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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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Parametric representation involves expressing variables xx and yy as functions of a common third variable, usually tt or θ\theta. For example, a circle of radius aa is represented as x=acos⁡θx = a \cos \theta and y=asin⁡θy = a \sin \theta. This allows us to track the position of a point on a curve as the parameter varies.

A unit circle in the Cartesian plane showing a point P defined by parameter theta.
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The Chain Rule is the foundation for differentiating parametric equations. To find the rate of change of yy with respect to xx (dydx\frac{dy}{dx}), we calculate the rate of change of both variables with respect to the parameter tt independently. The ratio of these rates, dy/dtdx/dt\frac{dy/dt}{dx/dt}, gives the required derivative, provided dx/dt≠0dx/dt \neq 0.

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Geometrically, dydx\frac{dy}{dx} represents the slope of the tangent to the curve at any point (x,y)(x, y). In parametric form, this slope is a function of the parameter (e.g., tt or θ\theta). If we need to find the slope at a specific point, we substitute the given value of the parameter into the expression for dydx\frac{dy}{dx}.

Graph of a curve with a tangent line representing the derivative at a specific parametric value.
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Higher-order derivatives like d2ydx2\frac{d^2y}{dx^2} require careful application of the chain rule twice. After finding dydx\frac{dy}{dx} as a function of tt, you must differentiate that result with respect to tt again, and then multiply by dtdx\frac{dt}{dx} (which is 1/dxdt1 / \frac{dx}{dt}). A common error is forgetting the final division by dxdt\frac{dx}{dt}.

Logic flow for finding the second derivative of a parametric function.

📐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.

Problem 3:

Find the slope of the tangent to the curve given by x=a(θ+sin⁡θ)x = a(\theta + \sin \theta) and y=a(1−cos⁡θ)y = a(1 - \cos \theta) at θ=π2\theta = \frac{\pi}{2}.

A graph of a cycloid-like curve with a tangent line at a specific point showing a slope of 1.

Solution:

  1. Find dxdθ\frac{dx}{d\theta}: dxdθ=a(1+cos⁡θ)\frac{dx}{d\theta} = a(1 + \cos \theta)
  2. Find dydθ\frac{dy}{d\theta}: dydθ=a(0−(−sin⁡θ))=asin⁡θ\frac{dy}{d\theta} = a(0 - (-\sin \theta)) = a \sin \theta
  3. Apply the formula for dydx\frac{dy}{dx}: dydx=asin⁡θa(1+cos⁡θ)=sin⁡θ1+cos⁡θ\frac{dy}{dx} = \frac{a \sin \theta}{a(1 + \cos \theta)} = \frac{\sin \theta}{1 + \cos \theta}
  4. Using trigonometric identities: dydx=2sin⁡(θ/2)cos⁡(θ/2)2cos⁡2(θ/2)=tan⁡(θ2)\frac{dy}{dx} = \frac{2 \sin(\theta/2) \cos(\theta/2)}{2 \cos^2(\theta/2)} = \tan\left(\frac{\theta}{2}\right)
  5. At θ=π2\theta = \frac{\pi}{2}: dydx=tan⁡(π4)=1\frac{dy}{dx} = \tan\left(\frac{\pi}{4}\right) = 1

Explanation:

We differentiate both parametric components with respect to the parameter θ\theta. The slope of the tangent is the ratio of these derivatives. Simplification using half-angle formulas helps evaluate the specific value easily.

Problem 4:

Find dydx\frac{dy}{dx} for the curve x=etcos⁡tx = e^t \cos t and y=etsin⁡ty = e^t \sin t at t=0t = 0.

Plot of a logarithmic spiral fragment showing the tangent at t=0.

Solution:

  1. Differentiate xx wrt tt (Product Rule): dxdt=et(−sin⁡t)+etcos⁡t=et(cos⁡t−sin⁡t)\frac{dx}{dt} = e^t(-\sin t) + e^t \cos t = e^t(\cos t - \sin t)
  2. Differentiate yy wrt tt (Product Rule): dydt=et(cos⁡t)+etsin⁡t=et(cos⁡t+sin⁡t)\frac{dy}{dt} = e^t(\cos t) + e^t \sin t = e^t(\cos t + \sin t)
  3. Calculate dydx\frac{dy}{dx}: dydx=et(cos⁡t+sin⁡t)et(cos⁡t−sin⁡t)=cos⁡t+sin⁡tcos⁡t−sin⁡t\frac{dy}{dx} = \frac{e^t(\cos t + \sin t)}{e^t(\cos t - \sin t)} = \frac{\cos t + \sin t}{\cos t - \sin t}
  4. Evaluate at t=0t = 0: dydx=cos⁡0+sin⁡0cos⁡0−sin⁡0=1+01−0=1\frac{dy}{dx} = \frac{\cos 0 + \sin 0}{\cos 0 - \sin 0} = \frac{1 + 0}{1 - 0} = 1

Explanation:

By applying the product rule to each parametric equation, we find the rates of change with respect to tt. Dividing them gives the derivative in Cartesian terms. At t=0t=0, the exponential terms cancel and trigonometric values simplify to 1.