]) х \[\frac{dt}{dx}\]


Solved Examples


Example 1:

If x = \[\frac{1-t^{2}}{1+t^{2}}\] and y = \[\frac{2at}{1+t^{2}}\], find dy/dx.


Solution:

Given,

x = \[\frac{1-t^{2}}{1+t^{2}}\] 

Differentiating x with respect to t (using u/v rule),

\[\frac{dx}{dt}\] = \[\frac{(1+t^{2})(0-2t)-(1-t^{2})(0+2t)}{(1+t^{2})^{2}}\]

Simplifying,

\[\frac{dx}{dt}\] = \[\frac{4t}{(1+t^{2})^{2}}\] 

Given,

y = \[\frac{2at}{1+t^{2}}\]

Differentiating y with respect to t (using u/v rule),

\[\frac{dy}{dt}\] = \[\frac{(1+t^{2})2a - 2at(2t)}{(1+t^{2})^{2}}\]

Simplifying,

\[\frac{dy}{dt}\] = \[\frac{2a(1-t^{2})}{(1+t^{2})^{2}}\]

Now, to find dy/dx, the parametric form is used, that is,

\[\frac{dy}{dx}\] = \[\frac{dy}{dt}\] x \[\frac{dt}{dx}\]

Substituting the values,

\[\frac{dy}{dx}\] = \[\frac{dy}{dt}\] x \[\frac{dt}{dx}\]

\[\frac{dy}{dx}\] = \[\frac{2a(1-t^{2})}{(1+t^{2})^{2}}\] x \[\frac{(1+t^{2})^{2})}{4t}\]

Cancelling out common factors and terms,

\[\frac{dy}{dx}\] = \[\frac{\not{2} a(1-t^{2})}{\not{(1+t^{2})^{2}}}\] x \[\frac{\not{(1+t^{2})^{2}}}{_{2}\not{4}t}\]

\[\frac{dy}{dx}\] = \[\frac{a(1-t^{2})}{2t}\]


Example 2:

Find the second derivative of a sin\[^{3}\]t with respect to a cos\[^{3}\]t at t = π/4.


Solution:

Let y = a sin\[^{3}\]t and x = a cos\[^{3}\]t . \[\frac{d^{2}y}{dx^{2}}\] is to be calculated.

First, differentiate y with respect to t.

\[\frac{dy}{dt}\] = 3a sin\[^{2}\]t cos t

Now, differentiate x with respect to t.

\[\frac{dx}{dt}\] = 3a cos\[^{2}\]t(-sin t)

To find dy/dx, the parametric form is used, that is,

\[\frac{dy}{dx}\] = \[\frac{dy}{dt}\] x \[\frac{dt}{dx}\]

Substituting the values,

\[\frac{dy}{dx}\] = \[\frac{3a sin^{2}t cos t}{3a cos^{2}t (-sin t)}\]

Cancelling out common terms and factors,

\[\frac{dy}{dx}\] = \[\frac{\not{3}a sin^{2}t cos t}{\not{3}a cos^{2}t(-sin t)}\]

\[\frac{dy}{dx}\] = - \[\frac{sin t}{cos t}\]

Since, \[\frac{sin x}{cos x}\] = tan x, therefore,

\[\frac{dy}{dx}\] = - tan t

To find the second-order derivative, dy/dx is differentiated again with respect to x,

\[\frac{d^{2}y}{dx^{2}}\] = - sec\[^{2}\] t \[\frac{dt}{dx}\]

Now, since dt/dx is the reciprocal of dx/dt, substituting the reciprocal of dx/dt in the above expression,

\[\frac{d^{2}y}{dx^{2}}\] = \[\frac{-sec^{2}t}{3a cos^{2}t(-sin t)}\]

Since, sec x = \[\frac{1}{cos x}\] ,

\[\frac{d^{2}y}{dx^{2}}\] = \[\frac{1}{3a}\](\[\frac{sec^{4}t}{sin t}\])

Now substituting π/4 for t,

\[\frac{d^{2}y}{dx^{2}}\] = \[\frac{1}{3a}\](\[\frac{sec^{4} \pi / 4 }{sin \pi / 4}\])

\[\frac{d^{2}y}{dx^{2}}\] = \[\frac{1}{3a}\](\[\frac{4}{1 / \sqrt{2}}\]) 

\[\frac{d^{2}y}{dx^{2}}\] = \[\frac{4\sqrt{2}}{3a}\]


Example 3:

If x = a(t-sin t) and y = a(1-cos t). Find dy/dx. Also, find the second-order derivative.


