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Q4:
If $x$ and $y$ are connected parametrically by the equations, without eliminating the parameter, Find $\frac{dy}{dx}$. $x = 4t, y = \frac{4}{t}$
If $x$ and $y$ are connected parametrically by the equations, without eliminating the parameter, Find $\frac{dy}{dx}$. $x = 4t, y = \frac{4}{t}$
Solution :
Step 1: Given Parametric Equations
The system is defined by the parameter $t$ where $t \in \mathbb{R} \setminus \{0\}$. The equations are:
- $x = 4t$
- $y = 4t^{-1}$
Step 2: Differentiating $x$ with respect to $t$
To find the derivative of the independent variable $x$ with respect to the parameter $t$:
$\frac{dx}{dt} = \frac{d}{dt}(4t)$
Applying the Power Rule of differentiation, $\frac{d}{dt}(at^n) = ant^{n-1}$ [Per the Power Rule of Calculus]:
$\frac{dx}{dt} = 4(1) = 4$
Step 3: Differentiating $y$ with respect to $t$
To find the derivative of the dependent variable $y$ with respect to the parameter $t$:
$\frac{dy}{dt} = \frac{d}{dt}(4t^{-1})$
Applying the Power Rule, $\frac{d}{dt}(ct^n) = cnt^{n-1}$:
$\frac{dy}{dt} = 4(-1)t^{-1-1}$
$\frac{dy}{dt} = -4t^{-2} = -\frac{4}{t^2}$
Step 4: Applying the Chain Rule for Parametric Differentiation
To determine $\frac{dy}{dx}$ without eliminating the parameter, we utilize the relationship between parametric derivatives [Per the Chain Rule for Parametric Equations]:
$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$
Substituting the derived expressions from Step 2 and Step 3:
$\frac{dy}{dx} = \frac{-\frac{4}{t^2}}{4}$
Step 5: Algebraic Simplification
Simplifying the complex fraction:
$\frac{dy}{dx} = -\frac{4}{t^2} \cdot \frac{1}{4}$
The constant $4$ cancels out:
$\frac{dy}{dx} = -\frac{1}{t^2}$
Visual Representation of the Curve
Final Solution: \frac{dy}{dx} = -\frac{1}{t^2}
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 $\frac{dy}{dx}$. $x = \sin t, y = \cos 2t$
- Q5: If $x$ and $y$ are connected parametrically by the equations, without eliminating the parameter, Find $\frac{dy}{dx}$. $x = \cos \theta - \cos 2\theta, y = \sin \theta - \sin 2\theta$
- Q6: If $x$ and $y$ are connected parametrically by the equations, without eliminating the parameter, Find $\frac{dy}{dx}$. $x = a(\theta - \sin \theta), y = a(1 + \cos \theta)$
- 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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