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Continuity and Differentiability - Derivative of functions expressed in parametric forms

Grade 12CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Definition of Parametric Equations: Sometimes, the relationship between two variables xx and yy is not expressed directly. Instead, both are defined as functions of a third variable, called a 'parameter' (usually tt or θ\theta). For example, x=f(t)x = f(t) and y=g(t)y = g(t). Visually, this means that as tt varies, the point (x,y)(x, y) moves across the Cartesian plane, tracing a curve.

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The Chain Rule Bridge: To find the derivative dydx\frac{dy}{dx} when given parametric equations, we use the chain rule. Since yy depends on tt and tt is linked to xx, we differentiate both with respect to the parameter. Mathematically, the derivative is the ratio of their individual rates of change: dydx=dy/dtdx/dt\frac{dy}{dx} = \frac{dy/dt}{dx/dt}.

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Geometric Interpretation: The value of dydx\frac{dy}{dx} at a specific parameter value t0t_0 represents the slope of the tangent line to the curve at the point (f(t0),g(t0))(f(t_0), g(t_0)). Even if the curve loops or crosses itself, the parametric derivative provides the precise steepness of the curve at that specific moment in the parameter's progression.

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The Non-zero Denominator Constraint: For the derivative dydx\frac{dy}{dx} to exist, the function x=f(t)x = f(t) must have a non-zero derivative with respect to tt (i.e., dxdt≠0\frac{dx}{dt} \neq 0). Visually, if dxdt=0\frac{dx}{dt} = 0 while dydt≠0\frac{dy}{dt} \neq 0, the curve is moving purely vertically, resulting in a vertical tangent line where the slope is undefined.

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Higher-Order Derivatives: Finding the second derivative d2ydx2\frac{d^2y}{dx^2} is a multi-step process. You cannot simply divide the second derivatives of yy and xx. Instead, you must differentiate the first derivative dydx\frac{dy}{dx} with respect to tt, and then multiply by dtdx\frac{dt}{dx} (the reciprocal of dxdt\frac{dx}{dt}) to account for the change in xx.

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Trigonometric Parameters: Many geometric shapes are best described using angles. For instance, a circle of radius aa is represented as x=acos⁡θx = a \cos \theta and y=asin⁡θy = a \sin \theta. As θ\theta sweeps from 00 to 2π2\pi, the derivative dydx\frac{dy}{dx} changes continuously, representing the slope at every point along the circular boundary.

📐Formulae

dydx=dydtdxdt, provided dxdt≠0\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}, \text{ provided } \frac{dx}{dt} \neq 0

dydx=dydθdxdθ, provided dxdθ≠0\frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}, \text{ provided } \frac{dx}{d\theta} \neq 0

d2ydx2=ddx(dydx)=ddt(dydx)dxdt\frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy}{dx}\right) = \frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{\frac{dx}{dt}}

💡Examples

Problem 1:

Find dydx\frac{dy}{dx} if x=acos⁡3θx = a \cos^3 \theta and y=asin⁡3θy = a \sin^3 \theta.

Solution:

Step 1: Differentiate xx with respect to θ\theta: dxdθ=a⋅3cos⁡2θ⋅(−sin⁡θ)=−3acos⁡2θsin⁡θ\frac{dx}{d\theta} = a \cdot 3 \cos^2 \theta \cdot (-\sin \theta) = -3a \cos^2 \theta \sin \theta Step 2: Differentiate yy with respect to θ\theta: dydθ=a⋅3sin⁡2θ⋅(cos⁡θ)=3asin⁡2θcos⁡θ\frac{dy}{d\theta} = a \cdot 3 \sin^2 \theta \cdot (\cos \theta) = 3a \sin^2 \theta \cos \theta Step 3: Use the parametric differentiation formula: dydx=dy/dθdx/dθ=3asin⁡2θcos⁡θ−3acos⁡2θsin⁡θ\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{3a \sin^2 \theta \cos \theta}{-3a \cos^2 \theta \sin \theta} Step 4: Simplify the expression: dydx=−sin⁡θcos⁡θ=−tan⁡θ\frac{dy}{dx} = -\frac{\sin \theta}{\cos \theta} = -\tan \theta

Explanation:

To solve this, we differentiate both the xx and yy functions using the chain rule (power rule combined with trigonometric derivatives) and then divide the results.

Problem 2:

If x=at2x = at^2 and y=2aty = 2at, find d2ydx2\frac{d^2y}{dx^2}.

Solution:

Step 1: Find the first derivative dydx\frac{dy}{dx}. dxdt=2at,dydt=2a\frac{dx}{dt} = 2at, \quad \frac{dy}{dt} = 2a dydx=dy/dtdx/dt=2a2at=1t\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{2a}{2at} = \frac{1}{t} Step 2: Differentiate dydx\frac{dy}{dx} with respect to xx to find d2ydx2\frac{d^2y}{dx^2}: d2ydx2=ddt(1t)⋅dtdx\frac{d^2y}{dx^2} = \frac{d}{dt}\left(\frac{1}{t}\right) \cdot \frac{dt}{dx} Step 3: Compute the derivatives: ddt(1t)=−1t2\frac{d}{dt}\left(\frac{1}{t}\right) = -\frac{1}{t^2} dtdx=1dx/dt=12at\frac{dt}{dx} = \frac{1}{dx/dt} = \frac{1}{2at} Step 4: Multiply the terms: d2ydx2=(−1t2)⋅(12at)=−12at3\frac{d^2y}{dx^2} = \left(-\frac{1}{t^2}\right) \cdot \left(\frac{1}{2at}\right) = -\frac{1}{2at^3}

Explanation:

The second derivative requires an extra step: differentiating the first derivative with respect to the parameter tt and then dividing by dxdt\frac{dx}{dt} according to the chain rule.