Solution:

Given,

x = a(t - sin t)

Differentiating x with respect to t,

\[\frac{dx}{dt}\] = a(1 - cos t)

Given,

y = a(1 - cos t)

Differentiating y with respect to t,

\[\frac{dy}{dt}\] = a sin t

To find dy/dx, parametric form is used, that is,

\[\frac{dy}{dx}\] = \[\frac{dy}{dt}\] х \[\frac{dt}{dx}\] 

Substituting the values,

\[\frac{dy}{dx}\] = \[\frac{a sin t}{a(1-cost)}\]

Cancelling out the common term from numerator and denominator,

\[\frac{dy}{dx}\] = \[\frac{sin t}{1-cos t}\] 

Substituting, sin t = 2sin\[\frac{t}{2}\]cos\[\frac{t}{2}\] and 1-cos t = 2 sin\[^{2}\] \[\frac{t}{2}\],

\[\frac{dy}{dx}\] = \[\frac{2sin\frac{t}{2}cos\frac{t}{2}}{2sin^{2}\frac{t}{2}}\]

Simplifying,

\[\frac{dy}{dx}\] = \[\frac{cos \frac{t}{2}}{sin \frac{t}{2}}\]

Since, \[\frac{cos x}{sin x}\] = cot x , therefore,

\[\frac{dy}{dx}\] = cot \[\frac{t}{2}\]

To find second-order derivative, dy/dx is differentiated again with respect to x,

\[\frac{d^{2}y}{dx^{2}}\] = \[\frac{1}{2}\] x -cosec\[^{2}\] \[\frac{t}{2}\] x \[\frac{dt}{dx}\]

Now, since dt/dx is the reciprocal of dx/dt, substituting the reciprocal of dx/dt in the above expression,

\[\frac{d^{2}y}{dx^{2}}\] = \[\frac{1}{2}\] x -cosec\[^{2}\] \[\frac{t}{2}\] x \[\frac{1}{a(1-cos t)}\] 


Example 4:

Find the second derivative of x = ln t and y = t\[^{3}\] + 1.


Solution:

Given,

x = ln t

Differentiating x with respect to t,

\[\frac{dx}{dt}\] = \[\frac{1}{t}\]

Given,

y = t\[^{3}\] + 1

Differentiating y with respect to t,

\[\frac{dy}{dt}\] = 3t\[^{2}\]

To find dy/dx, parametric form is used, that is,

\[\frac{dy}{dx}\] = \[\frac{dy}{dt}\] x \[\frac{dt}{dx}\] 

Substituting the values,

\[\frac{dy}{dx}\] = 3t\[^{2}\] x t

\[\frac{dy}{dx}\] = 3t\[^{3}\] 

To find second-order derivative, dy/dx is differentiated again with respect to x,

\[\frac{d^{2}y}{dx^{2}}\]  = 9t\[^{2}\] x \[\frac{dt}{dx}\]

Now, since dt/dx is the reciprocal of dx/dt, substituting the reciprocal of dx/dt in the above expression,

\[\frac{d^{2}y}{dx^{2}}\]  = 9t\[^{2}\] x t

\[\frac{d^{2}y}{dx^{2}}\]  = 9t\[^{3}\]

Since, x = ln t, therefore,

t = e\[^{x}\]

Substituting this value in \[\frac{d^{2}y}{dx^{2}}\],

\[\frac{d^{2}y}{dx^{2}}\] = 9e\[^{3x}\]


