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Q6:
If $x$ and $y$ are connected parametrically by the equations, without eliminating the parameter, Find $\frac{dy}{dx}$. $x = a(\theta - \sin \theta), y = a(1 + \cos \theta)$
If $x$ and $y$ are connected parametrically by the equations, without eliminating the parameter, Find $\frac{dy}{dx}$. $x = a(\theta - \sin \theta), y = a(1 + \cos \theta)$
Solution :
Step 1: Analyzing the Parametric Equations
The system is defined by the parameter $\theta$ as follows:
- $x = a(\theta - \sin \theta)$
- $y = a(1 + \cos \theta)$
[Per the Chain Rule for parametric differentiation: $\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta}$, provided $dx/d\theta \neq 0$.]
Step 2: Differentiating $x$ with respect to $\theta$
Applying the linearity of the derivative and the standard derivative formula $\frac{d}{d\theta}(\sin \theta) = \cos \theta$:
$\frac{dx}{d\theta} = \frac{d}{d\theta} [a(\theta - \sin \theta)]$
$\frac{dx}{d\theta} = a \left( \frac{d}{d\theta}(\theta) - \frac{d}{d\theta}(\sin \theta) \right)$
$\frac{dx}{d\theta} = a(1 - \cos \theta)$
Step 3: Differentiating $y$ with respect to $\theta$
Applying the standard derivative formula $\frac{d}{d\theta}(\cos \theta) = -\sin \theta$:
$\frac{dy}{d\theta} = \frac{d}{d\theta} [a(1 + \cos \theta)]$
$\frac{dy}{d\theta} = a \left( \frac{d}{d\theta}(1) + \frac{d}{d\theta}(\cos \theta) \right)$
$\frac{dy}{d\theta} = a(0 - \sin \theta) = -a \sin \theta$
Step 4: Calculating $\frac{dy}{dx}$
Substitute the results from Step 2 and Step 3 into the parametric derivative formula:
$\frac{dy}{dx} = \frac{-a \sin \theta}{a(1 - \cos \theta)}$
$\frac{dy}{dx} = \frac{-\sin \theta}{1 - \cos \theta}$
Step 5: Trigonometric Simplification
Utilize the half-angle identities to simplify the expression [Identity: $\sin \theta = 2 \sin(\theta/2) \cos(\theta/2)$ and $1 - \cos \theta = 2 \sin^2(\theta/2)$]:
$\frac{dy}{dx} = \frac{-2 \sin(\theta/2) \cos(\theta/2)}{2 \sin^2(\theta/2)}$
Cancel the common terms $2$ and $\sin(\theta/2)$:
$\frac{dy}{dx} = -\frac{\cos(\theta/2)}{\sin(\theta/2)}$
Since $\cot \theta = \frac{\cos \theta}{\sin \theta}$:
$\frac{dy}{dx} = -\cot\left(\frac{\theta}{2}\right)$
Final Solution: $\frac{dy}{dx} = -\cot\left(\frac{\theta}{2}\right)$
More Questions from Class 12 Mathematics Continuity and Differentiability EXERCISE 5.6
- Q1: If $x$ and $y$ are connected parametrically by the equations, without eliminating the parameter, Find $\frac{dy}{dx}$. $x = 2at^2, y = at^4$
- Q2: If $x$ and $y$ are connected parametrically by the equations, without eliminating the parameter, Find $\frac{dy}{dx}$. $x = a \cos \theta, y = b \cos \theta$
- Q3: If $x$ and $y$ are connected parametrically by the equations, without eliminating the parameter, Find $\frac{dy}{dx}$. $x = \sin t, y = \cos 2t$
- Q4: If $x$ and $y$ are connected parametrically by the equations, without eliminating the parameter, Find $\frac{dy}{dx}$. $x = 4t, y = \frac{4}{t}$
- Q5: If $x$ and $y$ are connected parametrically by the equations, without eliminating the parameter, Find $\frac{dy}{dx}$. $x = \cos \theta - \cos 2\theta, y = \sin \theta - \sin 2\theta$
- Q7: If $x$ and $y$ are connected parametrically by the equations, without eliminating the parameter, Find $\frac{dy}{dx}$. $x = \frac{\sin^3 t}{\sqrt{\cos 2t}}, y = \frac{\cos^3 t}{\sqrt{\cos 2t}}$
- Q8: If $x$ and $y$ are connected parametrically by the equations, without eliminating the parameter, Find $\frac{dy}{dx}$. $x = a\left(\cos t + \log \tan\left(\frac{t}{2}\right)\right), y = a \sin t$
- Q9: If $x$ and $y$ are connected parametrically by the equations, without eliminating the parameter, Find $\frac{dy}{dx}$. $x = a \sec \theta, y = b \tan \theta$
- Q10: If $x$ and $y$ are connected parametrically by the equations, without eliminating the parameter, Find $\frac{dy}{dx}$. $x = a(\cos \theta + \theta \sin \theta), y = a(\sin \theta - \theta \cos \theta)$
- Q11: If $x = \sqrt{a^{\sin^{-1} t}}, y = \sqrt{a^{\cos^{-1} t}}$, show that $\frac{dy}{dx} = -\frac{y}{x}$
CBSE Solutions for Class 12 Mathematics Continuity and Differentiability
- Continuity and Differentiability Class 12 EXERCISE 5.1
- Continuity and Differentiability Class 12 EXERCISE 5.2
- Continuity and Differentiability Class 12 EXERCISE 5.3
- Continuity and Differentiability Class 12 EXERCISE 5.4
- Continuity and Differentiability Class 12 EXERCISE 5.5
- Continuity and Differentiability Class 12 EXERCISE 5.7
Chapters in CBSE - Class 12 Mathematics
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