Alternative Method:

Differentiating x with respect to t,

x’ = \[\frac{1}{t}\]

Differentiating again,

x’’ = -\[\frac{1}{t^{2}}\]

Differentiating y with respect to t,

y’ = 3t\[^{2}\]

Differentiating again,

y’’ = 6t

Now,

y’’(x) = \[\frac{x’ y’’ - y’ x’’ }{(x’)^{3}}\]

Substituting values in the above equation,

y’’(x) = \[\frac{\frac{1}{t}(6t) - 3t^{2}(-\frac{1}{t^{2}})}{\frac{1}{t^{3}}}\]

y’’(x) = (6 + 3)t\[^{3}\]

y’’(x) = 9t\[^{3}\] 

y’’(x) = 9e\[^{3x}\]


Did you know

  • Parametric differentiation is used to derive many formulas of differentiation. For example:

Assuming,

            y = a\[^{x}\]

Taking log on both sides,

y = a\[^{x}\]

log\[_{e}\] y = log\[_{e}\] a\[^{x}\]

Applying the properties of logarithms,

log\[_{e}\] y = log\[_{e}\] a\[^{x}\]

log\[_{e}\] y = x log\[_{e}\] a

Differentiating both sides with respect to x,

\[\frac{d}{dx}\] log\[_{e}\]y = \[\frac{d}{dx}\] x log\[_{e}\]a

\[\frac{1}{y}\] \[\frac{dy}{dx}\] = 1 . log\[_{e}\] a + x . 0

\[\frac{dy}{dx}\] = y log\[_{e}\]a  

\[\frac{dy}{dx}\] = a\[^{x}\] log\[_{e}\] a        [a\[^{x}\] = y]

Therefore, \[\frac{d}{dx}\](a\[^{x}\]) = a\[^{x}\] log\[_{e}\] a 

  • If the equation of the curve be given in parametric form say x = f (t) and y = g (t) then,

\[\frac{dy}{dx}\] = \[\frac{dy}{dt}\] / \[\frac{dx}{dt}\] = \[\frac{g’(t)}{f’(t)}\]

             The equation of the tangent at any point t on the curve is given by,

             y - g(t) = \[\frac{g’(t)}{f’(t)}\] [x - f(t)]

  • The conversion of an equation to parametric form is called parameterization. For example, consider the equation, y = x\[^{3}\]. Let x = t. Hence, the equation can be described as, y = t\[^{3}\]. This provides greater efficiency when differentiating or integrating a function.

  • The parametric equations of a circle x\[^{2}\] + y\[^{2}\] = a\[^{2}\] are x = a cos θ and y = a sin θ. Therefore, the parametric coordinates of any point lying on the circle x\[^{2}\] + y\[^{2}\] = a\[^{2}\] are (a cos θ, a sin θ).

  • The parametric equation of a parabola y\[^{2}\] = 4ax are x = at\[^{2}\] and y = 2at where t is the parameter.

  • The parametric equations of an ellipse \[\frac{x^{2}}{a^{2}}\] + \[\frac{y^{2}}{b^{2}}\] = 1 are x = a cosθ and y = bsinθ where 0 ≤ θ ≤ 2π is the parameter.

FAQ (Frequently Asked Questions)

Q1. What is the Use of Higher-order Derivatives?

Ans. The derivative is the rate of change of a function. For example, the rate of change of distance with respect to time is termed as velocity. On differentiating distance with respect to time, velocity can be calculated. Now, acceleration can be calculated by differentiating velocity with respect to time. In other words, the second-order derivative of distance with respect to time gives acceleration.

Q2. What are the Advantages of Using Parametric Equations?

Ans. A parameter is used to represent something useful and thus provides important information about the graph. A plane curve traces the motion of an object over a certain interval of time. Let x = cos t and y = sin t. The position of the particle is represented by (x,y) where t represents time. The initial position of the particle is given by t = 0. Substituting this value, the initial position is (cos 0, sin 0) = (1,0). Hence, the position of the particle can be found at any time interval.